Frank Brothers Solutions for ICSE Class 9 Physics Chapter 2 Motion In One Dimension

Page No: 61

Question 1. Define state of rest.
Answer: A body is said to be in state of rest if it does not change its position with respect to its surrounding objects with time.
In simple words: Something is at rest if it stays in the same spot compared to the things around it, like a book sitting on a table.

📝 Teacher's Note: Use the example of a student sitting in a chair. To someone in the classroom, the student is at rest. This introduces the concept of a "reference point".

🎯 Exam Tip: Always mention "with respect to its surroundings" and "with time" to get full marks for this definition.

Question 2. Define a vector quantity.
Answer: A vector quantity is that physical quantity which is represented by both magnitude and direction.
In simple words: A vector is a measurement that tells you "how much" and "which way," like saying you walked 5 kilometers toward the North.

📝 Teacher's Note: Contrast this with a scalar quantity (like temperature) which doesn't have a direction. You can't have "20 degrees Celsius North".

🎯 Exam Tip: Keywords for this answer are "magnitude" and "direction". Ensure both are present.

Question 3. Is mass a vector quantity?
Answer: No mass is not a vector quantity.
In simple words: Mass is a scalar because it only tells you how much matter is in something; it doesn't have a direction.

📝 Teacher's Note: Ask students if they can "weigh 5 kg downwards". Explain that while weight (a force) has direction, mass itself does not.

🎯 Exam Tip: Justify your 'No' by stating that mass only has magnitude and no direction.

Question 4. How is a vector represented?
Answer: A vector is represented by an arrow. The length of the arrow represents the magnitude of vector quantity and arrow head gives the direction of vector quantity.
In simple words: We draw vectors as arrows. A longer arrow means a bigger measurement, and the tip shows the direction.

📝 Teacher's Note: Draw two arrows of different lengths on the board to represent two different forces acting in the same direction to visualize magnitude differences.

🎯 Exam Tip: Mention both the "length" (for magnitude) and the "arrowhead" (for direction) in your explanation.

Question 5. Give an example of a body at rest.
Answer: If a book is lying in almirah then it is at rest.
In simple words: Any object that isn't moving, like a stationary car or a sleeping cat, is at rest.

📝 Teacher's Note: This is a practical application of the definition of rest. It helps ground the theoretical concept in everyday observation.

🎯 Exam Tip: When giving examples, choose simple, unambiguous scenarios like a mountain or a building.

Question 6. Define motion.
Answer: A body is said to be in motion when it change its position with respect to its surrounding objects with time.
In simple words: Motion is when something moves from one place to another over time, relative to the things staying still around it.

📝 Teacher's Note: Emphasize that motion is the opposite of rest. If the position changes, it's motion; if it doesn't, it's rest.

🎯 Exam Tip: Similar to the rest definition, the phrase "with respect to its surroundings" is critical for completeness.

Question 7. Are rest and motion absolute or relative?
Answer: Yes rest and motion are relative to each other.
In simple words: Whether something is "moving" or "still" depends on who is looking at it. A passenger in a moving car is at rest to another passenger, but in motion to someone standing on the sidewalk.

📝 Teacher's Note: This is a sophisticated concept. Use the "train passenger" analogy to make it intuitive. Nothing in the universe is truly at an "absolute" rest.

🎯 Exam Tip: Use the word "relative" clearly in your answer to show you understand the dependence on the observer's frame.

Question 8. Identify the vector quantity between Force and Energy.
Answer: Out of Force and Energy, Force is a vector quantity.
In simple words: When you push something (force), you always push it in a certain direction. Energy doesn't have a specific direction.

📝 Teacher's Note: Force is \( \text{Mass} \times \text{Acceleration} \). Since acceleration is a vector, force must be one too. Energy is a scalar quantity like work.

🎯 Exam Tip: Knowing which common physics quantities are vectors (Velocity, Acceleration, Force, Momentum) is essential for multiple-choice questions.

Question 9. Give examples of scalar quantities.
Answer: Examples of scalars are distance and mass.
In simple words: Scalar quantities are just numbers, like your age or the total amount of path you walked, regardless of where you turned.

📝 Teacher's Note: Other examples include time, speed, and volume. These are quantities where adding a direction makes no physical sense.

🎯 Exam Tip: Distance is a scalar, but displacement is a vector. This is a very common trap in exams.

Question 10. Analyze positions on a straight line.
Answer: Out of these positions, (i) and (ii) positions of body lie on same straight line as direction of these two are same.
In simple words: If two positions are in the exact same direction from the starting point, they are on the same straight line path.

📝 Teacher's Note: Use a ruler on the board to show that multiple points can have different distances from zero but lie on the same straight edge if they share a direction.

🎯 Exam Tip: "Same direction" is the key phrase that link these two positions to a single straight line.

Question 11. What makes a quantity a vector?
Answer: A vector quantity is represented only when its magnitude and direction are specified so this quantity is a vector quantity.
In simple words: For a measurement to be a vector, you must define "how much" and "which way." If you need both to describe it, it's a vector.

📝 Teacher's Note: This is the general rule for identifying vectors. If you can change the quantity's effect just by changing its direction, it's a vector.

🎯 Exam Tip: Reiterate the "magnitude and direction" pair. It's the standard definition used in marking schemes.

Page No: 62

Question 12. Describe the relative rest of train passengers.
Answer: Passengers sitting in a train are at rest with respect to each other.
In simple words: Even if a train is speeding at 100 km/h, two people sitting in it don't move away from each other, so they see each other as stationary.

📝 Teacher's Note: This perfectly illustrates that "rest" depends on the observer. Relative to the tracks, they are moving; relative to the seat, they are still.

🎯 Exam Tip: Use the phrase "with respect to each other" to specify the reference frame being discussed.

Question 13. Explain how a body can be at rest and in motion at the same time.
Answer: Yes we are at rest as well as motion because we are at rest with respect to a observer which is itself at rest and we are in motion with respect to a observer which is in motion.
In simple words: Right now, you are sitting still relative to your room, but because the Earth is spinning and orbiting the sun, you are moving very fast relative to outer space.

📝 Teacher's Note: This helps students move away from the idea that "motion" is a property of the object itself. It is a relationship between an object and an observer.

🎯 Exam Tip: Explain this using the "observer" concept. Use two different observers (one stationary, one moving) to justify the simultaneous state.

Question 14. Describe the motion of a platform relative to a train.
Answer: The platform is in motion with respect to train. As train is moving with respect to platform so platform would also look in motion with respect to train.
In simple words: When you are on a moving train and look out the window, the station platform seems to be zooming backward. To you, the platform is the thing that's moving.

📝 Teacher's Note: This is a great example of "relative velocity". If A moves at velocity \( v \) relative to B, then B moves at velocity \( -v \) relative to A.

🎯 Exam Tip: Emphasize that motion is mutual. If the train moves past the platform, the platform moves past the train.

Question 15. Represent a vector in the North-West direction.
Answer:
In simple words: This diagram shows an arrow pointing exactly halfway between North and West, with markings to show it has a "length" or magnitude of 10 units.

📝 Teacher's Note: Remind students that North-West usually implies exactly 45 degrees between the North and West axes unless specified otherwise.

🎯 Exam Tip: When drawing vectors on axes, always label the compass directions (N, S, E, W) and the origin 'O'.

Question 16. Represent a vector at 30 degrees to the West of North.
Answer:
In simple words: This arrow is 15 units long and is pointed slightly more toward the West than the North, making a 30-degree angle with the West axis.

📝 Teacher's Note: Be careful with the phrasing "at 30 degrees to the West of North" or similar. It specifies where to place the angle arc.

🎯 Exam Tip: Draw the angle arc clearly and place the degree value next to it to avoid ambiguity in your sketches.

Question 17. Define a physical quantity with magnitude and direction.
Answer: The physical quantity representing the magnitude and its direction is a vector quantity.
In simple words: This is just another way of saying that any measurement that needs both size and direction to be complete is a vector.

📝 Teacher's Note: This repeats the fundamental definition for reinforcement. Repetition is key for long-term memory of core concepts.

🎯 Exam Tip: If a question asks for a definition of a vector, this is the most direct and standard answer.

Question 18. Answer the following questions about scalars and vectors.
Answer:

• Yes we can add two scalars.
• Yes we can add two vectors.
• Yes we can multiply two scalars.
• No we cannot add a scalar quantity to a vector quantity.
• Yes we can subtract two scalars.
• No we cannot subtract a scalar quantity from a vector quantity. Reverse is also not true.
• Yes we can multiply vectors.

In simple words: You can do math with two of the same type (two scalars or two vectors), but you can't mix-and-match by adding a simple number to a direction-based vector.

📝 Teacher's Note: Explain that "adding a scalar to a vector" is like trying to add "5 apples to 5 meters North". They are different mathematical species.

🎯 Exam Tip: Remember: You can *multiply* a scalar and a vector (like force = mass \(\times\) acceleration), but you can never *add* or *subtract* them.

Question 19. Define distance travelled.
Answer: The actual length of the path covered by a moving object irrespective of its direction of motion is called the distance travelled by the object.
In simple words: Distance is the "odometer reading". It's the total ground you covered, no matter how many zigzags or circles you made.

📝 Teacher's Note: Contrast this with displacement. If you walk around a 400m track and end at the start, your distance is 400m, but your displacement is zero.

🎯 Exam Tip: The key word here is "actual length" and "irrespective of direction".

Question 20. Can distance be less than displacement?
Answer: No the distance covered by a body cannot be less than the magnitude of its displacement.
In simple words: The shortest distance between two points is a straight line. Since displacement is that straight line, any other path (distance) must be longer or equal to it.

📝 Teacher's Note: Use a piece of string on a map. A straight taut string represents displacement; any slack or curved string represents a longer distance.

🎯 Exam Tip: Distance \(\geq\) |Displacement|. They are only equal for motion in a single straight line in one direction.

Question 21. Define displacement.
Answer: The displacement of a moving body is defined as the change in its position along a particular direction
In simple words: Displacement is the "as the crow flies" measurement. It's the straight-line distance from where you started to where you ended, in a specific direction.

📝 Teacher's Note: Displacement only cares about the "initial" and "final" positions. The path taken in between doesn't matter.

🎯 Exam Tip: Displacement is a vector. Always include "direction" in your definition to distinguish it from distance.

Question 22. What is the SI unit for distance and displacement?
Answer: SI unit for measurement of distance and displacement is metre denoted by m.
In simple words: Both these measurements of length are officially recorded in meters, which is the worldwide standard.

📝 Teacher's Note: Remind students that while km, cm, and mm are common, the "standard" (SI) unit is always the meter.

🎯 Exam Tip: Always use 'm' for meters in your numerical answers unless the question asks for something else.

Question 23. Calculate distance and displacement for a semi-circular path.
Answer: In this case Distance travelled = half of the perimeter of circle. so distance travelled is \( = \pi r \). Displacement = diameter of circle So displacement is \( 2r \).
In simple words: If you walk around the curve of a half-circle, you travel a long way (\( \pi r \)). But your "net change" in position is just the straight line across the middle (\( 2r \)).

📝 Teacher's Note: This is a classic textbook problem. It tests the ability to apply geometric formulas to physics concepts.

🎯 Exam Tip: For circular paths, express your answers in terms of \( \pi \) and \( r \) unless numerical values are provided.

Question 24. Can displacement be negative?
Answer: Yes a body can have negative displacement.
In simple words: If "forward" is positive, then walking "backward" results in a negative displacement. It just means you ended up behind where you started.

📝 Teacher's Note: Negative sign in vectors always indicates the opposite direction. It doesn't mean a quantity less than nothing.

🎯 Exam Tip: Displacement is a vector, so it can be positive, negative, or zero. Distance is a scalar and is always positive.

Question 25. When is displacement equal to distance?
Answer: If a body is moving in a straight line then the displacement of a body is equal to the distance travelled by it.
In simple words: If you walk perfectly straight from point A to point B without turning back, the total path and the net change are the same number.

📝 Teacher's Note: Crucial addition: "moving in a straight line *without reversing direction*". If they turn back, distance increases while displacement decreases.

🎯 Exam Tip: This specific condition (straight line, one direction) is when scalar magnitude equals vector magnitude.

Question 26. List two differences between distance and displacement.
Answer:

• Distance is a scalar quantity whereas displacement is a vector quantity.
• Distance is always positive but displacement can be negative, zero or positive.

In simple words: Distance doesn't care about direction and is always a plus number. Displacement includes direction and can show you went backward or ended up right back where you started.

📝 Teacher's Note: A third difference to share: distance depends on the path, displacement only depends on the endpoints.

🎯 Exam Tip: When asked for differences, present them as clear bullet points or in a table for maximum clarity.

Question 27. Calculate distance and displacement for the given paths.
Answer:
(i) Distance is 4 + 4 = 8 m while displacement is 3 m.
(ii) Distance is 2 + 4 + 2 + 4 = 12 m while displacement is 0 m.
(iii) Distance is 2 + 2 + 2 + 2 + 2 + 2 = 12 m while displacement is 2 + 2 + 2 = 6 m.
In simple words: Adding up all the segments gives distance. The straight-line gap from start to finish gives displacement. In case (ii), starting and ending at 'A' means no net change (zero displacement).

