GSEB Class 9 Science Solutions Chapter 8 Motion

Page 100

Question 1. An object has moved through a distance. Can it have zero displacements? If yes, support your answer with an example.
Answer: Yes, if the object's final and initial position is the same, meaning the object has returned to its starting position, then its displacement can be zero.
In simple words: An object can move a distance but end up back where it started, making its overall change in position (displacement) zero.

Exam Tip: Remember that distance is a scalar quantity (total path covered) while displacement is a vector quantity (shortest path between initial and final points).

Question 2. A farmer moves along the boundary of a square field of side 10 m in 40 sec. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?
Answer: The total distance covered in one round is 40 m in 40 seconds.
Time taken \( = \) 2 minutes 20 seconds \( = 2 \times 60 \) seconds \( + 20 \) seconds \( = 120 + 20 = 140 \) seconds.
The object will finish 3 rounds in 120 (\( 40 \times 3 \)) seconds, and in the next 20 seconds, it will be at the diametrically opposite point (point C if it started from point A).
Net displacement will be AC using the Pythagoras theorem.
\( (AC)^2 = (AB)^2 + (BC)^2 \)
\( (AC)^2 = (10)^2 + (10)^2 \)
\( (AC)^2 = 100 + 100 \)
\( (AC)^2 = 200 \)
\( AC = \sqrt{200} \)
\( AC = 10\sqrt{2} \)
Displacement \( = 14.14 \) m
In simple words: The farmer walks around a square field. After 2 minutes and 20 seconds, he will be exactly across the field from where he started. We use Pythagoras to find this straight-line distance, which is 14.14 meters.

Exam Tip: For square or circular paths, remember to determine the farmer's final position relative to the starting point after the given time, not just the total distance traveled.

Question 3. Which of the following is true for displacement?
(a) It cannot be zero.
(b) Its magnitude is greater than the distance traveled by the object.
Answer: (a) The statement is not true. Displacement can be zero.
(b) The statement is not true. The magnitude of displacement is never greater than the distance traveled by the object.
In simple words: Displacement can be zero if you end up where you started. Also, the straight-line distance (displacement) is always less than or equal to the total path covered (distance).

Exam Tip: Understand the key distinction between distance (scalar, total path) and displacement (vector, net change in position). Displacement can be zero or negative, but distance is always positive.

Page 102

Question 2. Distinguish between speed and velocity.
Answer:

• Speed is the ratio of distance and time.
• Speed is always positive.
• Speed is a scalar quantity.

• Velocity is the ratio of displacement and time.
• Velocity may be negative or positive.
• Velocity is a vector quantity.

Exam Tip: When distinguishing between concepts, always use clear, concise points and mention whether each is a scalar or vector quantity.

Question 3. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Answer: In a straight-line motion in one direction, the magnitude of the average velocity of an object is equal to its average speed. This occurs because, in such a scenario, the total distance traveled is exactly equal to the magnitude of the displacement, and the direction does not change.
In simple words: Average speed and average velocity are the same when an object moves in a straight line without changing its direction.

Exam Tip: Remember that average speed is total distance divided by total time, while average velocity is total displacement divided by total time. They are equal only when the path is a straight line and no change in direction occurs.

Question 4. What does the odometer of an automobile measure?
Answer: The odometer of an automobile measures the distance traveled by a vehicle.
In simple words: An odometer in a car simply tracks how far the car has driven.

Exam Tip: An odometer measures distance, not displacement, average speed, or instantaneous speed. It's a cumulative measure of the path covered.

Question 5. What does the path of an object look like when it is in uniform motion?
Answer: In uniform motion, the speed of a body stays constant, and its direction of motion may change. Therefore, the path of an object can be a straight line, a curved line, a zig-zag, or even a circle.
In simple words: If an object moves at a steady speed, its path can be straight, curved, zig-zag, or round.

Exam Tip: Uniform motion means constant speed. Uniform *velocity* means constant speed in a constant direction (always a straight line). Distinguish between these terms carefully.

