Selina Concise Solutions for ICSE Class 9 Physics Chapter 2 Motion In One Dimension

Exercise 2(A)

Question 1. Differentiate between Scalar and Vector quantities.
Answer:

In simple words: A scalar is just a number (like your weight), while a vector is a number with a specific direction (like a car driving North).

📝 Teacher's Note: Use the analogy of "distance" vs "displacement" to show that adding vectors isn't always as simple as \( 2 + 2 = 4 \).

🎯 Exam Tip: Remember to mention that vectors require a specific type of math (vector algebra) for addition and subtraction.

Question 2. Classify the following quantities as scalar or vector: (a) Pressure, (b) Momentum, (c) Weight, (d) Force, (e) Energy, (f) Speed.
Answer:
a) Pressure is a scalar quantity.
b) Momentum is a vector quantity.
c) Weight is a vector quantity.
d) Force is a vector quantity.
e) Energy is a scalar quantity.
f) Speed is a scalar quantity.
In simple words: Pressure and Energy are scalars because it doesn't matter which way they point. Force and Weight are vectors because they always push or pull in a specific direction.

📝 Teacher's Note: Students often mistake pressure for a vector. Remind them that at a point in a fluid, pressure acts equally in all directions, making it a scalar.

🎯 Exam Tip: Weight is always a vector because it is the force of gravity acting towards the center of the Earth.

Question 3. What do you understand by the term rest?
Answer: A body is said to be at rest if it does not change its position with respect to its immediate surroundings.
In simple words: If you are sitting on a chair and your distance from the table doesn't change, you are at rest.

📝 Teacher's Note: Rest is relative. A person in a moving train is at rest relative to their seat, but in motion relative to the platform.

🎯 Exam Tip: Always include the phrase "with respect to its surroundings" to get full marks.

Question 4. What do you understand by the term motion?
Answer: A body is said to be in motion if it changes its position with respect to its immediate surroundings.
In simple words: If you walk away from your desk, your distance from it is changing, so you are in motion.

📝 Teacher's Note: Like rest, motion is also relative. Everything in the universe is in motion relative to something else.

🎯 Exam Tip: Motion involves a continuous change of position over time.

Question 5. Define one-dimensional motion.
Answer: When a body moves along a straight line path, its motion is said to be one-dimensional motion.
In simple words: Think of a train on a perfectly straight track—it can only go forward or backward along that one line.

📝 Teacher's Note: In one-dimensional motion, only one coordinate (like x) changes with time.

🎯 Exam Tip: One-dimensional motion is often referred to as rectilinear motion.

Question 6. What is displacement? State its S.I. unit.
Answer: The shortest distance from the initial to the final position of the body is called the magnitude of displacement. It is in the direction from the initial position to the final position.
Its SI unit is metre (m).
In simple words: If you walk in a big circle and end up back where you started, your displacement is zero because you didn't actually go anywhere from your starting point.

📝 Teacher's Note: Draw a zig-zag path and then a straight dotted line from start to finish to show the difference between distance and displacement.

🎯 Exam Tip: Displacement is a vector quantity; don't forget to mention its direction.

Question 7. Differentiate between distance and displacement.
Answer: Distance is a scalar quantity, while displacement is a vector quantity. The magnitude of displacement is either equal to or less than the distance. The distance is the length of path travelled by the body so it is always positive, but the displacement is the shortest length in direction from initial to the final position so it can be positive or negative depending on its direction. The displacement can be zero even if the distance is not zero.
In simple words: Distance is how much ground you covered total. Displacement is how far you ended up from your starting line.

📝 Teacher's Note: Use the example of a 400m race on a track. The distance is 400m, but the displacement is 0m because the finish line is the start line.

🎯 Exam Tip: Displacement \( \le \) Distance. This is a very common true/false question.

Question 8. Can displacement be zero even if distance is not zero? Explain with an example.
Answer: Yes, displacement can be zero even if the distance is not zero. For example, when a body is thrown vertically upwards from a point A on the ground, after sometime it comes back to the same point A. Then, the displacement is zero, but the distance travelled by the body is not zero (it is 2h; h is the maximum height attained by the body).
In simple words: If you throw a ball up and catch it, the ball traveled up and down (distance), but since it ended up back in your hand, its total "move" (displacement) is zero.

📝 Teacher's Note: This occurs because the upward vector and downward vector cancel each other out in terms of displacement.

🎯 Exam Tip: Use a circular path or a "return trip" as your standard example for zero displacement.

Question 9. When is the magnitude of displacement equal to the distance?
Answer: The magnitude of displacement is equal to distance if the motion of the body is one-dimensional (along a straight line without changing direction).
In simple words: If you walk in a straight line from your house to the shop and don't turn back, your distance and displacement are exactly the same.

📝 Teacher's Note: This is the only case where displacement and distance are numerically identical.

🎯 Exam Tip: The condition is "straight line motion without reversing."

Question 10. Define velocity and state its S.I. unit.
Answer: The velocity of a body is the distance travelled per second by the body in a specified direction.
Its SI unit is metre/second (m/s).
In simple words: Velocity is just speed with a direction added to it.

📝 Teacher's Note: Formula: \( \text{Velocity} = \frac{\text{Displacement}}{\text{Time}} \). Make sure students distinguish it from Speed = Distance/Time.

🎯 Exam Tip: Always specify the unit as \( \text{m s}^{-1} \) or m/s.

Question 11. Define speed and state its S.I. unit.
Answer: The speed of a body is the rate of change of distance with time.
Its SI unit is metre/second (m/s).
In simple words: Speed tells you how fast you're going, like "60 miles per hour," but it doesn't say which way you're headed.

📝 Teacher's Note: Speed is a scalar quantity. Use the car speedometer as a real-world example—it shows speed, not velocity.

🎯 Exam Tip: Rate of change usually means dividing by time.

Question 12. Distinguish between speed and velocity.
Answer: Speed is a scalar quantity, while velocity is a vector quantity. The speed is always positive—it is the magnitude of velocity, but the velocity is given a positive or negative sign depending upon its direction of motion. The average velocity can be zero but the average speed is never zero.
In simple words: Velocity can be zero if you go out and come back, but your speed (since you were moving) will always be a positive number.

📝 Teacher's Note: Use a sign convention: moving right is positive velocity, moving left is negative velocity.

🎯 Exam Tip: Focus on the scalar vs vector difference and the fact that average velocity can be zero.

Question 13. What does velocity determine?
Answer: Velocity gives the direction of motion of the body.
In simple words: If you know the velocity, you know exactly where the object is going.

📝 Teacher's Note: Velocity is displacement per unit time. Because displacement has direction, velocity must also have direction.

Question 14. When is instantaneous velocity equal to average velocity?
Answer: Instantaneous velocity is equal to average velocity if the body is in uniform motion.
In simple words: If you are driving at a perfectly steady speed in one direction, your speed at 1:00 PM is the same as your average speed for the whole trip.

📝 Teacher's Note: "Uniform motion" means velocity remains constant throughout the interval.

🎯 Exam Tip: Uniform motion implies zero acceleration.

Question 15. Distinguish between uniform and variable velocity.
Answer: If a body travels equal distances in equal intervals of time along a particular direction, then the body is said to be moving with a uniform velocity. However, if a body travels unequal distances in a particular direction in equal intervals of time or it moves equal distances in equal intervals of time but its direction of motion does not remain same, then the velocity of the body is said to be variable (or non-uniform).
In simple words: Uniform velocity is like a car on cruise control going straight. Variable velocity is like a car turning a corner or slowing down for a light.

📝 Teacher's Note: Emphasize that velocity changes even if the speed is constant if the direction changes.

🎯 Exam Tip: A body in circular motion has variable velocity even if its speed is constant.

Question 16. Define average speed and average velocity. Can average velocity be zero?
Answer: Average speed is the ratio of the total distance travelled by the body to the total time of journey; it is never zero. If the velocity of a body moving in a particular direction changes with time, then the ratio of total displacement to the total time taken in entire journey is called its average velocity. Average velocity of a body can be zero even if its average speed is not zero.
In simple words: Average speed measures your whole "walk," while average velocity only cares about how far you ended up from where you started.

📝 Teacher's Note: Average velocity = (Total Displacement) / (Total Time). If Displacement is 0, Average Velocity is 0.

🎯 Exam Tip: This is a favorite comparison question in exams. Remember the "return trip" scenario.

Question 17. Why does a body in circular motion with uniform speed have variable velocity?
Answer: The motion of a body in a circular path with uniform speed has a variable velocity because in the circular path, the direction of motion of the body continuously changes with time.
In simple words: Imagine spinning a ball on a string. Even if it goes around at a steady speed, the ball is constantly changing direction, which means its velocity is never the same from one second to the next.

📝 Teacher's Note: This is a classic example of "accelerated motion" with constant speed.

🎯 Exam Tip: The change in direction at every point is the key reason here.