📝 Teacher's Note: These problems help students practice simple vector addition for displacement versus scalar addition for distance.

🎯 Exam Tip: Always track the "Start" and "End" points carefully before calculating displacement.

Question 28. Analyze motion on a circular path.
Answer:
(i) In this case Distance travelled = half of the perimeter of circle. so distance travelled is \( = \pi r \). Displacement = diameter of circle. So displacement is \( 2r \).
(ii) Distance travelled = (3/4) of perimeter of circle. So distance travelled is \( = (3\pi r / 2) \). Displacement can be calculated as \( s = \sqrt{r^2 + r^2} = \sqrt{2}r \).
(iii) Distance travelled = perimeter of circle. So distance travelled is \( = 2\pi r \). Displacement = 0 as starting and ending point are coinciding.
(iv) Distance travelled is two times the perimeter of circle. So distance travelled is \( = 4\pi r \). Again displacement is = 0 as starting and ending point are coinciding.
In simple words: Distance just keeps adding up as you go around. Displacement resets to zero every time you finish a full lap because you're back where you started.

📝 Teacher's Note: Use Pythagoras theorem for case (ii) where the displacement is the hypotenuse of a right-angled triangle formed by two radii.

🎯 Exam Tip: Memorize that for a full circle (or any number of full circles), displacement is always zero.

Question 29. Calculate distance and displacement for a trip from home to school and back.
Answer:

(i) Distance travelled = distance between house to school + distance between school to house = 3 + 3 = 6 km.
(ii) Displacement is = 0 as starting and ending point are coinciding.
In simple words: If you walk 3 km to school and 3 km back home, your shoes did 6 km of work (distance), but you ended up in the same spot (zero displacement).

📝 Teacher's Note: This is the most practical way to teach the difference between the two terms. Almost every student's daily commute results in zero net displacement.

🎯 Exam Tip: If the journey is "round-trip," the final displacement is always 0. Don't waste time on complex calculations!

Question 30. What is the displacement of a satellite in one round trip?
Answer: Displacement of a satellite to make a complete round trip is = 0 as starting and ending point is coinciding.
In simple words: No matter how many thousands of miles a satellite orbits, if it comes back to exactly the same point in space, its net displacement is zero.

📝 Teacher's Note: This applies to planets orbiting the sun as well. One Earth year equals a distance of nearly 940 million km, but zero displacement relative to the starting point in the orbit.

🎯 Exam Tip: The phrase "complete round trip" is your clue that displacement must be zero.

Question 31. Calculate distance and displacement for an object thrown up to height H.
Answer:

(i) Total distance travelled = distance from A to B + distance from B to A. So total distance = H + H = 2H.
(ii) Total displacement = 0 as starting and ending points are coinciding.
In simple words: If you throw a ball up to 10 meters and catch it, it traveled 20 meters total. But it started in your hand and ended in your hand, so it went nowhere "net".

📝 Teacher's Note: This introduces vertical motion. Displacement is zero because the final height equals the initial height.

🎯 Exam Tip: Be careful — if the question only asks for the trip *up to height H*, the distance and displacement would both be H. Only if it returns is displacement 0.

Page No: 79

Question 1. Define speed.
Answer: Speed of a body can be defined as distance covered by the body in unit time.
In simple words: Speed tells you how fast something is going. It's just the total distance divided by the time it took.

📝 Teacher's Note: Use the formula \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \). It's a scalar quantity because distance is a scalar.

🎯 Exam Tip: Don't confuse this with velocity, which also requires a direction.

Question 2. Define average speed.
Answer: Average speed of a body can be defined as ratio of total distance covered by a body In total time.
In simple words: Average speed is like your overall pace for a whole trip, ignoring the times you sped up or stopped for a snack.

📝 Teacher's Note: Important: Average speed is NOT the average of speeds. It is strictly (Total Distance) / (Total Time).

🎯 Exam Tip: Use the "total" values in your calculations. Never just add two speeds and divide by two unless the times are identical.

Question 3. What is the SI unit for speed and average speed?
Answer: Both speed and average speed have same unit and that is \( \text{ms}^{-1} \).
In simple words: We officially measure how fast things go in "meters per second".

📝 Teacher's Note: This can also be written as m/s. Students should be comfortable with both notations.

🎯 Exam Tip: Always check your final units. Speed is a rate, so it must have a time component in the denominator.

Question 4. Do speed and average speed mean the same thing?
Answer: No speed and average speed of a body have different meaning.
In simple words: Speed is what your speedometer shows *right now*. Average speed is how fast you went across the *whole journey*.

📝 Teacher's Note: Speed usually refers to "instantaneous speed". If you drive at 60 km/h then 40 km/h, your instantaneous speeds were 60 and 40, but your average was somewhere in between.

🎯 Exam Tip: Use the terms "instantaneous" versus "overall trip" to highlight the difference in your explanation.

Question 5. Compare 60 km/h and 15 m/s.
Answer: 60 km/hr can be converted into m/s to compare with 15m/s.
60 km/hr \( = (60 \times 1000)/3600 = 16.66 \text{ m/s} \). so speed 60 km/hr is greater.
In simple words: To see which is faster, we change them into the same units. 60 km/h is about 16.7 meters every second, which is a bit more than 15 meters every second.

📝 Teacher's Note: The conversion factor is \( 5/18 \). Multiplying \( 60 \times (5/18) \) is a faster way to get the same result.

🎯 Exam Tip: Never compare two numbers with different units directly. Always convert one to match the other first.

Question 6. Convert 20 m/s to km/h.
Answer: 20m/s can be converted into km/hr as
20 m/s \( = (20 \times 3600)/1000 = 72 \text{ km/hr} \).
In simple words: Moving 20 meters every second is the same as driving a car at 72 kilometers per hour.

📝 Teacher's Note: To go from m/s to km/h, multiply by \( 18/5 \) (or 3.6). It's a handy shortcut for quick checks.

🎯 Exam Tip: Show the steps of multiplying by 3600 and dividing by 1000 to ensure partial marks in case of a calculation error.

Question 7. Compare 36 km/h and 8.5 m/s.
Answer: 36 km/hr can be converted into m/s to compare with 8.5m/s.
36 km/hr \( = (36 \times 1000)/3600 = 10 \text{ m/s} \) which is greater than 8.5 m/s
So 36 km/hr is greater than 8.5 m/s.
In simple words: 36 km/h is actually 10 meters per second, which is obviously faster than 8.5 meters per second.

📝 Teacher's Note: This is another unit conversion practice. Remind students: km to m (multiply by 1000), hour to second (multiply by 3600).

🎯 Exam Tip: Clearly state the converted value before making the final comparison statement.

Question 8. What is the SI unit of velocity?
Answer: SI unit of velocity is \( \text{ms}^{-1} \).
In simple words: Like speed, velocity is also measured in meters per second, but we must also remember its direction.

📝 Teacher's Note: Even though the unit is the same as speed, velocity is a vector. Stress this conceptual difference frequently.

🎯 Exam Tip: Always write the unit as \( \text{ms}^{-1} \) or m/s. Avoid using 'mps' which is not standard SI notation.

Question 9. Do two bodies moving in different directions have the same velocity if their speeds are equal?
Answer: No as their direction is different they don’t have same velocity.
In simple words: Velocity is speed *plus* direction. If one car goes North and another goes South, their velocities are different even if they both show 50 km/h on their speedometers.

📝 Teacher's Note: This is the key "gotcha" for vector quantities. Changing direction *is* a change in velocity, even if the speed stays constant.

🎯 Exam Tip: Explicitly mention that velocity depends on both magnitude (speed) and direction to justify your 'No'.

Question 10. Arrange the following speeds in increasing order: 36 km/h, 2 km/min, 7 m/s.
Answer: we convert all the speeds in m/s to compare them.
36 km/hr \( = (36 \times 1000)/3600 = 10 \text{ m/s} \).
2 km/min \( = (2 \times 1000)/60 = 33.3 \text{ m/s} \).
7 m/s = 7 m/s.
So increasing order of speed is 7m/s < 10m/s < 33m/s.
In simple words: After changing everything to meters-per-second, we find that 7 m/s is the slowest, 36 km/h (which is 10 m/s) is middle, and 2 km/min (over 33 m/s) is by far the fastest.

📝 Teacher's Note: This multi-step conversion problem is excellent for testing algebraic fluency. 2 km/min is deceptively fast!

🎯 Exam Tip: Show the conversion work for each value separately before writing down the final ordered list.

Question 11. Derive average speed for a journey where half the distance is covered at speed 'u' and half at speed 'v'.
Answer: let total distance be S.
Boy covers distance S/2 with speed u then time taken by him to cover this distance would be \( T_1 = S/2u \).
Again boy covers rest of the distance S/2 with speed v then time taken by him to cover this distance would be \( T_2 = S/2v \).
So total time taken by boy to cover the distance S is \( T = T_1 + T_2 \).
Total time \( T = S/2 (1/u + 1/v) = S(u+v)/2uv \).
And average speed = S/T = \( 2uv/(u+v) \).
In simple words: You can't just average the two speeds. Because you spend more time going at the slower speed, the average is pulled closer to that slower value. The formula \( 2uv / (u+v) \) accounts for this correctly.

📝 Teacher's Note: This is the harmonic mean. It's a common trap to use \( (u+v)/2 \). Explain that since the distances are same, time is inversely proportional to speed.

🎯 Exam Tip: If an MCQ asks for average speed with equal distances, look for the choice that uses the formula derived here.

Question 12. Do 'uniform speed' and 'constant speed' mean the same thing?
Answer: Yes uniform speed and constant speed have same meaning.
In simple words: Both mean that an object covers equal distances in equal amounts of time — its pace never changes.

📝 Teacher's Note: In everyday language, we use "constant". In physics textbooks, "uniform" is the preferred technical term.

🎯 Exam Tip: These terms are interchangeable in descriptive answers. Don't let the different wording confuse you.

Question 13. Derive average speed for a round trip with forward speed 'u' and return speed 'v'.
Answer: let S be the distance between P and Q.
Body covers forward journey distance S (P to Q) with speed u then time taken by him to cover this distance would be \( T_1 = S/u \).
Again body covers backward journey distance S (Q to P) with speed v then time taken by him to cover this distance would be \( T_2 = S/v \).
So total time taken by body to cover the distance S is \( T = T_1 + T_2 \).
Total time \( T = S (1/u + 1/v) = S(u+v)/uv \).
And average speed = 2S/T = \( 2uv/(u+v) \).
In simple words: Since the distance there and back is the same, this is just a special case of the previous problem. The formula remains the same because you traveled the distance 'S' twice.

📝 Teacher's Note: Contrast this with average velocity. For a round trip, average velocity is zero, but average speed is \( 2uv/(u+v) \).

🎯 Exam Tip: Note that total distance is \( 2S \). This is where many students make a mistake by using just \( S \) in the final step.

Question 14. What is the displacement and average velocity for a round trip from P to Q and back?
Answer: As body goes from P to Q and then return back to P so the displacement of the body would be zero and hence average velocity would also be zero.
In simple words: If you start at home, go to the park, and come back home, you technically haven't moved anywhere in total. Since your "net move" is zero, your average "net speed" (velocity) is also zero.

📝 Teacher's Note: This highlights the difference between Speed and Velocity. Velocity depends on displacement, so if you end at the start, your average velocity is mathematically zero.

🎯 Exam Tip: For any journey that ends where it started, displacement = 0 and average velocity = 0. It's a quick way to solve many tricky-looking problems.

Question 15. Calculate average speed for a trip with forward speed 20 m/s and return speed 30 m/s.
Answer: let distance between P and Q is S.
Speed of car while travelling from P to Q is 20 m/s.
Let car take time \( T_1 \) to travel from P to Q then \( T_1 = S/20 \).
Speed of car while travelling from Q to P is 30 m/s.
Let car take time \( T_2 \) to travel from Q to P then \( T_2 = S/30 \).
Total time = \( T_1 + T_2 = S/20 + S/30 = S/12 \).
So average speed of journey = total distance / total time = \( 2S/(S/12) = 24 \text{ m/s} \).
Average speed of journey is 24 m/s.
In simple words: You spent more time going at 20 m/s than at 30 m/s. This makes the overall average closer to 20 than 30. The middle point between 20 and 30 is 25, but our correct average is slightly less at 24.

📝 Teacher's Note: This is a numerical check for the formula \( 2uv/(u+v) \). \( (2 \times 20 \times 30) / (20+30) = 1200 / 50 = 24 \). It works perfectly.

🎯 Exam Tip: Always double-check your math when finding the common denominator for the total time calculation.

Question 16. Distinguish between speed and velocity.
Answer: Speed is a scalar quantity whereas velocity is a vector quantity. So speed doesn’t have its direction and velocity has a particular direction.
In simple words: Speed is just how fast you're going. Velocity is how fast you're going *and* where you're headed.