Question 6. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, which is \( 3 \times 10^8 \) ms\(^{-1} \).
Answer: Given,
Speed \( u = 3 \times 10^8 \) m/s\(^{-1} \)
Time \( t = 5 \) min \( = 5 \times 60 = 300 \) s
Distance \( = ? \)
Distance \( = \) speed \( \times \) time
Distance, \( s = 3 \times 10^8 \times 300 \)
\( s = 9 \times 10^{10} \) m
In simple words: We know the speed of the signal (speed of light) and how long it took to reach us. By multiplying the speed by the time, we find the distance of the spaceship from Earth, which is 90 billion meters.

Exam Tip: Always ensure unit consistency. Convert all time measurements to seconds before multiplying by speed in meters per second.

Page 103

Question 1. When will you say a body is in
(i) Uniform acceleration?
(ii) Non-uniform acceleration?
Answer:

• A body is said to be in uniform acceleration when its velocity changes by equal amounts in equal time intervals. For instance, the motion of a body falling freely under gravity.
• A body is said to be in non-uniform acceleration when its velocity changes by unequal amounts in equal time intervals. For example, the motion of a cycle on a normal road.

Exam Tip: Provide clear definitions with common, relatable examples for both uniform and non-uniform acceleration to score full marks.

Question 2. A bus decreases its speed from 80 km h\(^{-1} \) to 60 km h\(^{-1} \) in 5 s. Find the acceleration of the bus.
Answer: Given,
Initial speed, \( u = 80 \) kmh\(^{-1} \)
Convert to m/s: \( u = 80 \times \frac{5}{18} = 22.22 \) m/s
Final speed, \( v = 60 \) kmh\(^{-1} \)
Convert to m/s: \( v = 60 \times \frac{5}{18} = 16.66 \) m/s
Time taken, \( t = 5 \) s
Acceleration, \( a = \frac{v - u}{t} = \frac{16.66 - 22.22}{5} \)
\( a = \frac{-5.56}{5} = -1.11 \) m/s\(^2 \)
The negative sign indicates retardation (deceleration).
In simple words: The bus slows down, so its acceleration is negative. We convert the speeds to meters per second, then use the formula for acceleration (change in velocity divided by time) to find it's -1.11 m/s\(^2 \).

Exam Tip: Always convert all units to SI (meters, seconds) before performing calculations. A negative acceleration indicates deceleration or retardation.

Question 3. A train starting from a railway station and moving with uniform acceleration attains speed 40 kmh\(^{-1} \) in 10 minutes. Find its acceleration.
Answer: Initial speed, \( u = 0 \)
Final speed, \( v = 40 \) km h\(^{-1} = 40 \times \frac{5}{18} = 11.11 \) m/s
Time taken \( t = 10 \) min \( = 10 \times 60 \) s \( = 600 \) s
Acceleration, \( a = \frac{v - u}{t} = \frac{11.11 - 0}{600} \)
\( a = \frac{11.11}{600} = 0.0185 \) m/s\(^2 \)
\( a = 1.85 \times 10^{-2} \) m/s\(^2 \)
In simple words: The train starts from a stop and gets faster. We convert its final speed and time to standard units (m/s and seconds), then divide the change in speed by the time it took to find its acceleration.

Exam Tip: Always remember that "starting from rest" implies an initial velocity of zero. Pay close attention to unit conversions between km/h and m/s, and minutes and seconds.

Page 107

Question 1. What is the nature of the distance-time graph for uniform and non-uniform motion of an object?
Answer:

• In uniform motion, the distance-time graph is a straight line with a certain slope.
• In non-uniform motion, the distance-time graph is a curved line.

Exam Tip: Understand that the slope of a distance-time graph represents speed. A constant slope means constant speed (uniform motion), while a changing slope means changing speed (non-uniform motion).

Question 2. What can you say about the motion of an object if its distance-time graph is a straight line parallel to the time axis?
Answer: The object is stationary because its position is not changing with time. A horizontal line on a distance-time graph indicates zero speed.
In simple words: If a distance-time graph is a flat line, the object is not moving; it's standing still.