Question 18. If a body starts from a point and returns to the same point, what are its displacement, average velocity, and average speed?
Answer: If a body starts its motion from a point and comes back to the same point after a certain time, then the displacement is zero, average velocity is also zero, but the total distance travelled is not zero, and therefore, the average speed is not zero.
In simple words: You walk to the shop and back. You did "work" (distance/speed), but your total "move" (displacement/velocity) from your house is zero because you're back in your bed.

📝 Teacher's Note: This reinforces the displacement concept. Displacement depends only on the starting and ending points.

Question 19. Define acceleration and state its S.I. unit.
Answer: Acceleration is the rate of change of velocity with time.
Its SI unit is metre/second\(^2\) (\( \text{m/s}^2 \)).
In simple words: Acceleration is how much you speed up or slow down every second.

📝 Teacher's Note: Formula: \( a = \frac{v - u}{t} \). Explain the unit \( \text{m/s}^2 \) as "(metres per second) per second."

🎯 Exam Tip: If velocity is constant, acceleration is zero.

Question 20. Distinguish between acceleration and retardation.
Answer: Acceleration is the increase in velocity per second, while retardation is the decrease in velocity per second. Thus, retardation is negative acceleration. In general, acceleration is taken positive, while retardation is taken negative.
In simple words: Stepping on the gas pedal is acceleration; stepping on the brakes is retardation.

📝 Teacher's Note: Retardation is also known as deceleration.

🎯 Exam Tip: When a car stops, the final velocity \( v = 0 \), and the acceleration value will be negative.

Question 21. What is the difference between uniform and variable acceleration?
Answer: The acceleration is said to be uniform when equal changes in velocity take place in equal intervals of time, but if the change in velocity is not the same in the same intervals of time, the acceleration is said to be variable.
In simple words: Uniform acceleration means speeding up by exactly the same amount (like 5 mph) every second. Variable means speeding up in jerks—sometimes a lot, sometimes a little.

📝 Teacher's Note: Gravity is a perfect example of uniform acceleration (\( 9.8 \text{ m/s}^2 \)).

Question 22. Define retardation and state its S.I. unit.
Answer: Retardation is the decrease in velocity per second.
Its SI unit is metre/second\(^2\) (\( \text{m/s}^2 \)).
In simple words: It's just acceleration with a minus sign because you are slowing down.

📝 Teacher's Note: The units for acceleration and retardation are identical.

Question 23. What determines the direction of motion?
Answer: Velocity determines the direction of motion.
In simple words: To know where something is heading, you look at its velocity vector.

Question 24. Give one example each for: (a) Uniform velocity, (b) Variable velocity, (c) Variable acceleration, (d) Uniform retardation.
Answer:
(a) Example of uniform velocity: A body, once started, moves on a frictionless surface with uniform velocity.
(b) Example of variable velocity: A ball dropped from some height is an example of variable velocity (its speed increases due to gravity).
(c) Example of variable acceleration: The motion of a vehicle on a crowded road is with variable acceleration.
(d) Example of uniform retardation: If a car moving with a velocity ‘v’ is brought to rest by applying brakes, then such a motion is an example of uniform retardation.
In simple words: Uniform motion is steady; variable motion changes; retardation is stopping.

📝 Teacher's Note: Most real-world motions are variable. Uniform motion is usually an idealization used for physics problems.

Question 25. A car leaves a trail of oil drops on the road at equal time intervals. If the drops are initially equidistant but later the distance between them decreases, what can you say about the car's motion?
Answer: Initially as the drops are equidistant, we can say that the car is moving with a constant speed but later as the distance between the drops starts decreasing, we can say that the car slows down.
In simple words: Equidistant drops mean the car traveled the same distance in each second. When the drops get closer together, it means the car is covering less ground each second—it's braking!

📝 Teacher's Note: This is a visual representation of a ticker-tape timer experiment used in physics labs.

🎯 Exam Tip: Decreasing distance between markers in equal time steps = Retardation.

Question 26. Define acceleration due to gravity. What is its average value?
Answer: When a body falls freely under gravity, the acceleration produced in the body due to the Earth’s gravitational acceleration is called the acceleration due to gravity (g). The average value of g is \( 9.8 \text{ m/s}^2 \).
In simple words: If you drop a stone, Earth's pull makes it faster by 9.8 meters per second every single second it falls.

📝 Teacher's Note: "Free fall" means the only force acting is gravity (no air resistance).

Question 27. Does the value of 'g' remain the same everywhere on Earth?
Answer: No. The value of ‘g’ varies from place to place. It is maximum at poles and minimum at the Equator on the surface of the Earth.
In simple words: Because the Earth isn't a perfect sphere (it's a bit squashed), you are actually closer to the center at the poles, so gravity pulls a bit harder there.

📝 Teacher's Note: The Earth is an oblate spheroid. Radius at poles < Radius at equator.

🎯 Exam Tip: Weight changes with location because 'g' changes, but mass stays the same.

Question 28. If a feather and a stone are dropped in a vacuum, which will hit the ground first?
Answer: In vacuum, both will reach the ground simultaneously because acceleration due to gravity is same (=g) on both objects.
In simple words: Without air to push back on the feather, gravity treats everything the same. A heavy stone and a light feather will fall at exactly the same speed!

📝 Teacher's Note: This is Galileo's famous principle. In air, the feather is slowed by air resistance, but in a vacuum, only gravity matters.

🎯 Exam Tip: The keyword here is "vacuum." If the question says "in air," the stone hits first.

Chapter 2. Motion in One Dimension

Question1. Identify the vector quantity among: speed, distance, mass, and velocity.
Answer: Velocity is a vector quantity. The others are all scalar quantities.
In simple words: Velocity is special because it tells you both how fast you are going and which way you are heading (like "50 km/h North"), whereas speed only tells you the fastness.

📝 Teacher's Note: Use the analogy of a compass to explain vector quantities. If you need a direction to make the information useful, it's likely a vector.

🎯 Exam Tip: Displacement, velocity, and force are the "big three" vector quantities you must memorize for this chapter.

Question 2M. State the S.I. unit of speed.
Answer: \( \text{m s}^{-1} \)
In simple words: In science, we measure speed by seeing how many meters an object covers in exactly one second.

📝 Teacher's Note: Remind students that \( \text{m s}^{-1} \) is mathematically the same as m/s. Get them comfortable with negative indices early.

🎯 Exam Tip: Always write the power clearly. A common mistake is writing \( ms^1 \) instead of \( ms^{-1} \).

Question 3M. State the S.I. unit of acceleration.
Answer: \( \text{m s}^{-2} \)
In simple words: Acceleration measures how many "meters per second" your speed increases every second.

📝 Teacher's Note: Explain that acceleration is the "rate of change of velocity." The "squared" second in the unit comes from dividing velocity (m/s) by time (s).

🎯 Exam Tip: Acceleration units always have "seconds squared" in the denominator.

Question 4M. What is the displacement of a body that returns to its starting point?
Answer: The displacement is zero.
In simple words: Displacement only cares about the gap between where you started and where you ended. if you end up back at the start, there is no gap at all!

📝 Teacher's Note: Contrast this with distance. If you walk 5km and return, your distance is 10km, but your displacement is 0.

🎯 Exam Tip: Displacement depends only on initial and final positions, regardless of the path taken.

Question 5M. Express a speed of 18 km h\(^{-1}\) in S.I. units.
Answer: \( 5 \text{ m s}^{-1} \)
In simple words: Moving 18 kilometers in one hour is the exact same speed as moving 5 meters in one second.

📝 Teacher's Note: Teach the shortcut: multiply km/h by \( \frac{5}{18} \) to get m/s instantly.

🎯 Exam Tip: S.I. units for speed must always be expressed in meters per second.

Question 1N. Speed of car \( = 72 \text{ km h}^{-1} \). Calculate speed of car in \( \text{m s}^{-1} \).
Answer: Speed of car \( = 72 \text{ km h}^{-1} \)
Speed of car in \( \text{m s}^{-1} = \frac{72 \times 1000}{3600} = 20 \text{ m/s} \)
In simple words: We turn the 72 kilometers into 72,000 meters and the one hour into 3,600 seconds to find the speed in standard units.

📝 Teacher's Note: Walk students through the fraction simplification: \( \frac{1000}{3600} = \frac{5}{18} \). This makes future calculations much faster.

🎯 Exam Tip: Show the conversion steps clearly to earn partial marks even if you make a calculation error.

Question 2N. Convert a speed of 15 m/s into km/h.
Answer: \( 15 \text{ m/s} = \frac{15}{1000} \times 3600 \text{ km/hr} \)
\( \implies \) \( 15 \text{ m/s} = 54 \text{ km/hr} \)
In simple words: To convert back to km/h, we multiply the speed by 3.6.

📝 Teacher's Note: This is the inverse of the previous conversion. Remind students that speed in km/h will always be a larger number than the same speed in m/s.

🎯 Exam Tip: Use the factor \( \frac{18}{5} \) for converting m/s to km/h.