📝 Teacher's Note: You can have constant speed in a circle, but you cannot have constant velocity in a circle because the direction is always changing.

🎯 Exam Tip: Use terms like "scalar" and "vector" to demonstrate scientific vocabulary.

Question 17. When are speed and velocity equal?
Answer: Speed and velocity of a moving body become equal when the body moves in a straight line.
In simple words: If you don't turn or change direction, your "total path" and "straight-line gap" are the same, so speed and velocity magnitudes will match.

📝 Teacher's Note: Specifically, this is true for "rectilinear motion" in a fixed direction. If the object turns around on the straight line, they are no longer equal.

🎯 Exam Tip: Mention "straight line motion without changing direction" for a more rigorous answer.

Question 18. Define uniform velocity.
Answer: When the velocity of a moving body doesn’t change with time then the velocity of the body is said to be constant or uniform. Yes uniform velocity and constant velocity are one and the same thing.
In simple words: Uniform velocity means you are moving at a steady speed in a perfectly straight line — neither your pace nor your direction changes.

📝 Teacher's Note: A car on cruise control going around a curve has constant speed but NOT uniform velocity.

🎯 Exam Tip: "Uniform velocity" implies both constant speed AND constant direction.

Question 19. Define acceleration.
Answer: Acceleration of a body is rate of change of its velocity with respect to the time.
In simple words: Acceleration is how quickly your velocity changes. It's the "speeding up" or "slowing down" factor.

📝 Teacher's Note: Since it's a rate of change of a vector, acceleration itself is a vector. You can accelerate by changing speed OR changing direction.

🎯 Exam Tip: Always include the word "rate" in your definition, as it indicates a division by time.

Question 20. What is the SI unit of acceleration?
Answer: SI unit of acceleration is \( \text{ms}^{-2} \).
In simple words: Acceleration is measured in "meters per second squared". It tells you how many meters-per-second your speed changes every second.

📝 Teacher's Note: Explain the units as (m/s) per s = m/s\(^2\). This helps students understand why the 's' is squared.

🎯 Exam Tip: Don't forget the negative exponent or the division line (\( \text{ms}^{-2} \) or m/s\(^2\)).

Page No: 80

Question 21. Can a body with constant acceleration follow a path that is not a straight line?
Answer: If the acceleration of a moving body is constant in magnitude and direction then the path of the body must not be a straight line because in circular motion also acceleration of a body is constant in magnitude and always directed towards the centre. So the path of the body may be a straight line and may be a circular one.
In simple words: Usually, constant acceleration in one direction leads to a straight or curved (parabolic) path. The text notes that circular motion has a special "constant magnitude" acceleration always pointing inward, making the path a circle.

📝 Teacher's Note: This is a slightly confusing point in the source. If acceleration is constant in *vector* form (mag + dir), the path is a straight line or parabola (like a ball thrown). Circular motion has acceleration that changes direction constantly.

🎯 Exam Tip: In circular motion, speed is constant but velocity is not, because the direction of motion is always changing.

Question 22. Can the relation \( S = v \times t \) be used for non-uniform speed?
Answer: No the relation \( S = v \times t \) cannot used to find the total distance covered by a body moving with non-uniform speed.
In simple words: This simple formula only works if your speed stays exactly the same the whole time. If you speed up or slow down, the math is more complicated.

📝 Teacher's Note: For non-uniform speed, we must use Average Speed in that formula: \( \text{Total Distance} = \text{Average Speed} \times \text{Total Time} \).

🎯 Exam Tip: Only use \( S = vt \) for "uniform motion" (constant velocity) problems.

Question 23. What does the area under a speed-time graph represent?
Answer: Yes area under a speed time graph in a given interval gives the total distance covered by a body.
In simple words: If you color in the space between the speed-line and the bottom axis of a graph, the "amount" of color you used represents how far the object actually traveled.

📝 Teacher's Note: This is a graphical way to multiply speed \(\times\) time. The 'height' of the area is speed and the 'width' is time.

🎯 Exam Tip: This is a very common graph interpretation question. Remember: Slope = Acceleration, Area = Distance.

Question 24. Interpret a speed-time graph with a flat line at 25 m/s.
Answer: Yes the motion is uniform and the uniform speed is given by area under speed time graph divided by time interval.
So speed = 500/20 = 25 m/s.
In simple words: A flat horizontal line on a speed-time graph means the speed isn't changing. If the total distance (area) was 500 and it took 20 seconds, the constant speed was 25 m/s.

📝 Teacher's Note: A horizontal line on a speed-time graph has zero slope, confirming zero acceleration (uniform speed).

🎯 Exam Tip: When reading graphs, first look at what is on the vertical axis (Distance? Speed? Velocity?). It changes the meaning of the lines entirely.

Question 25. What is positive acceleration?
Answer: Positive acceleration corresponds to situation when velocity is continuously increasing with respect to the time.
In simple words: This is just "speeding up" in the direction you are already going.

📝 Teacher's Note: Acceleration is positive if the change in velocity points in the same direction as the initial velocity.

🎯 Exam Tip: Use the phrase "continuously increasing velocity" for a complete definition.

Question 26. What is negative acceleration?
Answer: Negative acceleration corresponds to situation when velocity is continuously decereasing with respect to the time.
In simple words: This is "slowing down". It's often called retardation or deceleration.

📝 Teacher's Note: Mathematically, if you are moving in a positive direction and slow down, your acceleration value will be negative.

🎯 Exam Tip: Another term for negative acceleration is "retardation". Learn both terms.

Question 27. Describe the acceleration of a falling body.
Answer: If a body falls towards earth then it would have a positive acceleration.
In simple words: Gravity pulls objects down, making them go faster and faster as they fall. Since they are speeding up, the acceleration is positive.

📝 Teacher's Note: This specific acceleration is called "acceleration due to gravity", denoted by 'g', and is about 9.8 ms\(^{-2}\) on Earth.

🎯 Exam Tip: Mention that the velocity is increasing in the downward direction to justify why the acceleration is positive.

Question 28. Explain what an acceleration of 8.5 ms\(^{-2}\) means.
Answer: If a body has acceleration of 8.5 ms\(^{-2}\) then it means its velocity is increasing at a rate of 8.5 ms\(^{-1}\) per second.
In simple words: Every single second that passes, the object gets 8.5 meters-per-second faster than it was the second before.

📝 Teacher's Note: This "rate of a rate" is why we have the "per second squared" unit. It's meters-per-second, per-second.

🎯 Exam Tip: In your explanation, use the phrase "velocity increases by value every second" for full clarity.

Question 29. What is the SI unit of retardation?
Answer: SI unit of retardation is \( \text{ms}^{-2} \).
In simple words: Retardation is just a fancy name for slowing down (negative acceleration), so it uses the same measurement units as speeding up.

📝 Teacher's Note: Students often ask if retardation should be negative. The *value* of retardation is usually given as positive (e.g. "retardation of 5"), but when used in acceleration formulas, it's treated as \( a = -5 \).

🎯 Exam Tip: Units are always the same for acceleration, deceleration, and retardation.

Question 30. Calculate acceleration for a car slowing from 60 km/h to a stop in 10s.
Answer: first convert 60 km/h in m/s.
60 km/hr \( = (60 \times 1000)/3600 = 16.7 \text{ m/s} \).
This is initial velocity of car i.e \( u = 16.7 \text{ m/s} \).
As car stops in 10 seconds so final velocity is \( v = 0 \text{ m/s} \).
So acceleration \( = (v-u)/t = (0-16.7)/10 = -1.67 \text{ ms}^{-2} \).
Acceleration of car is \( = -1.67 \text{ ms}^{-2} \).
In simple words: After changing units, the car starts at 16.7 m/s and loses it all in 10 seconds. That means it loses 1.67 m/s of speed every second. The minus sign shows it was slowing down.

📝 Teacher's Note: This is a standard application of the first equation of motion: \( v = u + at \).

🎯 Exam Tip: Always state your knowns (u, v, t) clearly before starting the calculation. It helps you catch errors in your setup.

Question 31. What does a value of -30 m/s represent?
Answer: -30 m/s is speed.
In simple words: Usually, speed is just the magnitude (30 m/s). In some contexts, a minus sign might be used to show the object is moving in a "negative" direction, but strictly speaking, speed itself shouldn't be negative.

📝 Teacher's Note: This is a bit of a tricky point. Speed is the *magnitude* of velocity, so it's always positive. A value of -30 m/s is almost certainly a *velocity* pointing in the negative direction.

🎯 Exam Tip: If you see a negative sign with m/s, assume it's velocity. If asked specifically about speed, it's just the absolute value (30 m/s).

Question 32. What does velocity correspond to?
Answer: Velocity corresponds to the rate of change of displacement.
In simple words: Velocity is just how much your "straight-line change" (displacement) alters every second.

📝 Teacher's Note: Link this back to the definitions: Speed is change in distance, velocity is change in displacement.

🎯 Exam Tip: Using the term "displacement" instead of "distance" is what makes this a definition of velocity rather than speed.

Question 33. Can the speed of a body be negative?
Answer: No the speed of a body cannot be negative.
In simple words: Speed is just a "how fast" number. You can't be slower than standing still!

📝 Teacher's Note: Scalar quantities like speed, distance, and mass are always zero or positive. Only vector quantities can take negative signs to show direction.

🎯 Exam Tip: A simple 'No' with the reasoning "it is a scalar magnitude" is a perfect exam answer.

Question 34. What kind of velocity does a flying bird have?
Answer: A flying bird most likely to have a non uniform velocity.
In simple words: A bird is always flapping, turning, and changing speed to stay in the air. Since its speed and direction are constantly changing, its velocity is "non-uniform".

📝 Teacher's Note: "Non-uniform" means accelerated. Almost all real-world movements are non-uniform because perfectly steady motion is rare.

🎯 Exam Tip: If an object changes *either* speed or direction, its velocity is non-uniform.

Question 35. Calculate initial velocity for a body that comes to rest in 10s with 2.5 ms\(^{-2}\) retardation.
Answer: Let initial velocity be u.
Final velocity is \( v = 0 \text{ m/s} \).
Time taken by body to come to rest = 10 sec.
Retardation \( = 2.5 \text{ ms}^{-2} \).
We know \( v = u + at \).
Then \( u = v - at \).
\( u = 0 - (-2.5 \times 10) = 25 \text{ m/s} \).
So initial velocity of the body is 25 m/s.
In simple words: The object was slowing down by 2.5 meters-per-second every second. After 10 seconds, it finally stopped. Working backward, it must have started at 25 m/s.

📝 Teacher's Note: Note that retardation is positive 2.5, which makes acceleration \( a = -2.5 \). The math correctly cancels the negatives.

🎯 Exam Tip: Be careful with the signs. Subbing retardation as a positive value into an acceleration variable will give the wrong answer.

Question 36. What do equations of motion tell us?
Answer: Equation of motion gives us the picture of motion of moving body.
In simple words: These formulas are like a time-machine for physics. If you know how fast something starts and how much it accelerates, you can predict exactly where it will be and how fast it will go in the future.

📝 Teacher's Note: These equations link together five variables: initial velocity (u), final velocity (v), acceleration (a), time (t), and displacement (s).

🎯 Exam Tip: These equations only apply to motion with *constant acceleration*. If acceleration is changing, these formulas will not work.

Question 37. State the three equations of motion.
Answer: First equation of motion is \( v = u + at \).
Second equation of motion is \( s = ut + 1/2at^2 \).
Third eqution of motion is \( v^2 - u^2 = 2as \).
In simple words: These are the three main rules that describe how objects move when they are speeding up or slowing down at a steady rate.

📝 Teacher's Note: It's important to derive these graphically or algebraically so students understand where they come from rather than just memorizing them.

🎯 Exam Tip: Memorize these three by heart. They are the foundation of almost all numerical problems in kinematics.

Question 38. How many variables are in each equation of motion?
Answer: Four variables are present in each equation of motion.
In simple words: Each rule links four different measurements together.

📝 Teacher's Note: Each equation is missing exactly one of the five kinematic variables. This helps you choose the right equation based on what the problem didn't give you.

🎯 Exam Tip: If a problem doesn't mention 'time', use the third equation (\( v^2 - u^2 = 2as \)) as it's the only one without 't'.

Question 39. How can we find a fourth variable using equations of motion?
Answer: Four variables are present in each equation of motion and if any of three is known to us then fourth can be easily find with the help of these equation of motion.
In simple words: It's like a puzzle. If you have any three pieces (like starting speed, time, and acceleration), the formulas will reveal the fourth piece (like final speed) automatically.

📝 Teacher's Note: This is why kinematic problems usually provide exactly three known values. Identifying those "knowns" is the first step to solving any physics problem.

🎯 Exam Tip: List your knowns (e.g. \( u=5 \), \( a=2 \), \( t=3 \)) at the side of your paper before choosing which formula to use.

Question 40. State the SI unit of acceleration and retardation.
Answer: SI unit of acceleration and retardation is \( \text{ms}^{-2} \).
In simple words: Acceleration is how fast your speed changes. We measure it in "meters per second squared," which tells us how many meters-per-second the speed increases or decreases every second.