Exam Tip: A flat line on a distance-time graph means zero speed. A straight, sloped line means constant speed. A curved line means changing speed (acceleration).

Question 3. What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
Answer: The object is moving with a uniform speed because the speed is not changing with time. A constant speed, irrespective of direction, is uniform motion.
In simple words: If a speed-time graph is a flat line, the object is moving at a steady speed.

Exam Tip: On a speed-time graph, a horizontal line signifies constant speed (zero acceleration), whereas a sloped line means acceleration or deceleration.

Question 4. What is the quantity which is measured by the area occupied below the velocity-time graph?
Answer: The displacement of an object is measured by the area under the velocity-time graph. This area can represent the total change in position.
In simple words: The space under a velocity-time graph tells you how far an object has moved from its starting point (its displacement).

Exam Tip: Remember this fundamental principle: the area under a velocity-time graph gives displacement, while the slope gives acceleration.

Question 5. A stone is thrown in a vertically upward direction with a velocity of 5 ms\(^{-1} \). If the acceleration of the stone during its motion is 10 ms\(^{-2} \) in the downward direction, what will be height attained by the stone and how much time it will take to reach there?
Answer: Given, Initial velocity, \( u = 5 \) ms\(^{-1} \)
Final velocity, \( v = 0 \) ms\(^{-1} \) (at the highest point)
Acceleration, \( a = -10 \) ms\(^{-2} \) (negative because it's against the initial motion)
1. Height attained by the stone:
Using the equation: \( v^2 - u^2 = 2as \)
\( 0^2 - (5)^2 = 2 (-10)h \)
\( -25 = -20h \)
\( h = \frac{-25}{-20} \)
\( h = 1.25 \) m
2. Time taken:
Using the equation: \( v = u + at \)
\( 0 = 5 - 10 \times t \)
\( 10t = 5 \)
\( t = \frac{5}{10} \)
\( t = 0.5 \) s
In simple words: When a stone is thrown up, it slows down until it stops briefly at its highest point. We use physics formulas to find how high it went and how long it took to get there, considering gravity slows it down.

Exam Tip: For problems involving vertical motion under gravity, always consider the acceleration due to gravity (\( g \)) as negative when motion is upward and positive when motion is downward. The final velocity at the peak height is always zero.

Pages 109-110

Question 1. A bus starting from rest moves with a uniform acceleration of 0.1 ms\(^{-2} \) for 2 minutes. Find (a) the speed acquired (b) the distance traveled.
Answer: Given, initial speed, \( u = 0 \) m/s
Acceleration, \( a = 0.1 \) ms\(^{-2} \)
Time, \( t = 2 \) minutes \( = 2 \times 60 = 120 \) s
(a) Speed acquired:
Using equation: \( v = u + at \)
\( v = 0 + 0.1 \times 120 \)
\( v = 12 \) m/s
(b) Distance traveled:
Using equation: \( s = ut + \frac{1}{2} at^2 \)
\( s = (0 \times 120) + \frac{1}{2} \times 0.1 \times (120)^2 \)
\( s = 0 + \frac{1}{2} \times 0.1 \times 14400 \)
\( s = 0.05 \times 14400 \)
\( s = 720 \) m
In simple words: A bus starts moving and steadily gets faster for two minutes. We calculate its final speed and how far it has gone using the acceleration and time.

Exam Tip: Ensure to convert all units (especially time) to SI base units before starting calculations. Carefully apply the correct kinematic equations based on the given information (initial velocity, acceleration, time).

Question 2. A train is traveling at a speed of 90 kmh\(^{-1} \). Brakes are applied so as to produce a uniform acceleration of -0.5 ms\(^{-2} \). Find how far the train will go before it is brought to rest.
Answer: Given,
Initial speed, \( u = 90 \) km h\(^{-1} \)
Convert to m/s: \( u = 90 \times \frac{5}{18} = 25 \) m/s
Final speed, \( v = 0 \) m/s (as the train is brought to rest)
Acceleration, \( a = -0.5 \) m/s\(^2 \)
Using equation: \( v^2 - u^2 = 2as \)
\( 0^2 - (25)^2 = 2(-0.5)(s) \)
\( -625 = -1s \)
\( s = 625 \) m
In simple words: A train is moving fast, and then the brakes are applied. We figure out the distance it travels while slowing down until it completely stops.