Question 3N. Express the following in \( \text{ms}^{-1} \):
(a) 1 km h\(^{-1}\)
(b) 18 km min\(^{-1}\)
Answer:
(a) \( 1 \text{ km h}^{-1} \)
\( 1 \text{ km} = 1000 \text{ m} \)
\( 1 \text{ h} = 3600 \text{ s} \)
\( \therefore 1 \text{ km h}^{-1} = 1 \times \frac{1000}{3600} = \frac{5}{18} \text{ ms}^{-1} \)

(b) \( 18 \text{ km min}^{-1} \)
\( 1 \text{ km} = 1000 \text{ m} \)
\( 1 \text{ min} = 60 \text{ s} \)
\( \therefore 18 \text{ km min}^{-1} = 18 \times \frac{1000}{60} = 300 \text{ ms}^{-1} \)
In simple words: We break down the distance and time units separately and then divide them to get the answer in meters per second.

📝 Teacher's Note: Pay close attention to "per minute" in part (b). Students often default to dividing by 3600 even when it's not hours.

🎯 Exam Tip: Read the time unit carefully—"min" is different from "h".

Question 4N. Arrange the following in increasing order of speed: 18 km h\(^{-1}\), 10 m s\(^{-1}\), and 1 km min\(^{-1}\).
Answer: \( 18 \text{ km h}^{-1} < 10 \text{ m s}^{-1} < 1 \text{ km min}^{-1} \)
In simple words: To compare them fairly, we turn them all into m/s: 18 km/h is only 5 m/s, while 1 km/min is a very fast 16.6 m/s.

📝 Teacher's Note: Comparison is only possible when all quantities are in the same units. Converting everything to S.I. is the best practice.

🎯 Exam Tip: Use the "<" symbol to show increasing order clearly.

Question 5N. Total time taken \( = 3 \text{ hours} \). Speed of the train \( = 65 \text{ km/hr} \). Calculate the distance travelled.
Answer: Total time taken \( = 3 \text{ hours} \)
Speed of the train \( = 65 \text{ km/hr} \)
Distance travelled \( = \text{speed} \times \text{time} \)
\( \implies \) \( 65 \times 3 = 195 \text{ km} \)
In simple words: If a train covers 65 kilometers in every hour, in three hours it will cover three times that much.

📝 Teacher's Note: This is the simplest distance calculation. Ensure students know the "Speed Triangle": Distance on top, Speed and Time on the bottom.

🎯 Exam Tip: Always include the unit (km) in your final answer.

Question 6N. A car travels the first 30 km at 60 km/h and the next 30 km at 40 km/h. Calculate (i) total time and (ii) average speed.
Answer:
For the first 30 km travelled, speed \( = 60 \text{ km/h} \).
Thus time taken (\( t_1 \)) \( = \text{Distance} / \text{speed} \)
\( = (30/60) \text{ h} \)
\( \implies \) \( t_1 = 0.5 \text{ h} \) or 30 min.

For the next 30 km travelled, speed \( = 40 \text{ km/h} \)
Thus time taken (\( t_2 \)) \( = \text{Distance} / \text{speed} \)
\( = (30/40) \text{ h} \)
\( \implies \) \( t_2 = 0.75 \text{ h} \) or 45 min.

(i) Total time \( = (0.5 + 0.75) \text{ h} = 1.25 \text{ h} \).
(ii) Average speed of the car \( = \text{Total distance travelled} / \text{total time taken} \)
\( \implies \) \( \frac{60 \text{ km}}{1.25 \text{ hr}} = 48 \text{ km/h} \)
In simple words: You cannot just find the average of 60 and 40. You must find out how long the whole trip took and divide the total distance by that total time.

📝 Teacher's Note: Emphasize that Average Speed is NOT the average of speeds. It is a ratio of totals.

🎯 Exam Tip: The answer 50 km/h (average of 60 and 40) is a common trap. Always use the "Totals" formula.

Question 7N. A train travels 200 km and then returns back. The total time taken is 5 hours. Calculate (i) Average speed and (ii) Average velocity.
Answer:
Here, total distance \( = (200 + 200) \text{ km} = 400 \text{ km} \)
Total time taken \( = 5 \text{ h} \)
(i) Average speed \( = \text{Total distance travelled} / \text{total time taken} \)
\( \implies \) \( \frac{400 \text{ km}}{5 \text{ h}} = 80 \text{ km/h} \)

(ii) Average velocity of the train is zero because the train stops at the same point from where it starts, i.e. the displacement is zero.
In simple words: Speed counts the whole 400 km trip. But since the train ended up exactly where it started, its displacement is zero, and that makes the velocity zero too.

📝 Teacher's Note: This is a key conceptual point. Velocity is displacement-based. If you return home, your trip velocity is always zero.

🎯 Exam Tip: For any "return to start" problem, average velocity is always zero.

Question 8N. A car covers 1 km in 100 s towards east. Calculate: (i) Speed and (ii) Velocity.
Answer:
(i) Speed of the car \( = \text{Distance} / \text{time taken} \)
\( = \frac{1 \text{ km}}{100 \text{ s}} = \frac{1000 \text{ m}}{100 \text{ s}} = 10 \text{ m/s} \)

(ii) Velocity of car \( = \text{Speed with direction} \)
\( \implies \) \( 10 \text{ m/s} \) due east
In simple words: Speed is just the number (how fast), but velocity is the number plus the direction (how fast and where).

📝 Teacher's Note: Remind students to convert km to m before dividing by seconds.

🎯 Exam Tip: Velocity answers MUST include a direction (like "due east") to be correct.

Question 9N. Calculate the acceleration if a car reaches 10 m/s from rest in 2 s.
Answer: Here, final velocity \( = 10 \text{ m/s} \)
Initial velocity \( = 0 \text{ m/s} \)
Time taken \( = 2 \text{s} \)
Acceleration \( = (\text{Final Velocity} - \text{Initial Velocity}) / \text{time} \)
\( \implies \) \( (10/2) \text{ m/s}^2 \)
\( = 5 \text{ m/s}^2 \)
In simple words: Acceleration is how much the speed goes up every second. Here, it went up by 10 in 2 seconds, so that's 5 each second.

📝 Teacher's Note: "From rest" always means initial velocity \( u = 0 \).

🎯 Exam Tip: Use the formula \( a = \frac{v - u}{t} \) and state the units as \( \text{m/s}^2 \).

Question 10N. A car starts from rest and acquires a velocity of 180 m/s in 3 minutes. Calculate its acceleration.
Answer: Here, final velocity \( = 180 \text{ m/s} \)
Initial velocity \( = 0 \text{ m/s} \)
Time taken \( = 3 \text{ min} = 180 \text{ s} \)
Acceleration \( = (\text{Final Velocity} - \text{Initial Velocity}) / \text{time} \)
\( \implies \) \( (180 - 0) / 180 \text{ m/s}^2 \)
\( = 1 \text{ m/s}^2 \)
In simple words: The car gained 180 units of speed in 180 seconds, so it gained exactly 1 unit of speed every second.

📝 Teacher's Note: Never use minutes in the denominator for acceleration. Always convert to seconds first.

🎯 Exam Tip: Check your time conversion: \( 3 \text{ min} = 180 \text{ s} \). This is the most common place for errors.

Question 11N. A car moving at 50 m/s slows down to 20 m/s in 3 s. Calculate its acceleration and retardation.
Answer: Here, final velocity \( = 20 \text{ m/s} \)
Initial velocity \( = 50 \text{ m/s} \)
Time taken \( = 3 \text{ s} \)
Acceleration \( = (\text{Final Velocity} - \text{Initial Velocity}) / \text{time} \)
\( \implies \) \( (20 - 50) / 3 \text{ m/s}^2 \)
\( = -10 \text{ m/s}^2 \)
Negative sign here indicates that the velocity decreases with time, so retardation is 10 \( \text{m/s}^2 \).
In simple words: Acceleration is negative because the car is braking. Retardation is just the name for "slowing down" acceleration.

📝 Teacher's Note: Retardation is the absolute value of negative acceleration. If \( a = -10 \), then retardation \( = 10 \).

🎯 Exam Tip: If the question asks for "acceleration," include the minus sign. If it asks for "retardation," give the positive value.

Question 12N. A car starts from rest and reaches 18 km/h in 2 s. Find its acceleration.
Answer: Here, final velocity \( = 18 \text{ km/h} \) or 5 m/s
Initial velocity \( = 0 \text{ km/h} \)
Time taken \( = 2 \text{ s} \)
Acceleration \( = (\text{Final Velocity} - \text{Initial Velocity}) / \text{time} \)
\( \implies \) \( (5 - 0) / 2 \text{ m/s}^2 \)
\( = 2.5 \text{ m/s}^2 \)
In simple words: First we turn 18 km/h into 5 m/s. Since the car reached that speed in 2 seconds, it gained 2.5 meters per second every second.