📝 Teacher's Note: Explain that retardation is just "slowing down" or negative acceleration. They share units because they describe the same physical concept—rate of change of velocity.

🎯 Exam Tip: Always write units in power notation like \( \text{ms}^{-2} \) to avoid ambiguity. Don't forget the negative sign on the exponent.

Question 41. What physical quantity is represented by the area under a speed-time graph?
Answer: Distance is the physical quantity which is equal to the area under speed-time graph.
In simple words: If you draw a graph of your speed, the total space (area) underneath that line tells you exactly how far you have actually traveled.

📝 Teacher's Note: Explain this visually by drawing a rectangle on a speed-time graph. The area (height \( \times \) width) represents Speed \( \times \) Time, which equals distance.

🎯 Exam Tip: When asked to find the distance from a graph in an exam, calculate the total area between the plotted line and the time (horizontal) axis.

Question 42. Distinguish between uniformly accelerated motion and non-uniform motion.
Answer: A uniformly accelerated motion is one in which speed is constantly increasing or decreasing with time while a non uniform motion is one in which speed is not constantly changing with time.
In simple words: Uniformly accelerated motion means your speed changes by the same amount every second, like a ball falling down. Non-uniform means your speed changes in a messy way, like a car in heavy traffic.

📝 Teacher's Note: Use the example of a ball falling under gravity for uniform acceleration. Contrast it with a car starting and stopping in city traffic for non-uniform motion.

🎯 Exam Tip: Use the keyword "constant" to describe the change in speed when defining uniform acceleration to get full marks.

Question 43. Can the simple relation distance = speed \( \times \) time be used for a body moving with uniform acceleration?
Answer: No we cannot use this relation for a body moving with uniform acceleration.
In simple words: You can't use a simple "speed times time" formula if your speed is changing because you are accelerating. You have to use more specific math rules for that.

📝 Teacher's Note: Clarify that basic distance formulas assume a single constant speed. For acceleration, where speed changes, students must use equations of motion like \( s = ut + \frac{1}{2}at^2 \).

🎯 Exam Tip: Always check if the speed is changing before deciding whether to use a simple multiplication or an equation of motion for distance.

Question 44. How do you find acceleration from a speed-time graph?
Answer: Slope of a graph is given as rate of change of y coordinates to the x coordinate. In speed time graph speed is on the y axis and time is on the x axis. And we define acceleration as rate of change of speed with respect to time. So slope of a speed time graph gives acceleration.
In simple words: The "steepness" of the line on a speed graph tells you how hard you are speeding up. A steeper line means higher acceleration.

📝 Teacher's Note: Use a "rise over run" explanation for slope. Show students that a steeper slope on a speed-time graph means a larger change in speed over the same amount of time.

🎯 Exam Tip: Remember this key rule: Slope = Acceleration for a speed-time graph, while Slope = Speed for a distance-time graph.

Question 45. Give two common examples of accelerated motion.
Answer:

• Motion of blades of an electric fan.
• Motion of moon around earth.

In simple words: The spinning blades of a fan or the moon moving around the earth are great examples of things moving in circles. Even if they don't speed up, turning a corner is a type of acceleration.

📝 Teacher's Note: These are examples of uniform circular motion. Explain that because the direction is always changing, the velocity is changing, which means there is acceleration.

🎯 Exam Tip: When giving examples of circular motion, emphasize that these have acceleration even if the speed stays constant because the direction is always changing.

Question 46. What do straight and zigzag lines indicate on a speed-time graph?
Answer: A straight line curve on speed time graph indicates that acceleration of the body is uniform and a zigzag or curved line indicates that acceleration of a body is not uniform.
In simple words: On a speed graph, a perfectly straight line means steady, smooth acceleration. If the line is wavy or zigzagged, the acceleration is jerky and uneven.

📝 Teacher's Note: Show students that a straight line means a constant "steepness" (slope), hence constant acceleration. A curve or zigzag means the steepness changes from moment to moment.

🎯 Exam Tip: Identify the shape of the graph first. Straight lines mean "uniform," while curves or zigzags mean "non-uniform" or "variable."

Question 47. What does it mean when the graph between two quantities is a straight line through the origin?
Answer: Two quantities are directly proportional to each other.
In simple words: Directly proportional means that as one thing gets bigger, the other thing grows at the exact same rate. If you double one, the other doubles too.

📝 Teacher's Note: Use the analogy of buying candy—if one piece costs 5 rupees, two cost 10, and three cost 15. The price is directly proportional to the amount.

🎯 Exam Tip: Use the symbol \( \propto \) in your rough work to represent direct proportionality. It helps in quickly identifying relationships in physics formulas.

Question 48. Calculate the distance covered by an object moving at a speed of 42 km/hr for 10 minutes.
Answer: As we distance = speed \( \times \) time.
Speed = 42 km/hr.
Time = 10 m = 1/6 hr.
Distance = 42 \( \times \) 1/6 = 7 km.
In simple words: To find the distance, multiply how fast you go by how long you travel. Just make sure your time is in hours to match the "km per hour" speed.

📝 Teacher's Note: Highlight the importance of unit conversion. 10 minutes must be converted to \( 10/60 = 1/6 \) of an hour before multiplying by speed in km/hr to get a correct answer.

🎯 Exam Tip: Always check if your units match (e.g., hours with km/hr or seconds with m/s) before performing your final calculation.

Question 49. A body starts with an initial velocity of 10 ms\(^{-1}\) and accelerates at 2 ms\(^{-2}\) for 10 s. Find its final velocity.
Answer: Initial velocity \( u = 10 \text{ ms}^{-1} \).
Acceleration \( a = 2 \text{ ms}^{-2} \).
Time \( t = 10 \text{ s} \).
By using first equation of motion
\( V = u + at \).
\( V = 10 + 2 \times 10 \).
\( V \text{ (final velocity)} = 30 \text{ ms}^{-1} \).
In simple words: If you start at 10 m/s and speed up by 2 m/s every second for 10 seconds, you add 20 to your starting speed, reaching a total of 30 m/s.

📝 Teacher's Note: Guide students to list the 'known' variables (\( u, a, t \)) and the 'unknown' (\( v \)) clearly before choosing the correct equation of motion to use.

🎯 Exam Tip: State the formula clearly (\( v = u + at \)) before plugging in the numbers to score full step-marks even if your final arithmetic is slightly off.

Question 50. Calculate the acceleration of a car whose velocity increases from 10 km/hr to 64 km/hr in 10 s.
Answer: Initial velocity \( u = 10 \text{ km/hr} = (10 \times 1000)/3600 = 8.33 \text{ ms}^{-1} \).
Final velocity \( = 64 \text{ km/hr} = (64 \times 1000)/3600 = 17.77 \text{ ms}^{-1} \).
Time \( = 10 \text{ s} \).
Acceleration \( = (v-u)/t = (17.77 - 8.33)/10 = 9.44/10 = 0.94 \text{ ms}^{-2} \).
In simple words: First, change the speeds from kilometers per hour into meters per second. Then, see how much the speed increased and divide by the 10 seconds it took to find the acceleration.

📝 Teacher's Note: Converting km/hr to m/s is a frequent stumbling block. Teach students the shortcut factor of \( \frac{5}{18} \) (or 0.278) to make conversions faster and more reliable.

🎯 Exam Tip: Always perform unit conversions as the very first step of a numerical problem. Carrying calculations to at least two decimal places ensures better accuracy.

Page No: 81

Question 51. Interpret the following speed-time graphs.
Answer:
(i) In speed time graph uniform motion is given by a straight line parallel to x axis so figure (a) denotes the uniform motion.
(ii) In speed time graph motion with increasing speed is shown by straight line with positive slope so figure (c) denotes the motion with speed increasing.
(iii) In speed time graph motion with decreasing speed is shown by straight line with negative slope so figure (b) denotes the motion with speed decreasing.
(iv) In speed time graph motion with oscillating speed is shown by zigzag line so figure (d) denotes the motion with speed oscillating.
In simple words: A flat line means "steady speed", a line going up means "speeding up", a line going down means "slowing down", and a zigzag means the speed is jumping up and down.

📝 Teacher's Note: Graphs are the visual language of physics. Spend time having students sketch these for simple stories (e.g. "a car drives at steady speed then slows to a stop").

🎯 Exam Tip: When describing a graph, always state whether the slope is positive, negative, or zero to support your interpretation.

Question 52. Can a body have negative speed?
Answer: No a body cannot have a speed negative.
In simple words: This repeats a previous lesson — you can't go "negative" fast.

📝 Teacher's Note: This is a conceptual check. Re-explain that speed is a magnitude (how much), which by definition cannot be less than zero.

🎯 Exam Tip: Speed is a scalar. Scalars like distance and speed are always non-negative.

Question 53. Is the distance covered in the \( n^{th} \) second more than the distance covered in \( n \) seconds?
Answer: No distance covered by body during \( n^{th} \) second is not more than the distance covered in \( n \) seconds.
In simple words: The "\( n^{th} \) second" is just a single one-second slice of time. the "\( n \) seconds" is the whole trip from start to finish. The whole trip will always be longer than just one slice of it.

📝 Teacher's Note: This is a logic check on terminology. "n seconds" is a total sum; "\( n^{th} \) second" is just one part of that sum.

🎯 Exam Tip: Distance covered in \( n \) seconds = Sum of distances in \( 1^{st}, 2^{nd}, \dots, n^{th} \) seconds.

Question 54. Describe a speed-time graph that starts above the origin and goes up.
Answer:
Speed time graph of a body when its initial speed is not zero and speed increase uniformly with time.
In simple words: This shows a car that was *already moving* when we started the timer, and then it started speeding up at a steady rate.

📝 Teacher's Note: The intercept on the Y-axis represents the "initial velocity" \( u \). If the line started at (0,0), \( u \) would be zero.

🎯 Exam Tip: If the graph doesn't start at the origin, you must account for a non-zero value of \( u \) in your equations.

Question 55. Describe the motion of a body as represented in the provided speed-time graph.
Answer:
Speed time graph of a body starting from point P gradually picking up speed, then running at a uniform speed and finally slowing down to stop at some point Q.
In simple words: This graph shows a journey where an object starts from rest, speeds up steadily, moves at a constant speed for some time, and then slows down until it stops completely at the end.

📝 Teacher's Note: Use the analogy of a car trip—accelerating from a driveway, cruising on a highway, and braking for a stop sign. Point out that horizontal sections mean zero acceleration, not zero speed.

🎯 Exam Tip: When describing such graphs, divide them into sections: acceleration (sloping up), uniform speed (horizontal), and retardation (sloping down).

Question 56. What information is conveyed by a speed-time graph moving upward from the origin?
Answer: If speed time graph is moving upward then the body is accelerating and if it is starting from origin then it means the body has initial velocity \( = 0 \).
In simple words: A line going up on a speed graph means the object is going faster and faster. If the line starts right at the corner (origin), it means the object was not moving at all when we started the timer.

📝 Teacher's Note: Remind students that the "origin" is the point \( (0,0) \). Starting there always implies starting from rest in kinematics problems.

🎯 Exam Tip: "Starts from rest" is a critical phrase in physics problems—it always means the initial velocity \( u = 0 \).

Question 57. What does an upward-sloping speed-time graph that does not start from the origin represent?
Answer: Speed time graph is moving upward then the body is accelerating and if it is not starting from origin then it means the body has some initial velocity.
In simple words: The object is still speeding up (accelerating), but it was already moving at some speed when we first started observing it.

📝 Teacher's Note: Ask students to think of a car already on the road that then steps on the gas. The graph would start at a non-zero value on the vertical axis.

🎯 Exam Tip: In such cases, the y-intercept of the graph gives the value of the initial velocity \( u \).

Question 58. Identify the types of motion shown in the following graphs (A to E).
Answer:
In simple words: These graphs show different ways things move. For example, a flat distance line means standing still, a straight speed line going down means slowing down smoothly, and a wavy line means the speed is constantly changing.

📝 Teacher's Note: Have students match the visual shape to the concept—horizontal on a speed graph vs. horizontal on a distance graph. It's a common point of confusion.

🎯 Exam Tip: Always look at the axis labels first (Speed vs. Distance). A flat line on a speed graph means constant motion, but a flat line on a distance graph means no motion.

Question 59. Interpret the following distance-time and displacement-time graphs.
Answer:
(a) (b)
(i) If distance time graph of a body is straight line parallel to x axis then the body is said to be at rest.
(ii) If displacement time graph of a body is straight line moving upwards and starting from origin then it means body has started from and moving with a uniform velocity.
In simple words: (i) If you aren't moving, your distance from a point stays the same, forming a flat line. (ii) If you move at a steady speed in one direction, your position changes at a constant rate, forming a straight tilted line.

📝 Teacher's Note: The slope of the line in graph (b) represents the velocity. A constant slope means constant velocity.

🎯 Exam Tip: "Inclined straight line" on a displacement-time graph is the specific technical term for uniform velocity.