Exam Tip: Remember that "brought to rest" means the final velocity is zero. Negative acceleration implies deceleration. Use the appropriate kinematic equation to find the stopping distance.

Question 3. A trolley, while going down an inclined plane, has an acceleration of 2 cm s\(^{-2} \). What will be its velocity 3s after the start?
Answer: Given,
Acceleration, \( a = 2 \) cm s\(^{-2} \)
Convert to m/s\(^2 \): \( a = 0.02 \) m/s\(^2 \)
Time, \( t = 3 \) s
Initial velocity, \( u = 0 \) m/s (since it's going down from the start)
Using equation: \( v = u + at \)
\( v = 0 + (0.02 \times 3) \)
\( v = 0.06 \) m/s
(Or, in original units: \( v = 0 + (2 \times 3) = 6 \) cm s\(^{-1} \))
In simple words: A trolley rolls down a slope and speeds up. We calculate how fast it will be moving after 3 seconds, knowing its starting speed and how quickly it accelerates.

Exam Tip: Be consistent with units; if the answer is requested in cm/s, you can use cm/s\(^2 \), but converting to m/s and then back is safer if mixed units are present.

In-Text Activities Solved

Textbook Page 98

Activity 8.1 Discussion and conclusion
Answer: The walls of our classroom are at rest compared to the Earth. However, the Earth itself is moving; therefore, when classroom walls are viewed from outside the Earth, the walls are in motion. States of rest and motion are a relative phenomenon, meaning they depend on the observer's frame of reference.
In simple words: If you are inside a classroom, the walls seem still. But if you could see the classroom from space, it would be moving with the Earth. So, being "at rest" or "in motion" depends on where you are looking from.

Exam Tip: The concept of rest and motion is relative. Always specify the reference frame when describing the state of an object's motion.

Textbook Page 98

Activity 8.2 Discussion and conclusion
Answer: Yes, when we are seated in a train at rest and the train on the adjoining track begins to move, we often feel as if our train is moving in the opposite direction. This sensation occurs due to relative motion, where we perceive the adjoining train as stationary, leading our own train to appear to move in the opposite direction.
In simple words: When your train is still and the train next to you moves, sometimes it feels like your train is moving backward. This is because your brain uses the other train as a reference point.

Exam Tip: Relative motion is a common phenomenon. Be ready to explain how an object's motion is perceived differently based on the observer's motion.

Textbook Page 99

Activity 8.3 Discussion and conclusion
Answer: Let us consider the dimensions of a basketball court \( = x \times y \).
Distance traveled from A to C \( = x + y \)
Displacement from A to C \( = \sqrt{x^{2}+y^{2}} \)
The diagram illustrates the path from A to C along the sides of the court and the direct displacement.

In simple words: For a rectangular court, if you walk along the sides, the distance is the sum of the length and width. But your displacement, the straight line from start to end, is found using the Pythagorean theorem, like a shortcut across the court.

Exam Tip: Clearly distinguish between distance (actual path length) and displacement (straight-line change in position). For rectangular paths, displacement is the diagonal, found using Pythagoras' theorem.

Textbook Page 99

Activity 8.4 Discussion and conclusion
Answer: We can find the position of New Delhi and Bhubaneshwar from a map of India. After measuring the length of the line connecting New Delhi and Bhubaneshwar on the map, we can convert it using a suitable map scale (the given map scale). The actual distance will be less than 1850 km.
In simple words: To find the real distance between two cities on a map, we measure the line connecting them and then use the map's scale to convert that measurement into actual kilometers.

Exam Tip: When using maps, remember that the scale is crucial for converting map distances to real-world distances. The straight line between two points on a map represents the displacement, not necessarily the actual road distance.