Question 13N. Find the increase in velocity of a body if it accelerates at 2 m/s\(^2\) for 5 s.
Answer: Acceleration \( = \text{Increase in velocity} / \text{time taken} \)
Therefore, increase in velocity \( = \text{Acceleration} \times \text{time taken} \)
\( \implies \) \( (2 \times 5) \text{ m/s} \)
= 10 m/s
In simple words: If you gain 2 units of speed every second, after 5 seconds you have gained 10 units total.

📝 Teacher's Note: This is a simple rearrangement of the \( a = v/t \) formula.

Question 14N. A car traveling at 20 m/s is brought to a stop in 5 s. Calculate its acceleration.
Answer: Initial velocity of the car, u \( = 20 \text{ m/s} \)
Retardation (given by deceleration logic) \( = 2 \text{ m/s}^2 \)
Given time, t \( = 5 \text{ s} \)
Let ‘v’ be the final velocity.
We know that, Acceleration \( = \text{Rate of change of velocity} / \text{time} \)
\( = (\text{Final velocity} - \text{Initial velocity}) / \text{time} \)
Or, -2 \( = (v - 20) / 5 \)
Or, -10 \( = v - 20 \)
Or, v \( = -20 + 10 \text{ m/s} \)
Or, v \( = -10 \text{ m/s} \)
Negative sign indicates that the velocity is decreasing.
In simple words: The car was going at 20 and lost 10 units of speed, so it is now going at 10 meters per second.

Question 15N. A bicycle moves at 5 m/s and accelerates at 2 m/s\(^2\) for 5 s. Find its final velocity.
Answer: Initial velocity of the bicycle, u \( = 5 \text{ m/s} \)
Acceleration \( = 2 \text{ m/s}^2 \)
Given time, t \( = 5 \text{ s} \)
Let ‘v’ be the final velocity.
We know that, acceleration \( = \text{Rate of change of velocity} / \text{time} \)
\( = (\text{Final velocity} - \text{Initial velocity}) / \text{time} \)
Or 2 \( = (v - 5)/5 \)
Or, 10 \( = (v - 5) \)
Or, v \( = 5 + 10 \)
Or, v \( = 15 \text{ m/s} \)
In simple words: You start with 5 speed and gain 10 more speed from the acceleration, so you end up at 15.

Question 16N. A car at 18 km/h is brought to rest in 5 s. Calculate (i) its speed in m/s, (ii) its retardation, and (iii) its speed after 2 s.
Answer: Initial velocity of the bicycle, u \( = 18 \text{ km/hr} \)
Time taken, t \( = 5 \text{ s} \)
Final velocity, v \( = 0 \text{ m/s} \) (As the car comes to rest)
(i) Speed in m/s \( = \frac{18 \times 1000}{3600} = 5 \text{ m/s} \)

(ii) Retardation \( = (\text{Final velocity} - \text{Initial velocity}) / \text{time taken} \)
Or, Retardation \( = \frac{(0 - 5)}{5} \text{ ms}^{-2} = 1 \text{ ms}^{-2} \)

(iii) Let ‘V’ be the speed of the car after 2 s of applying the brakes.
Then, Acceleration \( = (V - 5) / 2 \)
Or, -1 \( = (V - 5) / 2 \)
Or, V \( = -2 + 5 \)
Or, V \( = 3 \text{ m/s} \)
In simple words: The car was doing 5 m/s. It loses 1 unit of speed every second. So after 2 seconds, it has lost 2 units, leaving it at 3 m/s.

📝 Teacher's Note: This multi-part problem is a classic exam question. It tests unit conversion, the definition of retardation, and the use of equations of motion.

🎯 Exam Tip: "Rest" or "stops" always means final velocity \( v = 0 \). Identifying this is the first step to solving the problem.

Exercise 2(B)

Question 1S. What is the relation between distance and time for uniform velocity?
Answer: For the motion with uniform velocity, distance is directly proportional to time.
In simple words: If you go at a steady speed, doubling your travel time will exactly double the distance you covered.

📝 Teacher's Note: This results in a straight line graph passing through the origin.

Question 2S. What information can be obtained from a displacement-time graph?
Answer: From displacement-time graph, the nature of motion (or state of rest) can be understood. The slope of this graph gives the value of velocity of the body at any instant of time, using which the velocity-time graph can also be drawn.
In simple words: The graph shows you if an object is still, moving steady, or speeding up. The steepness of the line tells you the speed.

📝 Teacher's Note: Slope = Velocity. This is a fundamental rule for graph analysis.

Question 3S. (a) What does the slope of a displacement-time graph represent? (b) Why can the displacement-time graph never be parallel to the displacement axis?
Answer: (a) Slope of a displacement-time graph represents velocity.
(b) The displacement-time graph can never be parallel to the displacement axis because such a line would mean that the distance covered by the body in a certain direction increases without any increase in time, which is not possible.
In simple words: A vertical line would mean you moved from one place to another in zero seconds—which is teleportation, not physics!

📝 Teacher's Note: Infinite slope means infinite velocity, which is physically impossible.

Question 4S. Describe the motion shown in a displacement-time graph if: (a) it is a horizontal line, (b) it is a straight line through origin, (c) it is a straight line towards origin, (d) it is a curve.
Answer:
(a) There is no motion, the body is at rest.
(b) It depicts that the body is moving away from the starting point with uniform velocity.
(c) It depicts that the body is moving towards the starting point with uniform velocity.
(d) It depicts that the body is moving with variable velocity.
In simple words: Flat line = sitting still; Straight slope = steady walking; Curve = running/slowing down.

Question 5S. Draw a displacement-time graph for a body moving with uniform velocity.
Answer: In simple words: A straight diagonal line shows that for every second that passes, the object moves the same number of meters.

📝 Teacher's Note: The slope of line OP gives the constant velocity.

Question 6S. What information can be obtained from a velocity-time graph regarding: (i) acceleration, (ii) distance, (iii) displacement?
Answer:
(i) The slope of the velocity-time graph gives the value of acceleration.
(ii) The total distance travelled by a body in a given time is given by the area enclosed between the velocity-time graph and X-axis (without any sign).
(iii) The displacement of a body in a given time is given by the area enclosed between the velocity-time graph and X-axis (with proper signs).
In simple words: Steepness of the line shows acceleration. The total "room" under the line shows how far you went.

📝 Teacher's Note: Explain that for distance, we add all areas as positive. For displacement, areas below the time-axis are subtracted (moving backwards).

🎯 Exam Tip: Area under graph = Displacement is the most important rule for V-T graphs.

Question 7S. In a displacement-time graph for two vehicles A and B, line A is steeper than line B. Which vehicle is faster?
Answer: Vehicle A is moving with a faster speed because the slope of line A is more than the slope of line B.
In simple words: The "mountain" of graph A is harder to climb because it's steeper—that means vehicle A is covering more distance in the same amount of time.

📝 Teacher's Note: Higher slope \( \implies \) Higher velocity.

Question 8S. What do these velocity-time graphs represent: (a) a horizontal straight line above time axis, (b) a straight line through origin, (c) a straight line towards time axis?
Answer:
(a) Represents uniform velocity (constant speed).
(b) Represents uniform acceleration (speeding up at a steady rate).
(c) Represents uniform retardation (slowing down at a steady rate).
In simple words: Flat speed = cruise control; uphill speed = zooming; downhill speed = braking.

Question 9S. Show how to derive the formula for distance using a velocity-time graph for uniform acceleration.
Answer:
In this graph, initial velocity = u, Velocity at time t = v. Let acceleration be ‘a’.
Then, distance travelled by the body in t s = area between the v-t graph and X-axis.
Or distance travelled by the body in t s = area of the trapezium OABD
\( = (1/2) \times (\text{sum of parallel sides}) \times (\text{perpendicular distance between them}) \)
\( \implies S = (1/2) \times (u + v) \times (t) \)
\( \implies S = \frac{(u + v)t}{2} \)
In simple words: To find the distance, you calculate the area of the shape (a trapezium) made by the starting speed, ending speed, and the time.

📝 Teacher's Note: This is a geometric way to find displacement before using the equations of motion.

Question 10S. What does the slope of a velocity-time graph represent?
Answer: The slope of the velocity-time graph represents acceleration.
In simple words: If the line on a speed graph is very steep, it means the car is speeding up very quickly.

Question 11S. If car B's line is steeper than car A's in a V-T graph, which car has more acceleration?
Answer: Car B has greater acceleration because the slope of line B is more than the slope of line A.
In simple words: The steeper line belongs to the car that changes its speed the fastest.

Question 12S. Describe the acceleration for: (A) horizontal line, (B) straight slope, (C) line with decreasing slope, (D) line with increasing slope in a V-T graph.
Answer:
For body A & B: The graph is a straight line. So, the slope gives constant velocity. Hence, the acceleration is zero.
For body C: The slope of the graph is decreasing with time. Hence, the acceleration is negative (retardation).
For body D: The slope of the graph is increasing with time. Hence, the acceleration is positive.
In simple words: If speed doesn't change, accel is zero. If you're slowing down, it's negative. If you're zooming faster, it's positive.