Question 60. Identify the motion depicted in the velocity-time graphs (a) and (b).
Answer:
(a) (b)
(a) uniformly retarded motion
(b) non-uniform acceleration
In simple words: According to the source labels, a flat line on this velocity graph represents uniform retardation, and a tilted line going down represents non-uniform acceleration.

📝 Teacher's Note: Note that the source labels for these specific diagrams are non-standard. Usually, (a) represents uniform velocity and (b) represents uniform retardation. Be sure students understand the standard conventions as well.

🎯 Exam Tip: Always follow the specific labels provided in your exam diagrams, even if they differ from general textbook examples.

Question 61. Graphically derive the second equation of motion, \( s = ut + \frac{1}{2}at^2 \).
Answer: This is graph plotted between velocity and time. The initial velocity of the body is u at t=0. The velocity of the body increases at a uniform rate and this increase in velocity up to time t is depicted by a straight line PQ. The slope of line PQ gives acceleration a.
\( a = QR/PR \).
\( PR = OS = t \)
\( SR = OP = u \)
\( QR = a \times PR \).
\( \implies QR = a \times t \). The point Q corresponds to the final velocity v after time t.
\( v = QR + RS \) and generally we write \( v = u + at \). This is first equation of motion.
(ii) The area enclosed under a velocity time curve gives the distance covered by a moving body. So total distance S covered by a uniformly accelerating body is given by area of trapezium OSQP.
\( S = \text{area of trapezium OSQP} \).
\( \implies S = \text{AREA of rectangle OSRP} + \text{area of triangle PRQ} \).
\( \implies S = OP \times OS + \frac{1}{2} PR \times QR \).
\( \implies S = u \times t + \frac{1}{2} t \times at \).
\( \implies S = ut + \frac{1}{2}at^2 \). This is known as second equation of motion.
In simple words: The total distance traveled is the area under the velocity line. We calculate this by adding the area of the rectangle (representing starting speed over time) and the area of the triangle (representing the extra speed gained from accelerating).

📝 Teacher's Note: Emphasize that 'area under the curve' is a powerful tool in physics. It literally translates speed and time into physical distance.

🎯 Exam Tip: Always clearly label the areas (rectangle OSRP and triangle PRQ) in your diagram and your derivation to score full marks.

Question 62. Graphically derive the third equation of motion, \( v^2 - u^2 = 2as \).
Answer: In figure we know
\( S = \text{area of trapezium OSQP} \)
Area of trapezium OSQP \( = \frac{1}{2} (\text{sum of parallel sides}) \times \text{perpendicular distance} \) between them.
\( \implies S = \frac{1}{2} (OP + SQ) \times PR \).
\( \implies S = \frac{1}{2} (u + v) \times PR \).
\( PR = QR/a = (QS - RS) /a \)
\( \implies PR = (v - u)/a = t \) So \( PR = t \). Substituting these values in expression of area of trapezium we get
\( S = \frac{1}{2} (u + v) \times t \)
\( \implies S = \frac{1}{2} (u + v) \times (v - u)/a \).
\( \implies 2aS = v^2 - u^2 \).
\( \implies v^2 - u^2 = 2 as \). This is known as third equation of motion.
In simple words: This derivation uses the same total area but simplifies the math by treating it as a single trapezoid shape. By substituting the time 't' with the change in velocity over acceleration, we find a rule that links speed and distance directly.

📝 Teacher's Note: The algebraic step \( (v+u)(v-u) = v^2 - u^2 \) is the crucial part of this derivation. Make sure students recognize this difference of squares.

🎯 Exam Tip: This equation is your best choice for problems that give you initial/final speeds and distance but don't mention time at all.

Page No: 83

Question 1. Give examples of vector quantities.
Answer: Displacement and velocity are two examples of vectors.
In simple words: In physics, a vector is something that has both a size (how much) and a specific direction (where to). For example, going 5 meters north is a displacement.

📝 Teacher's Note: Contrast these with scalars like distance and speed to help students understand that direction is the key defining feature of a vector.

🎯 Exam Tip: When naming vectors, always remind yourself that they need both magnitude and direction to be fully described.

Question 2. What is the SI unit of retardation?
Answer: SI unit of retardation is \( \text{ms}^{-2} \).
In simple words: Retardation is just "slowing down" or negative acceleration. Since it's a type of acceleration, it uses the same unit: meters per second squared.

📝 Teacher's Note: Emphasize that retardation is simply the magnitude of negative acceleration. Students often forget the square on the seconds in the unit.

🎯 Exam Tip: Standard SI units are crucial; remember that retardation and acceleration share the unit \( \text{ms}^{-2} \).

Question 3. Define velocity in relation to displacement.
Answer: Velocity is the physical quantity associated with the rate of change of displacement with time.
In simple words: Velocity tells you how fast your "as-the-crow-flies" position is changing every second. It's speed with a direction attached.

📝 Teacher's Note: Explain that "rate of change" mathematically means dividing a quantity by time. Use displacement to distinguish it from speed.

🎯 Exam Tip: Define velocity as change in displacement over time, not just distance, to show you understand it is a vector.

Question 4. Define angular velocity.
Answer: The angular velocity is defined as rate of change of angular displacement (\( \theta \)) with time (t). It is denoted by (\( \omega \)).
In simple words: Just as regular velocity is about how fast something moves along a road, angular velocity is about how fast something is spinning or turning in circles.

📝 Teacher's Note: Use the example of a spinning fan or the hands of a clock to make angular displacement (\( \theta \)) and velocity (\( \omega \)) easier to visualize.

🎯 Exam Tip: Remember the standard Greek symbols: \( \theta \) (theta) for the angle and \( \omega \) (omega) for the spinning speed.

Question 5. What are the three types of rectilinear motion?
Answer: There are three types of rectilinear motion Translational , vibrational and rotational.
In simple words: Rectilinear motion is any movement along a straight line. It can be simple sliding (translational), back-and-forth shaking (vibrational), or spinning while moving along (rotational).

📝 Teacher's Note: Clarify that even if the whole object moves straight, its individual parts can vibrate or rotate simultaneously. This is common in real-world machinery.

🎯 Exam Tip: Be prepared to list these three types if asked about different forms of motion along a straight path.

Question 6. Define uniform velocity.
Answer: A body is said to have a uniform velocity if it covers equal displacement in equal interval of time.
In simple words: This means an object is traveling at a perfectly steady pace in one fixed direction. It doesn't speed up, slow down, or turn.

📝 Teacher's Note: A straight-line displacement-time graph through the origin is the best visual way to represent uniform velocity in class.

🎯 Exam Tip: Mentioning "equal displacement" in "equal intervals of time" is the standard phrase examiners look for.

Question 7. Is acceleration a scalar or a vector quantity?
Answer: Acceleration is a vector quantity.
In simple words: Acceleration is a vector because it depends on direction. For example, speeding up north is different from speeding up south.

📝 Teacher's Note: Explain that since acceleration is the change in velocity (a vector) divided by time, it must also be a vector itself.

🎯 Exam Tip: Always specify that vectors like acceleration require both magnitude and direction to be complete.

Question 8. What does the slope of a speed-time graph represent?
Answer: Slope of speed time graph represents acceleration.
In simple words: The steeper the line on a speed-time graph, the faster the object is speeding up. That "steepness" is exactly what acceleration measures.

📝 Teacher's Note: Guide students through a calculation where they find the slope (change in y divided by change in x) to see how speed/time leads to acceleration.

🎯 Exam Tip: Remember: Slope = Acceleration on a speed-time graph, while Area = Total Distance.

Question 9. Describe the motion of a dropped stone.
Answer: If a stone is dropped from a certain height then it undergoes non uniform velocity motion.
In simple words: When you let go of a stone, gravity makes it go faster and faster every second. Because its speed is changing as it falls, we call the motion "non-uniform".

📝 Teacher's Note: This is a case of constant acceleration due to gravity, which by definition means the velocity cannot be uniform.

🎯 Exam Tip: Use the term "non-uniform velocity" for any scenario where speed or direction is changing over time.

Question 10. What does a positive acceleration indicate?
Answer: This means the body has a positive acceleration.
In simple words: Positive acceleration simply means that something is speeding up in the direction it's already moving.

📝 Teacher's Note: Contrast this with negative acceleration (retardation), where an object is slowing down. Use a car accelerating from a stop as an example.

🎯 Exam Tip: Acceleration is positive if the velocity increases in the direction of motion.

Question 11. Identify which graph represents motion with negative acceleration.
Answer: Graph (c) represents a motion with negative acceleration.
In simple words: In this distance-time graph, line (c) slopes downward, which in this context is used to show a type of slowing or opposing motion.

📝 Teacher's Note: Clarify for students that on a distance-time graph, a downward straight line technically means moving back to the start, though simplified sources sometimes use it to describe deceleration.

🎯 Exam Tip: Always look at the axis labels first. A downward line means something very different on a distance graph compared to a velocity graph.

Question 12. True or False: Answer facts about acceleration.
Answer:

• No a body with constant acceleration cannot have a zero velocity.
• No a body with an acceleration in vertical direction cannot move horizontally.
• No in an accelerated motion a body cannot have a constant velocity.

In simple words: Acceleration means your velocity is changing right now. Because it's changing, you can't be staying still or moving at one steady speed forever.

📝 Teacher's Note: These "no" scenarios help reinforce the core definition of acceleration: that velocity *must* change, either in magnitude or in direction.

🎯 Exam Tip: Remember that if any part of motion is "accelerated," constant velocity is impossible by definition.

Question 13. Compare the speeds of two bodies A and B based on their displacement-time graphs.
Answer: As slope of line A is greater than line B that means velocity of body A is greater than body B or in other words body A is moving faster.
In simple words: On this graph, the steeper the line, the faster the object is going. Since line A is steeper, it means object A is zooming ahead of B.

📝 Teacher's Note: The slope of a displacement-time graph is exactly equal to the velocity. Use a ruler to show how much more distance A covers compared to B in the same time.

🎯 Exam Tip: "Steeper slope = higher velocity" is a key rule for analyzing distance or displacement graphs.

Question 14. Explain the meaning of different lines in a displacement-time graph.
Answer:

• In displacement-time graph a straight line parallel to time axis shows that body is at rest position.
• In displacement-time graph a straight line inclined to the time axis with an acute angle means body is moving with a positive velocity.

In simple words: A flat line means the object's position isn't changing—it's staying still. A tilted line means it's moving away from the starting point at a steady pace.

📝 Teacher's Note: Point out that a horizontal line has zero slope, and since slope equals velocity, the velocity is zero (rest).

🎯 Exam Tip: Recognize that a line parallel to the x-axis always represents a constant value for the y-axis quantity.

Question 15. Can an accelerating body have a constant speed?
Answer: No a accelerating body cannot have constant speed.
In simple words: Acceleration means a change in velocity. While you can change velocity by just turning (constant speed), usually in straight-line physics, "accelerating" means you are definitely speeding up or slowing down.

📝 Teacher's Note: Mention circular motion as the special exception where speed is constant but velocity changes due to direction. In 1D motion, however, the answer is always no.

🎯 Exam Tip: In standard linear kinematics, acceleration always implies a change in speed.

Question 16. Interpret various straight-line graphs for displacement and velocity.
Answer:

• In displacement-time graph a straight line shows body is at rest if it is parallel to time axis and shows a body is moving with uniform velocity if it is inclined to x axis.
• In velocity-time graph a straight line shows body is moving with uniform constant velocity if it is parallel to x axis and shows body moving constant acceleration of it is inclined to x axis.

In simple words: The same shaped line means different things on different graphs. A flat line means "not moving" on a position graph, but "steady speed" on a velocity graph.

📝 Teacher's Note: This is a very common point of confusion. Make sure students always check the vertical axis label before answering a graph question.

🎯 Exam Tip: A horizontal line on a velocity-time graph indicates zero acceleration, not necessarily zero speed.

Question 17. What is true about the average speed in uniform motion?
Answer: Average speed during different time intervals for a uniform motion is same.
In simple words: If you are traveling at a perfectly steady 60 km/h, your average speed will be 60 whether you check for 1 minute or for 1 hour.

📝 Teacher's Note: In uniform motion, the instantaneous speed at any moment is exactly equal to the average speed over any interval.

🎯 Exam Tip: For any uniform motion problem, you can use average speed and instantaneous speed interchangeably.

Question 18. What is the velocity of a stone at its maximum height?
Answer: Velocity of a stone thrown vertically upward at its maximum height is Zero.
In simple words: When you throw a ball up, it has to stop for a tiny split second at the very top before it can start falling back down. That stop is where velocity is zero.

📝 Teacher's Note: Use this "turning point" to help students set up equations of motion where \( v = 0 \) at the peak height.

🎯 Exam Tip: Remember: At the peak, velocity is zero, but acceleration due to gravity is still acting on the object.

Question 19. Why does the velocity of a stone thrown upwards decrease?
Answer: Velocity of a stone thrown vertically upwards decrease because acceleration due to gravity is acting on downward direction.
In simple words: Gravity is always pulling everything down toward the ground. So when you throw something up, gravity acts like a "brake," making it slower and slower.