Textbook Page 100

Activity 8.5 Discussion and conclusion
Answer: Object A travels the same distance in equal time intervals, which means the motion of object A is uniform motion. Object B travels unequal distances in equal time intervals, which means the motion of object B is non-uniform motion.
In simple words: Object A moves at a steady pace, covering the same ground in the same amount of time. Object B's movement is uneven; it covers different distances in the same amount of time.

Exam Tip: Uniform motion implies constant speed over equal time intervals. Non-uniform motion implies varying speed, covering different distances in equal time intervals.

Textbook Page 101

Activity 8.6 Discussion and conclusion
Answer: Suppose we take 10 minutes to walk from our house to the bus stop or school.
Distance \( = \) speed \( \times \) time
Speed \( = 4 \) kmh\(^{-1} \) (given)
Distance \( = 4 \) \( \frac{\text{km}}{\text{h}} \times \frac{10}{60} \text{h} = \frac{40}{60} \) km \( = \frac{2}{3} \) km \( \approx 0.67 \) km
In simple words: If you walk at a speed of 4 kilometers per hour for 10 minutes, you will cover about 0.67 kilometers.

Exam Tip: Always ensure unit consistency; convert minutes to hours when working with speed in km/h to get distance in km.

Textbook Page 102

Activity 8.7 Discussion and conclusion
Answer: The speed of light is greater than the speed of sound. Therefore, thunder takes more time to reach us, while the lightning is seen almost instantly. This is why we see lightning before we hear thunder.
In simple words: Light travels much faster than sound. That's why we see lightning flash before we hear the thunder boom.

Exam Tip: This is a classic example illustrating the difference in speeds of light and sound. Understand that sound travels much slower than light.

Textbook Page 103

Activity 8.8 Discussion and conclusion
Answer:

• When the speed of the car is increasing (acceleration).
• When a bus is retarding (applying brakes).
• When a body is falling freely (constant acceleration due to gravity).
• The motion of a car on a normal road (often involves changes in speed and direction, leading to acceleration).

Exam Tip: Remember that acceleration isn't just about speeding up; it also includes slowing down (negative acceleration or deceleration) and changing direction, even if speed is constant.

Textbook Page 106

Activity 8.9 Discussion and conclusion
Answer: Distance-time graph: The average speed of the train from A to B is \( \frac{120 \text{ km}}{3 \text{ h}} = 40 \) km/h. The average speed of the train from C to D is \( \frac{60 \text{ km}}{1.5 \text{ h}} = 40 \) km/h.
In simple words: Looking at the distance-time graph, the train's average speed from point A to point B is 40 km/h, and its average speed from point C to point D is also 40 km/h.

Exam Tip: For distance-time graphs, the slope of the line indicates speed. A steeper slope means higher speed. For uniform motion, the slope is constant.

Textbook Page 107

Activity 8.10 Discussion and conclusion
Answer: Feroz travels faster than Sania. This implies that Feroz covers more distance in the same amount of time, or covers the same distance in less time, compared to Sania.
In simple words: Feroz moves quicker than Sania, meaning he gets further in the same time, or reaches a point faster.

Exam Tip: To compare who travels faster, you need to compare their speeds, which can be determined by either distance covered per unit time or time taken per unit distance.

Textbook Page 111

Activity 8.11 Discussion and conclusion
Answer:

1. The moment the stone is released, it moves tangent to the circular path at that instant.
2. By releasing the stone at different positions on the circular path, we will discover that the direction in which the stone moves is always different, but it is always tangent to the circular path.

Exam Tip: Understand that in circular motion, velocity is always tangential. When the object is released, it continues in a straight line along the tangent at the point of release due to inertia.