Question 13S. Sketch a velocity-time graph for a body moving with uniform velocity and uniform acceleration.
Answer:
In simple words: A straight diagonal line starting from the corner shows the speed going up by the same amount every second.

Question 14S. How is retardation calculated from a V-T graph?
Answer: Retardation is calculated by finding the negative slope of the velocity-time graph.
In simple words: You measure how fast the speed line is dropping down towards zero.

Question 15S. What does the area under an acceleration-time graph represent?
Answer: The area enclosed between the straight line and time axis for each interval of time gives the value of change in speed in that interval of time.
In simple words: If you know how hard you're speeding up and for how long, you can find out exactly how much your speed has increased.

📝 Teacher's Note: Change in velocity \( = \text{Acceleration} \times \text{Time} \). This is exactly what the area of the rectangle represents.

Question 16S. Describe the displacement-time squared (\( S - t^2 \)) graph for uniform acceleration.
Answer: For motion under uniform acceleration, such as the motion of a freely falling body, distance is directly proportional to the square of the time (\( S \propto t^2 \)). This results in a straight line graph.
In simple words: Instead of plotting against 1, 2, 3 seconds, we plot against 1, 4, 9. This makes the curved falling line turn into a perfectly straight one.

📝 Teacher's Note: This is a linearization technique used to find the value of acceleration 'a' easily from the slope.

Question 17S. How can the value of acceleration due to gravity (g) be obtained from an \( S - t^2 \) graph?
Answer: The value of acceleration due to gravity (g) can be obtained by doubling the slope of the \( S - t^2 \) graph for a freely falling body.
In simple words: If you find out how steep the line is on this special graph and multiply that number by 2, you get the strength of Earth's gravity.

📝 Teacher's Note: From \( S = \frac{1}{2} gt^2 \), the slope is \( g/2 \). Therefore, \( g = 2 \times \text{slope} \).

Question 18S. A body starts from rest and falls freely. Its displacement is directly proportional to:
(a) time, (b) square of time, (c) square root of time, (d) inverse of time.
Answer: (b) square of time
In simple words: If you fall for twice as long, you don't fall twice as far—you fall four times as far!

Question 1M. If a body's displacement is directly proportional to the square of time, then its:
(a) Velocity is uniform
(b) Acceleration is uniform
(c) Speed is decreasing
(d) None of the options
Answer: (b) Acceleration is uniform.
In simple words: This means the object is speeding up at a perfectly steady rate every second, like an apple falling to the ground.

📝 Teacher's Note: Explain that uniform acceleration means the velocity changes by the same amount in every equal interval of time.

🎯 Exam Tip: This is a standard conceptual MCQ. Direct proportionality with \( t^2 \) always indicates constant (uniform) acceleration.

Question 2M. What is the shape of a speed-time graph for a body moving with uniform velocity?
(a) A curve
(b) A straight line inclined to the time axis
(c) A straight line parallel to the time axis
(d) A vertical line
Answer: (c) A straight line parallel to the time axis.
In simple words: Since the speed isn't changing as time goes by, the line on the graph stays flat, like a long horizontal road.

📝 Teacher's Note: Use a graph to show that at \( t = 1s, 2s, 3s \), the speed value on the Y-axis remains exactly the same.

🎯 Exam Tip: A flat line on a *displacement-time* graph means the body is at rest, but on a *speed-time* graph, it means the body is moving at a constant speed.

Question 3M. What is the shape of a velocity-time graph for a body moving with uniform acceleration?
(a) A horizontal line
(b) A curve
(c) A vertical line
(d) A straight line inclined to the time axis
Answer: (d) A straight line inclined to the time axis.
In simple words: Because the speed is going up steadily, the line on the graph climbs up like a perfectly straight mountain slope.

📝 Teacher's Note: The "inclination" or steepness of this line directly represents the magnitude of the acceleration.

🎯 Exam Tip: If the line passes through the origin (0,0), it implies the body started from rest.

Question 1N. From a given velocity-time graph, find the velocity of the body at t = 1s, 2s, and 3s.
Answer:
Velocity of body at t = 1s is 2 m/s
Velocity of body at t = 2s is 4 m/s
Velocity of body at t = 3s is 6 m/s
In simple words: By looking at the graph, we can see exactly how fast the object was moving at each second. In this case, it gains 2 m/s of speed every second.

📝 Teacher's Note: This is a simple exercise in coordinate reading. Encourage students to draw dotted "normals" from the time axis to the line and then to the velocity axis.

🎯 Exam Tip: Always include the units (m/s) in your answer even if the question is simple reading from a graph.

Question 2N. For the given displacement-time graph:
(i) Calculate the average velocity from the part AB of the graph.
(ii) Find the displacement of the car at 2.5s and 4.5s.
Answer:
(i) From the part AB of the graph,
Average velocity = (Displacement at B – Displacement at A) / Time taken
\( \implies V = \frac{30 - 20}{6 - 4} \text{ m/s} \)
\( \implies V = \frac{10}{2} \text{ m/s} \)
\( V = 5 \text{ m/s} \)
(ii) (a) From the graph, the displacement of car at 2.5 s is 12.5 m.
(b) From the graph, the displacement of car at 4.5 s is 22.5 m.
In simple words: To find the average speed for a specific section, we check how much distance was covered and divide it by how much time it took. To find a spot measurement, we just look at the line on the grid.

📝 Teacher's Note: Average velocity over a section is the "Slope" of that section. Here, between 4s and 6s, the car covered 10m, so its speed was 5 m/s.

🎯 Exam Tip: When calculating slope, always use the formula \( \frac{y_2 - y_1}{x_2 - x_1} \) to avoid confusion.

Question 3N. From the provided displacement-time graph, calculate:
(i) Total distance travelled in the interval 1s to 5s.
(ii) Average velocity in the same interval.
Answer:
(i) Total distance travelled in interval 1s to 5s = \( 18\text{m} - 6\text{m} = 12 \text{ m} \).
(ii) Average velocity = Total displacement / Total time taken
\( \implies V = \frac{12 \text{ m}}{4 \text{ s}} \)
\( V = 3 \text{ m/s} \).
In simple words: In that 4-second gap, the object moved from the 6-meter mark to the 18-meter mark. That is a total movement of 12 meters, which averages out to 3 meters every second.

📝 Teacher's Note: Point out that the time taken is the difference between final and initial time (\( 5s - 1s = 4s \)).

🎯 Exam Tip: Note that "Total distance" and "Total displacement" are identical here because the body moves in a single direction without turning back.

Question 4N. From a displacement-time graph, calculate:
(a) Velocity during the intervals: (i) 0 to 5s, (ii) 5s to 7s, (iii) 7s to 9s.
(b) Average velocity between 5s and 9s.
Answer:
(a) (i) Velocity from 0 to 5 s = Displacement / time
\( = (3/5) \text{ m/s} \)
\( = 0.6 \text{ m/s} \)
(ii) Velocity from 5 s to 7 s = Displacement / time
\( = (0/2) \text{ m/s} \)
\( = 0 \text{ m/s} \).
(iii) Velocity from 7 s to 9 s = Displacement / time
\( = \frac{7 - 3}{9 - 7} \text{ m/s} \)
\( = \frac{4}{2} \text{ m/s} \)
\( = 2 \text{ m/s} \)
(b) From, 5 s to 9 s, displacement = \( 7\text{m} - 3\text{m} = 4\text{m} \).
Time elapsed between 5 s to 9 s = 4 s
Average velocity = Displacement / time
\( = (4/4) \text{ m/s} \)
\( = 1 \text{ m/s} \)
In simple words: The object moved at three different speeds. It stood still between 5 and 7 seconds, which is why its speed there was zero. For the whole second half of the trip, it averaged out to 1 meter per second.

📝 Teacher's Note: A horizontal line on a displacement-time graph always means the velocity is zero (the body is at rest).

🎯 Exam Tip: "Average velocity" is always \( \text{Total Displacement} / \text{Total Time} \). Do not just average the three individual velocities (\( 0.6, 0, 2 \)); that will give the wrong answer!

Question 5N. A cyclist's motion is recorded on a graph. Find:
(i) Average velocity in the first 4 seconds.
(ii) The final position and displacement at the end of 10s.
(iii) When does the cyclist reach his starting point?
Answer:
(i) Displacement in first 4s = 10 m
Therefore, the average velocity = Displacement / time
\( = (10/4) \text{ m/s} \)
\( = 2.5 \text{ m/s} \)
(ii) Initial position = 0 m
Final position at the end of 10 s = -10m
Displacement = Final position – Initial position
\( = (-10) \text{ m} - 0 \)
\( = -10 \text{ m} \)
(iii) At 7 s and 13 s, the cyclist reaches his starting point.
In simple words: The cyclist goes out 10 meters, then turns around and goes past his house to 10 meters in the other direction. He passes his front door twice: at 7 seconds and again at 13 seconds.

📝 Teacher's Note: Explain that a negative displacement means the object is on the opposite side of the starting reference point.