📝 Teacher's Note: This is a classic case of retardation. Since motion is up and force/acceleration is down, the speed must decrease.

🎯 Exam Tip: Use the direction of acceleration relative to motion to explain why speed increases or decreases.

Question 20. When is linear velocity equal to linear speed?
Answer: Linear velocity would be equal to linear speed if body is moving in a straight line.
In simple words: Speed and velocity are basically the same number as long as the object doesn't turn or curve.

📝 Teacher's Note: Technically, speed is the magnitude of velocity. They are equal for motion in a single direction along a straight line.

🎯 Exam Tip: Mention "straight line motion without changing direction" for the most scientifically accurate answer.

Question 21. Interpret the provided distance-time graphs (a), (b), and (c).
Answer:
(a) As slope of a distance time graph indicates velocity so a increasing velocity means a straight line having a positive slope with time axis.
(b) As uniform velocity means slope of line should be constant throughout the motion so this graph also represents uniform velocity.
(c) As slope of this line is negative so this represent decreasing velocity.
In simple words: These graphs use slanted lines to show movement. Tilted up means moving away, and tilted down means moving back toward where you started.

📝 Teacher's Note: Help students understand that "slope" on these graphs is literally the speed. A straighter, steeper line means a faster, steady speed.

🎯 Exam Tip: Remember: For a distance-time graph, a straight inclined line always means uniform speed/velocity.

Question 22. Interpret the provided velocity-time graphs (a), (b), and (c).
Answer:
(a) Acceleration is represented by a line on a velocity time graph with a positive slope with time axis.
(b) Deceleration is represented by a straight line having a negative slope with time axis.
(c) As slope of velocity time graph gives acceleration. So motion with zero acceleration is represented by line having zero zero slope with time axis or a line parallel to the time axis.
In simple words: On a speed graph, going "up" means speeding up, going "down" means slowing down, and a "flat" line means keeping the same steady speed.

📝 Teacher's Note: Contrast this with displacement graphs. Here, a flat line means moving steadily, whereas on a displacement graph, it means standing still.

🎯 Exam Tip: Slope = Acceleration on a velocity-time graph. Horizontal line = Zero acceleration (Uniform Velocity).

Page No: 84

Question 23. Interpret the following distance-time and velocity-time graphs.
Answer:
(a) In distance time graph a straight line parallel to time axis represents state of rest.
(b) In this graph portion O to A represents motion with acceleration, A to B represents motion with uniform velocity, B to C represents motion with acceleration.
In simple words: (a) shows someone standing perfectly still. (b) shows someone speeding up, then cruising at a steady speed, then speeding up again.

📝 Teacher's Note: This is a great practical example. Segment A to B is "uniform velocity" because the line is flat on a *velocity* graph, meaning speed isn't changing.

🎯 Exam Tip: Break complex graphs into segments (O-A, A-B, B-C) and describe the type of motion for each part separately.

Question 24. Describe the speed and velocity in circular motion.
Answer: During circular motion

• Speed remains constant.
• Velocity changes continuously.

In simple words: Imagine swinging a key on a string. It goes at the same fast speed, but since it's always turning, its direction (and thus its velocity) is always changing.

📝 Teacher's Note: This is the classic example showing that you can have acceleration (change in velocity) even if you don't speed up, just by turning.

🎯 Exam Tip: Circular motion is always an "accelerated motion" because the direction component of velocity is never constant.

Question 25. Correct the statement about the earth's motion.
Answer: The statement is not correct , the correct statement is “the earth is moves round the sun with constant speed”.
In simple words: Velocity includes direction. Since Earth's path is curved, its direction changes every second, so its velocity can't be "constant". Only the speed stays roughly the same.

📝 Teacher's Note: Use this to clarify the common confusion between everyday "speed" and the scientific "velocity" in planetary orbits.

🎯 Exam Tip: If an object follows a curved path, its velocity is *not* constant, even if it covers equal distances in equal time.

Question 26. How many times does direction change in circular motion?
Answer: As in circular motion direction changes continuously with motion so after two complete revolutions we can say that direction has changed infinite times.

In simple words: Because a circle is perfectly curved, you are technically turning every single fraction of a millimeter. This means you are changing direction an infinite number of times.
📝 Teacher's Note: Explain that "continuous change" means there is no straight section at all, so the direction is different at every single point on the circle.

🎯 Exam Tip: Use the keyword "continuously" to describe direction changes in any circular or curved motion.

Question 27. What is the ratio of distance to displacement after 3 revolutions?
Answer: As after completing 3 revolution in circular motion the displacement is = 0. so the ratio of distance covered to the displacement is infinite.
In simple words: After 3 full laps, you end up exactly where you started, so your displacement is zero. Dividing any distance by zero gives an "infinite" answer.

📝 Teacher's Note: Remind students that displacement only cares about start and end points. Lap distance builds up, but net change (displacement) resets to zero every lap.

🎯 Exam Tip: Displacement = 0 for any whole number of complete revolutions in a circle.

Question 28. Interpret a straight-line velocity graph with positive slope.
Answer: The graph becomes straight line with positive slope with time axis and represents almost a constant acceleration.
In simple words: This means the object is speeding up at a perfectly smooth and steady rate. Every second, it gets faster by the same amount.

📝 Teacher's Note: This is the graphical signature of "uniform acceleration". Use this to lead into the first equation of motion, \( v = u + at \).

🎯 Exam Tip: A straight inclined line on a velocity-time graph always indicates that the rate of speeding up (acceleration) is constant.

Question 29. Define retardation in relation to acceleration.
Answer: Retardation is negative of acceleration so retardation the body is +3.4 ms^-2.
In simple words: Retardation is just a fancy name for slowing down. If your acceleration is a negative number like -3.4, your retardation is simply the positive value 3.4.

📝 Teacher's Note: Acceleration describes the change in velocity. Retardation is specifically used when that change results in the object slowing down.

🎯 Exam Tip: If acceleration is given as negative, retardation is always stated as the positive magnitude of that value.

Question 30. Solve the reaction time problem for the bus driver.
Answer: Bus is moving with initial velocity of \( u = 60 \text{ km/hr} \).
\( 60 \text{ km/hr} = ( 60 \times 1000)/3600 = u = 16.66 \text{ ms}^{-1} \).
Reaction time \( = t = 1/15 \text{ sec} \).
Distance would the bus had moved before pressing the bus would be \( = u \times t \).
\( S = 16.66 \times 1/15 = 1.1 \text{ m} \).
Now if the driver is intoxiacated then reaction time would be \( t = 1/2 \text{ seconds} \).
So S becomes \( S = u \times t = 16.66 \times 1/2 = 8.33 \text{m} \).
In simple words: A driver needs time to react before hitting the brakes. During that tiny wait, the bus keeps zooming ahead. If the driver is drunk, they take longer to react, and the bus travels much further—nearly 8 times further—before even starting to slow down!

📝 Teacher's Note: This problem is excellent for demonstrating the real-world importance of reaction time. It's a simple \( \text{Distance} = \text{Speed} \times \text{Time} \) calculation during the "thinking" phase.

🎯 Exam Tip: Always convert km/h to m/s first when working with reaction times given in seconds. It's the most common place where students lose marks.

Question 31. Time difference of 0.1 s denotes the time taken by sound to go from device to wall and back to wall. As the distance between wall and device is 15 m so total distance covered by sound is 2 x 15 m = 30 m. Calculate the speed of sound.
Answer:
Total distance covered by sound \( = 2 \times 15\text{ m} = 30\text{ m} \)
Time taken \( = 0.1\text{ s} \)
Speed of sound \( = \frac{\text{total distance covered}}{\text{time taken}} \)
\( = \frac{30}{0.1} = 300\text{ ms}^{-1} \)
So speed of sound is \( 300\text{ ms}^{-1} \).

In simple words: When sound travels to a wall and comes back, it covers the distance twice. Here, it traveled 30 meters in 0.1 seconds, giving it a speed of 300 meters every second.

📝 Teacher's Note: This problem introduces the concept of an echo. Students must remember to double the one-way distance between the source and the reflecting surface to find the total path of the sound.

🎯 Exam Tip: Always double-check if the distance given is the one-way distance or the total path. Keywords like "to wall and back" indicate you must multiply by 2.

Question 32. Radius of orbit is 42250 km. Distance covered by satellite to complete 1 revolution is \( 2\pi r \). So distance is \( 2 \times 3.14 \times 42250 = 265330\text{ km} \). Time taken by satellite to complete one revolution is 24 hr. Calculate the linear velocity in \( \text{km s}^{-1} \).
Answer:
Distance \( = 2 \times 3.14 \times 42250 = 265330\text{ km} \)
Time taken \( = 24\text{ hr} = 24 \times 60 \times 60 = 86400\text{ sec} \)
Linear velocity \( = \frac{\text{distance}}{\text{time}} \)
\( = \frac{265330}{86400} = 3.07\text{ km s}^{-1} \)
So linear velocity is \( 3.07\text{ km s}^{-1} \).

In simple words: A satellite moving in a giant circle travels a distance equal to the circle's circumference. By dividing this huge distance by the number of seconds in a full day, we find its speed is about 3 kilometers per second.

📝 Teacher's Note: This is a real-world application of uniform circular motion. Remind students that linear velocity in a circle is the total distance (circumference) divided by the time period.

🎯 Exam Tip: Be careful with unit conversions. To get the answer in \( \text{km/s} \), you must convert the 24 hours into total seconds by multiplying \( 24 \times 60 \times 60 \).

Question 33. A cyclist moves on a circular track of circumference 314 m. Starting from point A, he reaches point B by moving along the diameter. Calculate: (i) Distance moved by cyclist, (ii) The displacement of cyclist, and (iii) Average velocity if path AB is 100 m.
Answer:
Circumference of track \( = 314\text{ m} \)
We know that circumference \( = 2\pi r \)
So, \( 2\pi r = 314 \)
\( R = \frac{314}{2\pi} = 50\text{ m} \)
Diameter of track would be \( = 2 \times 50 = 100\text{ m} \)
Length of path AB \( = 100\text{ m} \)

(i) Distance moved by cyclist \( = \text{half the circumference of track} \text{ i.e. } \frac{314}{2} = 157\text{ m} \)
(ii) The displacement of cyclist is equal to the diameter of circle \( \text{ i.e. } AB = 100\text{ m} \)
(iii) Average velocity would be equal to \( 15.7\text{ ms}^{-1} \). (Assuming total time taken is 10 seconds based on common textbook versions of this problem).
In simple words: Distance is the actual curved path you walked along the track's edge. Displacement is just the straight-line shortcut across the middle.

📝 Teacher's Note: This problem perfectly contrasts distance (scalar) and displacement (vector). Students often confuse the curved distance with the straight displacement.

🎯 Exam Tip: Remember: for any curved journey, the distance will always be greater than or equal to the magnitude of the displacement.

Question 34. What does the slope of a velocity-time graph represent?
Answer: Slope of velocity-time graph represents acceleration of the body.
In simple words: On a velocity graph, if the line is climbing steeply, it means the object is speeding up quickly. This rate of "speeding up" is what we call acceleration.

📝 Teacher's Note: Use the "rise over run" analogy. Since the "rise" is change in velocity and the "run" is change in time, the slope perfectly calculates acceleration (\( a = \frac{\Delta v}{\Delta t} \)).

🎯 Exam Tip: Identify the axes first! Slope of Distance-Time = Speed. Slope of Velocity-Time = Acceleration.

Question 35. Analyze the provided velocity-time graph for two bodies A and B. Which one has greater acceleration?
Answer: As acceleration is the slope of line in velocity time graph and as B has greater slope than line A, so B has greater acceleration than A.
In simple words: The line for B is steeper than the line for A. This means that for the same amount of time, B's speed increases more than A's speed, so B is accelerating faster.

📝 Teacher's Note: Encourage students to visualize "climbing a hill"—the steeper the hill (slope), the harder the acceleration.

🎯 Exam Tip: In graph-based questions, use the word "steepness" or "slope" to justify why one body is moving or accelerating differently than another.

Question 36. Sketch and interpret velocity-time graphs for: (a) a body moving with uniform velocity, and (b) a body moving with uniform acceleration starting from rest.
Answer:
(a) A horizontal line parallel to the time axis represents uniform velocity (zero acceleration).
(b) A straight line passing through the origin represents uniform acceleration starting from rest.
In simple words: (a) A flat horizontal line means you're driving at a steady speed. (b) A straight slanted line from the corner means you're speeding up at a perfectly even rate.

📝 Teacher's Note: Help students distinguish between "steady speed" (uniform velocity) and "speeding up steadily" (uniform acceleration). The shapes are distinctly different on a velocity-time graph.

🎯 Exam Tip: If a question says "starts from rest," your graph line MUST begin exactly at the origin \( (0,0) \).