Question 1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Answer: Time taken \( = 2 \) min \( 20 \) sec \( = 140 \) sec.
Radius, \( r = 100 \) m (since diameter is 200 m)
In 40 seconds, the athlete completes one round.
In 140 seconds, the athlete completes \( = \frac{140}{40} = 3.5 \) rounds.
So, distance traveled in 140 seconds \( = 3.5 \times 2\pi r \)
\( = 3.5 \times 2 \times \frac{22}{7} \times 100 \)
\( = 3.5 \times \frac{44}{7} \times 100 \)
\( = 0.5 \times 44 \times 100 \)
\( = 2200 \) m
At the end of this motion, after 3.5 rounds, the athlete will be at a diametrically opposite position relative to the starting point.
So, Displacement \( = \) diameter \( = 200 \) m.
In simple words: The athlete runs around a circular track. After 140 seconds, they have completed three and a half laps. We calculate the total distance covered by adding up all the laps. Since they are halfway around the track on the last lap, their final position is directly opposite their starting point, making their displacement equal to the track's diameter.

Exam Tip: For circular motion, distance is \( n \times 2\pi r \) (where n is the number of rounds). Displacement depends on the final position relative to the start; it's zero after full rounds and the diameter after half rounds.

Question 2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph's average speed and velocities in jogging (a) from A to B and (b) from A to C?
Answer:(a) For motion from A to B.
Distance covered \( = 300 \) m
Displacement \( = 300 \) m
Time taken \( = 2 \) min \( 30 \) s \( = (2 \times 60) + 30 = 120 + 30 = 150 \) s
Average speed from A to B \( = \frac{\text{Distance}}{\text{Time}} = \frac{300}{150} = 2 \) m/s
Average velocity from A to B \( = \frac{\text{Displacement}}{\text{Time}} = \frac{300}{150} = 2 \) m/s
(b) For motion from A to C.
Total distance covered \( = 300 \text{ m (A to B)} + 100 \text{ m (B to C)} = 400 \) m
Displacement from A to C \( = 300 \text{ m (forward)} - 100 \text{ m (backward)} = 200 \) m
Total time taken \( = 2 \) min \( 30 \) s \( + 1 \) min \( = 150 \) s \( + 60 \) s \( = 210 \) s.
Average speed from A to C \( = \frac{\text{Distance}}{\text{Time}} = \frac{400}{210} = 1.90 \) m/s
Average velocity from A to C \( = \frac{\text{Displacement}}{\text{Time}} = \frac{200}{210} = 0.95 \) m/s
In simple words: Joseph runs down a road and then turns back. For each part of his run, we calculate his average speed (total distance over total time) and average velocity (total displacement over total time). When he turns around, his displacement gets smaller while his distance keeps growing.

Exam Tip: Crucially distinguish between average speed (scalar, total path) and average velocity (vector, net change in position). For motion that involves turning back, displacement will be less than distance.

Question 3. Abdul, while driving to school, computes the average speed for his trip to be 20 km h\(^{-1} \). On his return trip along the same route, there is less traffic and the average speed is 30 km h\(^{-1} \). What is the average speed for Abdul's trip?
Answer: Let the distance while driving to school be \( x \) km.
Time taken for the forward trip (to school) at a speed of 20 km/h: \( t_1 = \frac{x}{20} \) h
Time taken for the return trip (from school) at a speed of 30 km/h: \( t_2 = \frac{x}{30} \) h
Total distance traveled \( = x + x = 2x \) km
Total time taken for the whole trip \( = t_1 + t_2 = \frac{x}{20} + \frac{x}{30} \)
To add the fractions, find a common denominator (60):
\( = \frac{3x}{60} + \frac{2x}{60} = \frac{5x}{60} \) h
Average speed for the entire trip \( = \frac{\text{Total distance traveled}}{\text{Total time taken}} \)
\( = \frac{2x}{\frac{5x}{60}} \)
\( = 2x \times \frac{60}{5x} \)
\( = \frac{120x}{5x} \)
\( = 24 \) km/h
In simple words: Abdul drives to school and back using the same route but at different speeds. To find his overall average speed, we calculate the total distance he traveled and divide it by the total time he spent driving.