🎯 Exam Tip: To find when a body returns to the start, look for where the graph line crosses the X-axis (where Y = 0).

Question 6N. Two cars A and B are moving. From their graph, determine:
(i) How far ahead was B at the start?
(ii) Their individual velocities.
(iii) When does A catch B?
(iv) Where does A catch B?
Answer:
(i) Initially, the car B was 40 km ahead of car A.
(ii) Straight line depicts that cars A and B are moving with uniform velocities.
For car A:
Displacement at t = 1 h is 40 m
Velocity = Displacement / time
\( = (40/1) \text{ km/h} \)
\( = 40 \text{ km/h} \)
For car B:
Displacement at t = 4 h is \( (120 - 40) \text{ km} \), i.e. 80 km
Velocity = Displacement / time
\( = (80/4) \text{ km/h} \)
\( = 20 \text{ km/h} \)
(iii) Car A catches car B in 2 hours.
(iv) After starting, car A will catch car B at 80 km from the origin.
In simple words: Car B had a head start, but Car A was twice as fast. They met up at the 80km mark exactly 2 hours after they started moving.

📝 Teacher's Note: The point of intersection on a displacement-time graph shows exactly when and where two moving bodies meet.

🎯 Exam Tip: To find the velocity of B, remember to subtract the 40km head start from the final distance. The distance B covered was 80km, not 120km.

Question 8N. A body moves such that its velocity at t = 1s, 2s, 3s is 1 m/s, 2 m/s, and 3 m/s respectively. Find the displacements at these times.
Answer:
Velocity of the body at t = 1 s is 1 m/s.
Displacement of the body at t = 1 s is velocity \( \times \) time = \( (1) \times (1) \text{ m} \) or 1 m.
Velocity of the body at t = 2s is 2 m/s.
Displacement of the body at t = 2 s is velocity \( \times \) time = \( (2) \times (2) \text{ m} \) or 4 m.
Velocity of the body at t = 3 s is 3 m/s.
Displacement of the body at t = 3 s is velocity \( \times \) time = \( (3) \times (3) \text{ m} \) or 9 m.
In simple words: Because the body is speeding up, it covers more ground each second than it did the previous one. This is why the displacement grows as 1, 4, 9.

📝 Teacher's Note: This is a numerical proof of the \( s \propto t^2 \) rule for uniform acceleration.

Question 9N. For the provided velocity-time graph, find:
(i) How to determine distance travelled from the graph.
(ii) Ratio of distance in part BC to distance in part AB.
(iii) The types of motion in BC, AB, and CD.
(iv) Which part has lower acceleration magnitude?
Answer:
(i) Distance travelled in any part of the graph can be determined by finding the area enclosed by the graph in that part with the time axis.
(ii) Distance travelled in part BC = Area of the rectangle tBC2t = base \( \times \) height.
\( = (2t - t) \times v_o \)
\( = v_o t \)
Distance travelled in part AB = Area of the triangle ABt
\( = (1/2) \times \text{base} \times \text{height} \)
\( = (1/2) \times t \times v_o \)
\( = (1/2) v_o t \)
Therefore, distance travelled in part BC : distance travelled in part AB :: 2 : 1.
(iii) (a) BC shows motion with uniform velocity. (b) AB shows motion with uniform acceleration. (c) CD shows motion with uniform retardation.
(iv) (a) The magnitude of acceleration is lower as the slope of line AB is less than that of line CD.
(b) Slope of line AB = \( v_o/t \)
Slope of line CD = \( v_o/0.5t \)
Slope of line AB / Slope of line CD = \( (v_o /t) / (v_o /0.5t) \)
Slope of line AB : Slope of line CD :: 1 : 2.
In simple words: We find distance by calculating the area of shapes under the line. Part BC covers more ground because it stays at top speed the whole time. Part CD is steeper than AB, so the car braked harder than it accelerated.

📝 Teacher's Note: Help students recognize the geometric shapes (Triangle for acceleration, Rectangle for constant velocity) under the V-T graph.

🎯 Exam Tip: "Magnitude of acceleration" refers only to the numerical value, ignoring the negative sign of retardation.

Question 10N. From the given graph, calculate:
(i) Acceleration in parts AB, BC, and CD.
(ii) Displacement for each part.
(iii) Total displacement.
Answer:
(i) Acceleration in the part AB = Slope of AB
\( = 30 / 4 \text{ ms}^{-2} \)
\( = 7.5 \text{ ms}^{-2} \)
Acceleration in the part BC = 0 \( \text{ms}^{-2} \) (horizontal line)
Acceleration in the part CD = slope of CD
\( = -(30/2) \text{ ms}^{-2} \)
\( = -15 \text{ ms}^{-2} \) (Retardation)
(ii) Displacement of part AB = Area of \( \Delta AB4 = (1/2) (4) (30) \)
\( = 60 \text{ m} \)
Displacement of part BC = Area of rectangle 4BC8
\( = (30) \times (4) = 120 \text{ m} \)
Displacement of part CD = Area of \( \Delta C8D = (1/2) (2) (30) \)
\( = 30 \text{ m} \)
(iii) Total displacement = Displacement AB + Displacement BC + Displacement CD
\( = 60 + 120 + 30 = 210 \text{ m} \)
In simple words: The car speeds up for 4 seconds, drives steady for 4 seconds, and then slams on the brakes for 2 seconds. In total, it traveled 210 meters.

📝 Teacher's Note: Point out that the time for part CD is \( 10s - 8s = 2s \). A common mistake is using the full 10s on the x-axis.

🎯 Exam Tip: Remember: Acceleration = Slope, Displacement = Area. These are the two most important tools for graph problems.

Question 11N. A ball is thrown up and returns to the starting point in 12s. If its velocity is 10 m/s throughout (constant speed), calculate:
(i) Total distance travelled.
(ii) Total displacement.
Answer:
Distance travelled in first 6 s = velocity \( \times \) time
\( = 10 \text{ m/s} \times 6 = 60 \text{ m} \)
Distance travelled in next 6 s = velocity \( \times \) time
\( = 10 \text{ m/s} \times 6 = 60 \text{ m} \)
Total distance travelled in 12 s = \( (60 + 60) \text{ m} = 120 \text{ m} \)
Total displacement = 0, as the ball returns to its starting point.
In simple words: The ball went up 60 meters and came back 60 meters. Its total trip was 120 meters long, but since it's back where it started, its final "jump" (displacement) is zero.

📝 Teacher's Note: This scenario assumes an ideal constant speed. In reality, gravity would change the speed, but the problem focuses on the distance vs displacement definitions.

🎯 Exam Tip: Always look for phrases like "returns to its starting point" to immediately identify that total displacement is zero.

Question 12N. Analyze a motion graph from 0 to 6 seconds.
Answer:
(i) From 0 to 4 seconds, the motion is uniformly accelerated and from 4 to 6 seconds, the motion is uniformly retarded.
(ii) Displacement of the particle at 6 s = \( (1/2) (6) (2) = 6 \text{ m} \)
(iii) The particle does not change its direction of motion.
(iv) Distance travelled by the particle from 0 to 4s (\( D_1 \)) = \( (1/2) (4) (2) = 4 \text{ m} \)
Distance travelled by the particle from 4 to 6s (\( D_2 \)) = \( (1/2) (2) (2) = 2 \text{ m} \)
\( D_1 : D_2 :: 4 : 2 \)
\( D_1 : D_2 :: 2 : 1 \)
(v) Acceleration from 0 to 4 s = \( (2/4) \text{ ms}^{-2} \) or \( 0.5 \text{ ms}^{-2} \)
Retardation from 4 s to 6 s = \( (2/2) \text{ ms}^{-2} \) or \( 1 \text{ ms}^{-2} \).
In simple words: The object sped up slowly for 4 seconds and then braked quickly to a stop in 2 seconds. It covered twice as much ground while speeding up as it did while stopping.

📝 Teacher's Note: Since the entire graph is above the time axis, the velocity is always positive, meaning the direction never flipped.

Exercise 2(C)

Question 1S. List the three equations of uniformly accelerated motion.
Answer:
1. \( v = u + at \)
2. \( s = ut + \frac{1}{2} at^2 \)
3. \( v^2 = u^2 + 2as \)
In simple words: These three magic formulas can solve almost any problem involving objects speeding up or slowing down.

📝 Teacher's Note: Remind students what each letter stands for: \( u = \) start speed, \( v = \) end speed, \( a = \) acceleration, \( t = \) time, and \( s = \) distance.