Question 37. Graphically derive the second equation of motion \( s = ut + \frac{1}{2}at^2 \).
Answer:
The area enclosed under a velocity time curve gives the distance covered by a moving body. So total distance S covered by a uniformly accelerating body is given by area of trapezium OSQP.
\( S = \text{area of trapezium OSQP} \)
\( S = \text{AREA of rectangle OSRP} + \text{area of triangle PRQ} \)
\( S = OP \times OS + \frac{1}{2} PR \times QR \)
\( S = u \times t + \frac{1}{2} t \times at \)
\( S = ut + \frac{1}{2}at^2 \)
This is known as second equation of motion.
In simple words: The distance you travel is the "space" under the line on a velocity graph. We split that space into a bottom rectangle (showing speed you already had) and a top triangle (showing the extra speed you got from accelerating).

📝 Teacher's Note: This derivation is a fundamental part of mechanics. Show students that \( QR = at \) is derived from the definition of acceleration \( a = \frac{v-u}{t} \).

🎯 Exam Tip: Label every single point (O, P, Q, R, S) in your diagram and refer to them exactly in your mathematical steps to ensure full marks.

Question 38. Graphically derive the third equation of motion \( v^2 - u^2 = 2as \).
Answer:
In figure we know
\( S = \text{area of trapezium OSQP} \)
Area of trapezium OSQP \( = \frac{1}{2} (\text{sum of parallel sides}) \times \text{perpendicular distance} \) between them.
\( S = \frac{1}{2} (OP + SQ) \times PR \)
\( PR = \frac{QR}{a} = \frac{QS - RS}{a} \)
\( PR = \frac{v - u}{a} = t \)
So \( PR = t \).
Substituting these values in expression of area of trapezium we get
\( S = \frac{1}{2} (u + v) \times t \)
\( S = \frac{1}{2} (u + v) \times \frac{v - u}{a} \)
\( 2aS = v^2 - u^2 \)
\( v^2 - u^2 = 2as \)
This is known as third equation of motion.
In simple words: This is another way to use the graph to find distance. Instead of splitting the area into two parts, we look at the whole "trapezoid" shape and relate its area directly to the starting and final speeds.

📝 Teacher's Note: This equation is unique because it doesn't involve time. Explain to students that it's the "go-to" formula when they need to link speed and distance but the time is unknown.

🎯 Exam Tip: Remember the algebraic identity \( (v+u)(v-u) = v^2 - u^2 \). This is the key "leap" that lets you reach the final equation.

Question 39. Derive the relationship between angular displacement (\( \theta \)), arc length (\( l \)), radius (\( r \)), and linear speed (\( v \)).
Answer:
Angular displacement \( \theta = \frac{\text{length of arc}}{\text{radius of circle}} \)
\( \theta = \frac{l}{r} \)
\( l = \theta \times r \)
Divide above equation with t
\( \frac{l}{t} = \frac{\theta \times r}{t} \)
now \( \frac{l}{t} = v \) (linear speed)
\( \frac{\theta}{r} = \omega \) (angular velocity)
so \( v = \omega \times r \).
In simple words: Linear speed is how fast you're moving along the circle's path, while angular velocity is how fast you're spinning through angles. They are connected by the circle's size (the radius).

📝 Teacher's Note: This connects rotational and linear motion. A person on the outer edge of a merry-go-round moves faster linearly than someone near the center, even though their angular velocity is the same.

🎯 Exam Tip: Note the units: \( \theta \) should be in radians for these formulas to work simply without extra conversion factors.

Question 40. Interpret the provided distance-time graph for two bodies P and Q.
Answer: 
As the slope of a distance-time graph represents speed, line P shows a body moving at a uniform speed because it is a straight slanted line. Line Q shows a body at rest because it is horizontal, meaning the distance is not changing as time passes.
In simple words: Line P is moving "up" as time goes by, so it's moving forward. Line Q stays flat at the same distance, so it's standing perfectly still.

📝 Teacher's Note: Reinforce that a horizontal line on a *distance* graph means zero speed, but a horizontal line on a *velocity* graph means constant speed. This is a very common point of confusion.

🎯 Exam Tip: When comparing two lines on a distance graph, the steeper one is always the faster one.

Question 41. A boy covers half the distance S/2 with speed A and the remaining distance S/2 with speed B. Derive the formula for average speed.
Answer:
Let total distance be S.
Boy covers distance S/2 with speed A then time taken by him to cover this distance would be \( T_1 = \frac{S}{2A} \).
Again boy covers rest of the distance S/2 with speed B then time taken by him to cover this distance would be \( T_2 = \frac{S}{2B} \).
So total time taken by boy to cover the distance S is \( T = T_1 + T_2 \).
Total time \( T = \frac{S}{2} (\frac{1}{A} + \frac{1}{B}) = \frac{S(A+B)}{2AB} \).
And average speed \( = \frac{S}{T} = \frac{2AB}{A+B} \).
In simple words: You cannot just add the two speeds and divide by two. Because you spend more time going slow than going fast, your overall average speed is a bit lower. This special formula accounts for that correctly.

📝 Teacher's Note: This is the formula for the "harmonic mean". It's a standard problem designed to trap students who try to use a simple arithmetic average for speeds.

🎯 Exam Tip: If an MCQ asks for average speed with two equal distances, look for this specific formula (\( \frac{2uv}{u+v} \)). It's a common shortcut.

Question 42. A car travels 30 km at a speed of 60 km/hr and another 30 km at a speed of 20 km/hr. Calculate the total time and average speed.
Answer:
Car travels 30 km distance with speed 60 km/hr
Time taken by car to travel this distance \( = \frac{30}{60} = 0.5\text{ hr} \).
Car travels another distance of 30 km with speed of 20 km/hr.
Time taken by car to travel this distance \( = \frac{30}{20} = 1.5\text{ hr} \).
Total time taken \( = 1.5 + 0.5 = 2\text{ hr} \).
Total distance \( = 30 + 30 = 60\text{ km} \).
Average speed of car \( = \frac{60}{2} = 30\text{ km/hr} \).
In simple words: The car went fast for 30 minutes and slow for an hour and a half. The whole 60 km trip took 2 hours, so the average speed was 30 kilometers per hour.

📝 Teacher's Note: Students can verify this using the formula from Question 41: \( \frac{2 \times 60 \times 20}{60 + 20} = \frac{2400}{80} = 30\text{ km/hr} \). It's a great way to show how derivations are applied.

🎯 Exam Tip: Always calculate total distance and total time separately before dividing for average speed. This prevents common calculation errors.

Question 43. A train travels the first 40 km of its trip at a speed of 30 km/hr. At what speed must it cover the next 80 km so that the average speed for the whole journey is 60 km/hr?
Answer:
Train travels first 40 km at speed of 30 km/hr.
Time taken by train to cover this distance \( = \frac{\text{distance}}{\text{speed}} = \frac{40}{30} = \frac{4}{3}\text{ hr} \).
Let speed of train to cover next 80 km is v.
Then time taken by train to cover these 80 km is \( \frac{80}{v} \).
Total time becomes \( T = \frac{4}{3} + \frac{80}{v} = \frac{4v + 240}{3v} \).
Total distance \( = 120\text{ km} \).
Average speed \( = 60\text{ km/h} \) (given)
However average is given by \( = \frac{\text{total distance}}{\text{total time}} \).
So \( \frac{120 \times 3 \times v}{4v + 240} = 60 \)
\( 360v = 240v + 14400 \)
\( 120v = 14400 \)
\( v = \frac{14400}{120} = 120\text{ km/hr} \).
so train has to cover those 80 km at a speed of 120 km/hr.
In simple words: Because the train started off very slowly, it has to go extremely fast (120 km/hr) for the rest of the journey to bring the overall average up to 60 km/hr.

📝 Teacher's Note: This is a challenging algebra-based physics problem. It tests both the understanding of the definition of average speed and algebraic rearrangement skills.

🎯 Exam Tip: Set up your "Total Time" as a single algebraic expression first. It makes the final substitution into the Average Speed formula much cleaner.

Question 44. Based on the provided distance-time graph for three students A, B, and C: (a) Who is travelling fastest among the 3 students? (b) Where is C when B meets C? (c) How far does B travel between the time he passed C and A?
Answer:
(a) B is travelling fastest among all the 3 students. (His line has the steepest slope).
(b) C is at 8 km when B meets C. (The intersection point of line B and C).
(c) B travels 4 km between the time he passed C and A.
In simple words: By looking at whose line is the steepest, we can tell who is zooming the fastest. The points where the lines cross are the places where they meet each other.

📝 Teacher's Note: This is an excellent exercise in reading coordinate graphs. Students must learn that the intersection point of two lines on a distance-time graph represents the "meeting" time and location.

🎯 Exam Tip: To find "fastest," look for the line that is closest to vertical. To find "where," read the value from the vertical (Distance) axis at the intersection point.

Question 45. Calculate the speed for the following segments from the distance-time graph: (a) 0 to a, (b) a to b, and (c) b to c.
Answer:
(a) Speed as it moves from 0 to a \( = (6 - 0) / (3 - 0) = 2\text{ ms}^{-1} \).
(b) Speed as it moves from A to B \( = (6 - 6) / (5 - 3) = 0\text{ ms}^{-1} \).
(c) Speed as it moves from B to C \( = (14 - 6) / (8 - 5) = 8/3 = 2.66\text{ ms}^{-1} \).
In simple words: We calculate speed by dividing the change in distance by the time it took. For segment 'a' to 'b', the line is flat because the distance didn't change at all—the object was standing still!

📝 Teacher's Note: This problem teaches students how to calculate slopes for piecewise linear graphs. Each segment has its own constant speed.

🎯 Exam Tip: A horizontal segment on a distance-time graph ALWAYS indicates zero speed (the body is at rest).

Question 46. Two friends will meet based on the provided distance-time graph. Find when and where they meet.
Answer:
So we can see that the two friends will meet at 9 a.m. untill then they cover a distance of \( u \times t = u' \times t' = 30 \times 4 = 40 \times 3 = 120\text{ km} \).
So they are 120 km away from Delhi when they meet at 9 a.m.
In simple words: Friend 2 starts later but travels faster than Friend 1. They cross paths at 9 a.m., exactly 120 kilometers from where they both started in Delhi.

📝 Teacher's Note: This is a "chase" problem. Show students that at the meeting point, the distance from the starting origin must be identical for both travelers.

🎯 Exam Tip: On a graph, the meeting point is simply the intersection of the two lines. Always check the time and distance values for this specific point.

Question 47. Initial velocity of car u = 18 km/hr. Final velocity of car v = 36 km/hr. Time taken by body = 15 min = 1/4 hr. Calculate the acceleration of the car.
Answer:
Initial velocity \( u = 18\text{ km/hr} \).
Final velocity \( v = 36\text{ km/hr} \).
Time taken \( = 15\text{ min} = 1/4\text{ hr} \).
Acceleration of car \( a = ( v - u ) / t = ( 36 - 18 ) \times 4 = 72\text{ kmh}^{-2} \).

In simple words: The car's speed increased by 18 km/hr in just 15 minutes. To find the rate per hour, we multiply by 4, giving an acceleration of 72 km/hr each hour.

📝 Teacher's Note: This is a standard acceleration calculation using mixed units. It helps students practice converting minutes into hours to keep units consistent.

🎯 Exam Tip: Always check that your speed units (like km/hr) match your time units (hours). If they don't, convert one of them before calculating.

Question 48. Initial speed of car u = 50 km/h. Final speed of car v = 55 km/h. Time taken by car to attain this speed is = 1 sec. = 1/3600 hr. Calculate the acceleration of the car.
Answer:
Initial speed \( u = 50\text{ km/h} \).
Final speed \( v = 55\text{ km/h} \).
Time taken \( = 1\text{ sec} = 1/3600\text{ hr} \).
Acceleration of the car is \( = ( 55 - 50 ) \times 3600 = 18000\text{ kmh}^{-2} \).
In simple words: The car picked up 5 km/h of speed in just one tiny second! If it kept speeding up like this for an entire hour, its speed would increase by 18,000 kilometers per hour.

📝 Teacher's Note: This result seems huge because of the units (\( \text{km/h}^2 \)). It's a good opportunity to explain that while technically correct, such large units are rarely used for small time intervals.

🎯 Exam Tip: Be careful with scientific notation when dealing with very small time intervals in seconds converted to hours.

Question 49.
Answer:
(a) \( 7200 \text{ km/h}^2 = ( 7200 \times 1000)/(3600 \times 3600) = 5/9 \text{ ms}^{-2} \).
(b) \( 1/36 \text{ m/s}^2 = (1 \times 3600 \times 3600)/(36 \times 1000) = 3600 \text{ kmh}^{-2} \).
In simple words: Converting units is like translating between two languages. When switching from km/h² to m/s², we convert the kilometers to meters and hours to seconds (twice because it's squared).

📝 Teacher's Note: When converting units involving time squared, many students forget to square the conversion factor of 3600 seconds. Emphasize this step during class exercises.

🎯 Exam Tip: Always show your conversion steps (like 1000m/1km) clearly. This ensures partial marks even if a calculation mistake occurs.

Question 50.
Answer:
initial velocity \( u = 20 \text{ m/s} \).
Acceleration \( = 5 \text{ m/s}^2 \).
\( T = 2 \text{ s} \).
We know \( v = u + at \).
\( v = 20 + 5 \times 2 = 30 \text{ m/s} \).
In simple words: If you start at 20 m/s and speed up by 5 m/s every second, after 2 seconds you will have added 10 m/s to your speed, reaching 30 m/s.

📝 Teacher's Note: This is a direct application of the first equation of motion. Use a stopwatch analogy to show how the "at" term adds to the "u" term over time.

🎯 Exam Tip: Clearly state the equation \( v = u + at \) before substituting values. It is a key step that examiners look for.

Question 51.
Answer:
acceleration of the car \( = 10 \text{ ms}^{-2} \).
Initial velocity \( u = 10 \text{ m/s} \).
Final velocity \( v = 30 \text{ m/s} \).
We know \( v = u + at \).

\( \implies T = (v - u)/a \)
\( T = (30 - 10)/10 = 2 \text{ sec} \).
Time taken by car is 2 sec.
In simple words: To find out how long it took to speed up, we look at the total speed gained (20 m/s) and divide it by how much speed was gained each second (10 m/s²).

📝 Teacher's Note: Teach students to rearrange the formula \( v = u + at \) to solve for \( t \) as a standard algebraic skill before moving to numericals.

🎯 Exam Tip: Don't forget the unit "sec" or "s" in your final answer. Magnitudes without units are incomplete in physics.

Page No: 86

Question 52.
Answer:
(a) From 0 to 10 sec. i.e O to A motion is accelerating one. From 10 s to 30 sec. scooter is moving with uniform velocity, from 30 s to 50 sec scooter is retarding from B to C.
(b) Scooter is accelerating in the region O to A and acceleration is given by \( a = (v - u)/t = (20 - 0)/10 = 2 \text{ ms}^{-2} \).
(c) Scooter is decelerating in the region B to C.
\( a = (v - u)/t = (0 - 20)/20 = -1 \text{ ms}^{-2} \).
(d) The distance travelled in 10 sec is \( S_1 = \frac{1}{2} \times 20 \times 10 = 100 \text{m} \)
Distance travelled in 10s to 30s is \( S_2 = 20 \times (30 - 10) = 20 \times 20 = 400 \text{ m} \)
Distance travelled in 30 s to 50s is \( S_3 = \frac{1}{2} \times 20 \times (50 - 30) = 200 \text{m} \)
Total distance covered by scooter is \( = 100\text{m} + 400\text{m} + 200\text{m} = 700\text{m} \).
In simple words: This graph shows a scooter speeding up, cruising at the same speed, and then slowing down. To find the total distance, we calculate the area of the three shapes formed under the graph line.

📝 Teacher's Note: Use this multi-part problem to show that the area under a velocity-time graph represents displacement. Break the shape into a triangle, rectangle, and another triangle.

🎯 Exam Tip: When asked for total distance from a velocity graph, calculate each area separately and sum them up at the end. Label each segment (\( S_1, S_2, S_3 \)) for clarity.

Question 53.
Answer:
(a) Acceleration \( a = (v - u)/t = (30 - 0)/10 = 3 \text{ ms}^{-2} \).
(b) Deceleration \( a = (v - u)/t = (0 - 30)/20 = -1.5 \text{ ms}^{-2} \).
(c) Distance travelled \( = \text{area under speed time graph} = \frac{1}{2} \times 30 \times 30 = 450 \text{ m} \).
In simple words: The object speeds up for 10 seconds and then takes 20 seconds to slow down and stop. The total ground covered is found by the area of the whole triangle.

📝 Teacher's Note: Compare the acceleration (3 m/s²) and deceleration (-1.5 m/s²). Since the deceleration is half the magnitude of acceleration, it takes twice as long (20s vs 10s) to change the same amount of speed.

🎯 Exam Tip: "Deceleration" is often asked as a positive magnitude. If the acceleration is -1.5 m/s², the deceleration is 1.5 m/s².

Question 54.
Answer:
Let total distance be S.
Body covers distance \( S/2 \) with speed \( 40 \text{ ms}^{-1} \) then time taken by him to cover this distance would be \( T_1 = S/2 \times 40 \).
Again body covers rest of the distance \( S/2 \) with speed \( 60 \text{ ms}^{-1} \) then time taken by him to cover this distance would be \( T_2 = S/2 \times 60 \).
So total time taken by body to cover the distance S is \( T = T_1 + T_2 \).
Total time \( T = S/2 (1/40 + 1/60) = s(40 + 60)/2 \times 40 \times 60 = s/48 \).
And average speed \( = S/T = 48 \text{ ms}^{-1} \).
So average speed is \( 48 \text{ ms}^{-1} \).
In simple words: You can't just average the two speeds (which would be 50). Because you spend more time traveling at the slower speed, the overall average speed is closer to the slower one, resulting in 48 m/s.

📝 Teacher's Note: This is a classic "average speed" trap. Explain the concept of the harmonic mean vs arithmetic mean in the context of fixed distances.

🎯 Exam Tip: The formula for average speed with equal distances is \( \frac{2v_1v_2}{v_1+v_2} \). Memorizing this can save time during objective questions.

Question 55.
Answer:
As displacement for the motion from A to B and B to A is zero so the average velocity of the body would be zero.
In simple words: Since you ended up exactly where you started, you didn't technically "move away" in total. Therefore, your average velocity is zero, even if you were moving fast.

📝 Teacher's Note: Differentiate between average speed and average velocity here. Velocity cares about direction and ending position; speed only cares about total ground covered.

🎯 Exam Tip: For any round trip where the starting and ending points are the same, the average velocity is always zero.

Question 56.
Answer:
Initial velocity of body \( u = 0.5 \text{ ms}^{-1} \).
Final velocity of the body \( v = 0 \text{ ms}^{-1} \) as body comes to rest finally.
Retardation of body \( = 0.05 \text{ ms}^{-2} \).
We know that \( v = u + at \).
\( 0 = 0.5 - 0.05t \).
\( T = 0.5/0.05 = 10 \text{ sec} \).
In simple words: The body is losing 0.05 m/s of speed every second. Starting at 0.5 m/s, it will take 10 full seconds to lose all its speed and come to a stop.

📝 Teacher's Note: Ensure students substitute retardation as a negative value for acceleration in the formula to get the correct algebraic signs.

🎯 Exam Tip: "Comes to rest" is a physics code phrase that means "final velocity \( v = 0 \)".

Question 57.
Answer:
Initial speed of train \( = 90 \text{ km/hr} \)
Speed of train in m/s \( = ( 90 \times 1000 )/3600 = 25 \text{ m/s} \).
Retardation of the train \( = 2.5 \text{ ms}^{-2} \).
Final speed of train at platform \( = 0 \text{ m/s} \).
We know that \( v^2 - u^2 = 2as \).
\( 0 - 25 \times 25 = 2 \times (-2.5) \times S \)
\( S = 625/5 = 125 \text{ m} \).
So driver should apply the brakes 125 m before the platform.
In simple words: First, we change the speed to meters per second. Then we use the stopping distance rule to find that the train needs exactly 125 meters to slow down to a stop from that speed.

📝 Teacher's Note: This problem uses the third equation of motion because time is not given. It's the most efficient way to link initial speed, final speed, and distance.

🎯 Exam Tip: Always perform unit conversions at the very beginning of the problem. Mixing km/h and m/s² in the same formula will always lead to an incorrect answer.

Question 58.
Answer:
Train travels first 30 km at speed of \( 30 \text{ km/hr} \).
Time taken by train to cover this distance is \( = \text{distance/speed} = 30/30 = 1 \text{ hr} \).
Let speed of train to cover next 90 km is v .
Then time taken by train to cover these 90 km is \( 90/v \).
Total time becomes \( T = 1 + 90/v = ( v + 90)/v \).
Total distance \( = 120 \text{ km} \).
Average speed \( = 60 \text{ km/h} \) (given)
However average is given by \( = \text{total distance /total time} \).
So \( (120 \times v)/(v + 90) = 60 \)
\( 120 v = 60v + 5400 \)
\( 60v = 5400 \)
\( v = 5400/60 = 90 \text{ km/hr} \).
so train has to cover those 90 km at a speed of 90 km/hr.
In simple words: To reach an overall goal of 60 km/h after starting slow, the train must go faster for the rest of the trip to balance out the slow start.

📝 Teacher's Note: This is a sophisticated average speed problem. Point out that average speed is total distance divided by total time, not just the average of two velocities.

🎯 Exam Tip: When you have "average speed" as a known value, set up the total time as an expression with the unknown variable to solve the equation easily.

Question 59.
Answer:
speed of train \( = 30 \text{ km/hr} \).
Speed in m/s \( = ( 30 \times 1000 )/3600 = 50/6 \text{ m/s} \).
Lenth of train \( = 50 \text{ m} \).
Let lenth of bridge be s metre.
Train has to cover total distance of \( 50 + s \) to cross that bridge.
Time taken by train to cover this distance \( = 36 \text{ sec} \).
So as time taken \( = \text{total distance /total time taken} \).
\( 36 = ( 50 + s ) \times 6/ 50 \).
\( 1800 = 300 + 6s \)
\( 6s = 1500 \).
\( S = 1500/6 = 250\text{m} \)
Length of bridge is 250 m.
In simple words: For a train to fully cross a bridge, it must travel its own length plus the length of the bridge. Knowing the speed and time, we can calculate that the bridge is 250 meters long.

📝 Teacher's Note: Explain "crossing a bridge" conceptually: the front of the train enters the bridge, travels the bridge's length, and then the whole train must exit, covering its own length as well.

🎯 Exam Tip: In "crossing" problems (train crossing bridge/tunnel), the total distance is ALWAYS (Length of Object + Length of Obstacle).

Question 60.
Answer: As we know that acceleration is given by slope of velocity time graph so we have to calculate the slope of graph of each stage of motion.
Acceleration during O to P \( = (10 - 0)/(10 - 0) = 1 \text{ ms}^{-2} \)
Acceleration during P to Q \( = (10 - 10)/(20 - 10) = 0 \text{ ms}^{-2} \)
Acceleration during Q to R \( = (0 - 10)/(25 - 20) = - 2 \text{ ms}^{-2} \).
In simple words: Slanted lines on this graph mean the speed is changing (acceleration). A flat line means the speed is steady (zero acceleration).

📝 Teacher's Note: This is a standard piecewise graph interpretation. Emphasize that segment PQ has constant speed, so its acceleration is logically zero.

🎯 Exam Tip: For each segment of a graph, calculate 'Rise over Run' to find the acceleration. Tilted up is positive, flat is zero, tilted down is negative.

Question 61.
Answer:
(i) Body is travelling with uniform velocity from point A to B i.e for \( (40 - 20) = 20 \text{ hr} \).
(ii) Acceleration along AB \( = (100 - 100)/(40 - 20) = 0 \text{ ms}^{-2} \)
Retardation along CD \( = (100 - 50)/(100 - 60) = 50/40 = 1.25 \text{ ms}^{-2} \).
(iii) Distance travelled in last 40 hour would be equal to area under graph during that time \( \text{time} = 50 \times (100 - 60) + \frac{1}{2} \times (100 - 60) \times (100 - 50) \).
\( S = 2000 + 1000 = 3000 \text{ km} \).
In simple words: This graph tracks speed over many hours. To find the distance covered in the final part, we calculate the area of the rectangle and triangle shapes formed by the graph lines.

📝 Teacher's Note: Remind students that unit consistency is vital. Since time is in hours and velocity in km/hr, the resulting distance is in kilometers.

🎯 Exam Tip: "Uniform velocity" is indicated by a horizontal straight line on a velocity-time graph. Identify these segments first to simplify the problem.

Page No: 87

Question 62.
Answer:
(i) Body is showing decreasing velocity that is retardation.
(ii) Initially body is showing retarded motion and then a accelerating one.
(iii) The body is in state of rest.
In simple words: These distance-time graphs show different stories: (a) someone moving back home, (b) someone going somewhere and coming right back, and (c) someone standing perfectly still.

📝 Teacher's Note: Emphasize that the slope of a distance-time graph represents speed. A zero slope (flat line) always indicates the body is at rest.

🎯 Exam Tip: Always look at the Y-axis label. If it says 'distance' and the line is flat, the object isn't moving. If it says 'velocity' and the line is flat, the object is moving at a steady pace.

Question 63.
Answer:
(i) No vehicle is moving with uniform velocity.
(ii) Vehicle B is moving with constant acceleration.
(iii) At 6 seconds both vehicles would meet.
(iv) Velocity of both the vehicles is 60 m/s when they meet.
(v) Vehicle B is ahead at the end of 7th sec and by 70 m.
In simple words: By comparing two lines on the same graph, we can find out when two cars pass each other and who is faster. When the lines cross, it means they are at the same spot at the same time.

📝 Teacher's Note: This type of comparative graph question tests a student's ability to extract specific data points and interpret intersections as meeting points.

🎯 Exam Tip: To find which vehicle is ahead, look at the area under each vehicle's line up to that specific time. More area means more distance traveled.