Question 2S. Derive the first and second equations of motion.
Answer:
First equation of motion:
Consider a particle moving along a straight line with uniform acceleration ‘a’. At \( t = 0 \), let the particle be at A and u be its initial velocity, and at \( t = t \), let v be its final velocity.
Acceleration \( = \text{Change in velocity} / \text{Time} \)
\( \implies a = (v – u)/t \)
\( \implies at = v – u \)
\( \implies v = u + at \)

Second equation of motion:
Average velocity \( = \text{Total distance traveled} / \text{Total time taken} \)
\( \implies \text{Average velocity} = s/t \) ... (1)
Average velocity can also be written as \( (u+v)/2 \) ... (2)
From equations (1) and (2): \( s/t = (u+v)/2 \) ... (3)
Substitute \( v = u + at \) into equation (3):
\( s/t = (u + u + at)/2 \)
\( s = (2u + at) \times \frac{t}{2} = \frac{2ut}{2} + \frac{at^2}{2} \)
\( \implies s = ut + \frac{1}{2} at^2 \)
In simple words: The first formula just comes from the definition of acceleration. The second one combines the idea of "average speed" with the first formula.

Question 3S. Write the relation for distance covered when initial velocity is zero.
Answer: \( s = \frac{1}{2} at^2 \)
In simple words: If you start from a complete stop, the distance you travel depends entirely on how hard you accelerate and for how long.

📝 Teacher's Note: This is just the second equation where \( u = 0 \). Point out how the first term \( ut \) vanishes.

🎯 Exam Tip: This specific form of the equation is frequently used for "free fall" problems where an object is dropped from rest.

Question 1M. Write the formula for final velocity.
Answer: \( v = u + at \)
In simple words: Your final speed is your start speed plus the speed you gained while accelerating.

📝 Teacher's Note: This is a fundamental definition of acceleration rearranged into a linear equation.

🎯 Exam Tip: Always check if the speed is given in km/h or m/s; convert to m/s before using this formula in a problem.

Question 2M. What is the distance covered?
Answer: 5 km
In simple words: The result of the calculation based on the given problem data is 5 kilometers.

📝 Teacher's Note: Without the original question text, ensure that students know how to convert km to m (\( \times 1000 \)) if needed for other parts of a physics problem.

🎯 Exam Tip: Always double-check if the answer options ask for the value in meters or kilometers.

Question 1N. Initial velocity \( u = 0 \), Acceleration \( a = 2 \text{ m/s}^2 \), Time \( t = 2 \text{ s} \). Let ‘S’ be the distance covered.
Using the second equation of motion,
\( S = ut + \frac{1}{2} at^2 \)
\( \implies S = 0 + \frac{1}{2} (2) (2)^2 \)
Answer: \( S = 4 \text{ m} \)
In simple words: If you start from zero and gain 2 m/s of speed every second, after 2 seconds you will have moved exactly 4 meters.

📝 Teacher's Note: Show students how the \( \frac{1}{2} \) and the acceleration of \( 2 \) cancel each other out in this specific problem to make the math easier.

🎯 Exam Tip: Don't forget the power of 2 on the time. Squaring time is the most frequent calculation error in this topic.

Question 2N. Initial velocity \( u = 10 \text{ m/s} \), Acceleration \( a = 5 \text{ m/s}^2 \), Time \( t = 5\text{s} \). Let ‘S’ be the distance covered.
Using the second equation of motion,
\( S = ut + \frac{1}{2} at^2 \)
\( \implies S = (10)(5) + \frac{1}{2} (5) (5)^2 \)
\( \implies S = 50 + 62.5 \)
Answer: \( S = 112.5 \text{ m} \)
In simple words: The object was already moving at 10 m/s and then it sped up. In 5 seconds, it covered over 100 meters.

📝 Teacher's Note: Here, because the initial speed (\( u \)) is not zero, the object covers 50 meters just by its initial momentum, plus another 62.5 meters because it was accelerating.

🎯 Exam Tip: Always show the addition of the two terms (\( ut \) and \( \frac{1}{2}at^2 \)) separately to ensure you get partial marks if there is a calculation error.

Question 3N. Calculate acceleration when speed increases from 30 km/h to 33.6 km/h in 2 seconds, and then to 37.2 km/h in the next 2 seconds.
Acceleration \( = \) Change in velocity/time taken
In the first two seconds,
Acceleration \( = \frac{33.6 - 30}{2} \text{ km/h}^2 \)
\( \implies = 1.8 \text{ km/h}^2 \)
\( \implies = 0.5 \text{ m/s}^2 \) ... (i)
In the next two seconds,
Acceleration \( = \frac{37.2 - 33.6}{2} \text{ km/h}^2 \)
\( \implies = 1.8 \text{ km/h}^2 \)
\( \implies = 0.5 \text{ m/s}^2 \) ... (ii)
Answer: From (i) and (ii), we can say that the acceleration is uniform.
In simple words: The car sped up by the exact same amount in the first 2 seconds as it did in the next 2 seconds. This means the speeding-up process was smooth and steady (uniform).

📝 Teacher's Note: To convert \( \text{km/h}^2 \) to \( \text{m/s}^2 \), you multiply by \( 1000 \) and divide by \( (3600 \times 3600) \). However, for basic comparison, checking if the difference in speeds is equal is enough.

🎯 Exam Tip: If the change in velocity is equal for equal intervals of time, the acceleration is uniform. State this conclusion clearly to get the final mark.

Question 4N. Initial velocity \( u = 0 \text{ m/s} \), Acceleration \( a = 2 \text{ m/s}^2 \), Time \( t = 5 \text{ s} \). Calculate:
(i) Final velocity
(ii) Distance travelled
Answer:
(i) Let ‘\( v \)’ be the final velocity.
\( v = u + at \)
\( \implies v = 0 + 2(5) \)
\( \implies v = 10 \text{ m/s} \)
(ii) Let ‘\( s \)’ be the distance travelled.
Using the third equation of motion,
\( v^2 - u^2 = 2as \)
\( \implies (10)^2 - (0)^2 = 2(2)(s) \)
\( \implies 100 = 4s \)
\( \implies s = 100/4 = 25 \text{ m} \)
In simple words: The object reached a speed of 10 meters per second and covered a total distance of 25 meters during that 5-second zoom.

📝 Teacher's Note: You could also use the second equation \( s = ut + \frac{1}{2}at^2 \) to find distance. Both will give the same answer of 25 meters. Let students try both to gain confidence.

🎯 Exam Tip: Always specify what you are calculating (Velocity or Distance) with the appropriate unit (m/s or m).

Question 5N. Initial velocity \( u = 20 \text{ m/s} \), Final velocity \( v = 0 \), Distance travelled \( s = 10 \text{ cm} = 0.1 \text{ m} \). Calculate retardation.
Using the third equation of motion,
\( v^2 - u^2 = 2as \)
\( \implies (0)^2 - (20)^2 = 2(a)(0.1) \)
\( \implies -400 = 0.2a \)
\( \implies a = -400/0.2 = -2000 \text{ m/s}^2 \)
Answer: retardation = 2000 \( \text{m/s}^2 \)
In simple words: This is an incredibly sudden stop! The object was going fast and stopped in a tiny distance (just 10 cm), so the braking power (retardation) had to be huge.

📝 Teacher's Note: Remind students to convert \( \text{cm} \) to \( \text{m} \) before solving SI problems. 10 cm \( = \) 0.1 m.

🎯 Exam Tip: "Retardation" is the negative of acceleration. If your acceleration comes out as \( -2000 \), your retardation is \( 2000 \). Don't write a negative sign for a value already called retardation.

Question 6N. Initial velocity \( u = 20 \text{ m/s} \), Final velocity \( v = 0 \), Time taken \( t = 5 \text{ s} \). Calculate retardation.
Using the first equation of motion,
\( v = u + at \)
\( \implies 0 = 20 + 5a \)
\( \implies 5a = -20 \)
\( \implies a = -4 \text{ m/s}^2 \)
Answer: retardation = 4 \( \text{m/s}^2 \)
In simple words: The car was going at 20 m/s and braked to a stop over 5 seconds. It lost 4 meters per second of speed every second.

📝 Teacher's Note: This is a standard braking problem. Contrast this with 5N to show how the distance or time given changes the intensity of the stop.

🎯 Exam Tip: "Brought to rest" or "comes to a stop" always means final velocity \( v = 0 \).

Question 7N. A train travels from station A to B at 60 km/h and returns at 80 km/h. Calculate (i) Average speed (ii) Average velocity.
Let ‘\( s \)’ be the distance between stations A and B.
(i) Average speed \( = \) Total distance / total time taken
Here, total distance \( = s + s = 2s \)
Total time taken \( = \) Time \( \text{A} \rightarrow \text{B} + \) Time \( \text{B} \rightarrow \text{A} \).
\( \text{Time} = \frac{s}{60} + \frac{s}{80} = \frac{4s + 3s}{240} = \frac{7s}{240} \)
\( \text{Average speed} = \frac{2s}{\frac{7s}{240}} = \frac{2 \times 240}{7} \)
Answer: Average speed = 68.57 km/h
(ii) Average velocity \( = \) Displacement / total time taken
Answer: Because the train starts and ends at the same station, the displacement is zero. Thus the average velocity is zero.
In simple words: Even if you go fast on your trip, if you end up back in your own bed, your average velocity for the whole trip is zero. For average speed, we calculate the whole distance moved.

📝 Teacher's Note: Average speed is NOT the average of the two speeds \( (60+80)/2 \). This is the most common trap for students. Always use Total Distance / Total Time.

🎯 Exam Tip: For any "return to starting point" question, always write "Displacement is zero, hence average velocity is zero" to get full marks immediately.

Question 8N. Initial velocity \( u = 90 \text{ km/h} \), Final velocity \( v = 0 \text{ m/s} \), Acceleration \( a = -0.5 \text{ m/s}^2 \).
(i) Find velocity after 10 s.
(ii) Find time taken to come to rest.
\( u = 90 \text{ km/h} = 90 \times \frac{5}{18} \text{ m/s} = 25 \text{ m/s} \)
(i) Let ‘\( V \)’ be the velocity after \( t = 10 \text{ s} \)
\( V = u + at \)
\( \implies V = 25 + (-0.5) (10) \)
Answer: V = 20 m/s
(ii) Let \( t' \) be the time taken to come to rest (\( v=0 \)).
\( 0 = 25 + (-0.5)t' \)
\( \implies 0.5t' = 25 \)
Answer: t' = 50 s
In simple words: The car starts at 25 m/s and loses half a meter of speed every second. After 10 seconds, it's down to 20 m/s. It will take 50 seconds total to stop.

📝 Teacher's Note: Help students with the unit conversion. \( 90 \text{ km/h} \) is a very common speed in physics problems—students should learn that it equals \( 25 \text{ m/s} \) by heart.

🎯 Exam Tip: Be very careful when dividing by 0.5; it is the same as multiplying by 2.

Question 10N. Distance travelled \( s = 100 \text{ m} \), Average velocity \( V = 20 \text{ m/s} \), Final velocity \( v = 25 \text{ m/s} \).
(i) Find initial velocity (\( u \)).
(ii) Find acceleration (\( a \)).
(i) Average velocity \( = \frac{\text{Initial velocity} + \text{Final velocity}}{2} \)
\( \implies 20 = \frac{u + 25}{2} \)
\( \implies 40 = u + 25 \)
Answer: u = 15 m/s
(ii) Using the third equation of motion,
\( v^2 - u^2 = 2as \)
\( \implies (25)^2 - (15)^2 = 2(a)(100) \)
\( \implies 625 - 225 = 200a \)
\( \implies 400 = 200a \)
Answer: a = 2 \( \text{m/s}^2 \)
In simple words: We used the average speed to figure out how fast the car was going at the start (15 m/s). Then we looked at the distance to see how hard it was accelerating (2 m/s²).

📝 Teacher's Note: This problem shows how different equations of motion can be used together. The average velocity formula is often the bridge to find \( u \) or \( v \) quickly.

🎯 Exam Tip: Note the difference between "Average velocity" and "Final velocity." One is for the whole trip, the other is just for the end.

Question 11N. Initial velocity \( u = 0 \text{ m/s} \), Distance travelled \( s = 270 \text{ m} \), Time taken \( t = 3 \text{ s} \).
(i) Find uniform acceleration.
(ii) Find velocity after 10 s.
(i) Using the second equation of motion,
\( S = ut + \frac{1}{2} at^2 \)
\( \implies 270 = 0 + \frac{1}{2} a (3)^2 \)
\( \implies 270 = 4.5a \)
Answer: a = 60 \( \text{m/s}^2 \)
(ii) Let \( v \) be the velocity of the body 10 s after the start.
Using the first equation of motion,
\( v = u + at \)
\( \implies v = 0 + (60)(10) \)
Answer: v = 600 m/s
In simple words: This object is accelerating extremely fast—like a rocket! In just 10 seconds, it would be going 600 meters every second.

📝 Teacher's Note: A common point of confusion is using the time from part (i) in part (ii). Remind students that part (ii) is a new scenario starting from \( t = 0 \).

🎯 Exam Tip: Always state the formula you are using before doing the math to avoid losing marks on logic.

Question 12N. A body travels 3 m in the first second (\( t_1=1 \)) and 8 m in the first two seconds (\( t_2=2 \)).
(i) Find initial velocity (\( u \)).
(ii) Find acceleration (\( a \)).
Using \( S = ut + \frac{1}{2} at^2 \):
\( S_1 = u(1) + \frac{1}{2} a(1)^2 \implies 3 = u + 0.5a \) ... (i)
\( S_2 = u(2) + \frac{1}{2} a(2)^2 \implies 8 = 2u + 2a \) ... (ii)
Divide (ii) by 2: \( 4 = u + a \) ... (iii)
Subtract (i) from (iii):
\( (4 - 3) = (u - u) + (a - 0.5a) \)
\( \implies 1 = 0.5a \)
Answer: a = 2 \( \text{m/s}^2 \)
Substitute \( a = 2 \) in (iii):
\( 4 = u + 2 \)
Answer: u = 2 m/s
In simple words: We used two different measurements to solve a simultaneous equation (like a puzzle) to find that the object started at 2 m/s and gained 2 m/s of speed every second.

📝 Teacher's Note: This is a higher-level physics problem. It requires simultaneous equations. Help students see that they need two different "snapshots" of motion to find two unknowns (\( u \) and \( a \)).

🎯 Exam Tip: When given distances for two different time intervals, always write down two equations using \( S = ut + \frac{1}{2}at^2 \).

Question 13N. Initial velocity \( u = 25 \text{ m/s} \), Final velocity \( v = 0 \).
(i) Distance covered in 5s before brakes are applied.
(ii) Acceleration after brakes are applied and it stops in 15m.
(iii) Distance to stop if it takes 10s after brakes.
(i) \( S = \text{Speed} \times \text{time} = 25 \times 5 \)
Answer: S = 125 m
(ii) Acceleration \( = \frac{v - u}{t} = \frac{0 - 25}{15} \)
Answer: retardation = 2.5 \( \text{ms}^{-2} \)
(iii) If time to stop is 10 s:
\( v^2 - u^2 = 2as \)
\( \implies 0 - (25)^2 = 2(-2.5)S' \)
\( \implies 625 = 5S' \)
Answer: S' = 125 m
In simple words: The car covered 125 meters while cruising. When the brakes were hit, it took another 125 meters to come to a full stop.

📝 Teacher's Note: Part (ii) in the OCR has a calculation mismatch (\( -25/15 \)). Usually, these problems are designed so the math works out cleanly. Based on the "2.5" answer, the time was likely meant to be 10 seconds.

Question 14N. Initial velocity \( u = 75 \text{ km/s} \), Final velocity \( v = 120 \text{ km/s} \), Time taken \( = 6 \text{ s} \). Calculate:
(i) Acceleration
(ii) Distance travelled in first 10 s.
(i) Acceleration \( = \frac{v - u}{t} = \frac{120 - 75}{6} = \frac{45}{6} \)
Answer: a = 7.5 \( \text{kms}^{-2} \)
(ii) Distance covered in 10 s: \( S = ut + \frac{1}{2} at^2 \)
\( \implies S = (75)(10) + \frac{1}{2}(7.5)(10)^2 \)
\( \implies S = 750 + 375 \)
Answer: S = 1125 km
In simple words: This is a high-speed aircraft or rocket. It gained speed very quickly and covered over a thousand kilometers in just 10 seconds!

📝 Teacher's Note: Note the units here are \( \text{km/s} \). This is much faster than a standard car. Ensure students don't accidentally convert to m/s unless asked, as the km values are easier to work with here.

🎯 Exam Tip: If acceleration is in \( \text{km/s}^2 \), the distance will come out in kilometers.

Question 15N. (i) Starts from rest (\( u=0 \)), acceleration \( a = 2 \text{ m/s}^2 \), time \( t=10\text{s} \). Find max velocity reached.
(ii) Slows down to stop in 50s. Find retardation.
(iii) Calculate total distance travelled.
(i) \( v = u + at = 0 + 2(10) \)
Answer: V = 20 m/s
(ii) Acceleration \( = \frac{0 - 20}{50} = -0.4 \text{ m/s}^2 \).
Answer: retardation = 0.4 \( \text{m/s}^2 \)
(iii) Distance first 10s: \( S_1 = \frac{1}{2}(2)(10)^2 = 100 \text{ m} \).
Distance next 200s (at steady 20 m/s): \( S_2 = 20 \times 200 = 4000 \text{ m} \).
Distance last 50s: \( S_3 = (20)(50) + \frac{1}{2}(-0.4)(50)^2 = 1000 - 500 = 500 \text{ m} \).
Total distance \( = 100 + 4000 + 500 \).
Answer: Total distance = 4600 m
(iv) Average velocity \( = \) Total distance / total time \( = \frac{4600}{260} \).
Answer: Vavg = 17.69 m/s
In simple words: The trip had three stages: speeding up, cruising, and slowing down. We find the distance for each part and add them all up to find the total journey.

📝 Teacher's Note: This is a complex multi-part problem. Encourage students to solve each segment of the motion independently and then combine the results.

🎯 Exam Tip: For "total time," remember to add all intervals: \( 10 + 200 + 50 = 260 \text{ seconds} \).