Exam Tip: When calculating average speed for a round trip with different speeds, it's essential to calculate the time taken for each segment and the total distance. Avoid simply averaging the two speeds directly, as that is often incorrect.

Question 4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms\(^{-2} \) for 8.0 s. How far does the boat travel during this time?
Answer: Here, initial velocity, \( u = 0 \) m/s (starts from rest)
Acceleration, \( a = 3 \) ms\(^{-2} \)
Time, \( t = 8 \) s
Using the equation: \( s = ut + \frac{1}{2} at^2 \)
\( s = (0 \times 8) + \frac{1}{2} \times 3 \times (8)^2 \)
\( s = 0 + \frac{1}{2} \times 3 \times 64 \)
\( s = \frac{192}{2} \)
\( s = 96 \) m
In simple words: A motorboat begins from a stop and steadily speeds up for 8 seconds. We use a formula to figure out how far it has moved during that time.

Exam Tip: When "starting from rest," the initial velocity \( u \) is 0. Always choose the correct kinematic equation based on the known and unknown variables.

Question 5. A driver of a car traveling at 52 km/h applies brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 34 km/h in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars traveled farther after brakes were applied?
Answer: Given, initial speed of the first car, \( u_1 = 52 \) km/h \( = 52 \times \frac{5}{18} = 14.4 \) m/s
Time, \( t_1 = 5 \) s; final speed, \( v_1 = 0 \)
Acceleration of first car: \( a_1 = \frac{v_1 - u_1}{t_1} = \frac{0 - 14.4}{5} = -2.88 \) m/s\(^2 \)
Initial speed of second car, \( u_2 = 34 \) km/h \( = 34 \times \frac{5}{18} = 9.4 \) m/s
Time, \( t_2 = 10 \) s; final speed, \( v_2 = 0 \)
Acceleration of second car: \( a_2 = \frac{v_2 - u_2}{t_2} = \frac{0 - 9.4}{10} = -0.94 \) m/s\(^2 \)
Distance traveled by first car: \( s_1 = \frac{v_1^2 - u_1^2}{2a_1} = \frac{0^2 - (14.4)^2}{2(-2.88)} = \frac{-207.36}{-5.76} = 36 \) m
Distance traveled by second car: \( s_2 = \frac{v_2^2 - u_2^2}{2a_2} = \frac{0^2 - (9.4)^2}{2(-0.94)} = \frac{-88.36}{-1.88} = 47 \) m
The second car traveled farther (47 m) after brakes were applied.
In simple words: Two cars apply brakes and stop. We calculate how quickly each car slowed down and how far it traveled before stopping. The car that was slower but took longer to stop actually covered more ground during braking.

Exam Tip: For problems involving stopping distance, remember to convert all speeds to m/s. The area under the speed-time graph represents the distance traveled. A graphical representation would show two triangles, and the area of each triangle gives the distance. Calculate the areas (distances) to compare which car traveled farther.

Question 6. Given figure shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:
(a) Which of the three is traveling the fastest?
(b) Are all three ever at the same point on the road?
(c) How far has C traveled when B passes A?
(d) How far has B traveled by the time it passes C?
Answer:(a) B is traveling fastest because its slope on the distance-time graph is the steepest.
(b) All three are never at the same point on the road at the same time because their lines on the graph never intersect at a single point.
(c) When B passes A, C has traveled approximately 6.86 km.
(d) When B passes C, B has traveled approximately 5.71 km.
In simple words: Looking at the graph, the steepest line shows the fastest object, which is B. The lines for A, B, and C never cross at the exact same spot, so they are never all at the same place at once. We also find out how far C and B have traveled at the specific moments when they pass other objects.

Exam Tip: For distance-time graphs:

• The steepest slope indicates the fastest speed.
• Lines crossing mean objects are at the same position at that time.
• To find distance traveled at a specific point, read the distance value on the y-axis corresponding to the time when the event occurs.

Question 7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s\(^{-2} \), with what velocity will it strike the ground? After what time will it strike the ground?
Answer: Given,
Initial velocity, \( u = 0 \) m/s (gently dropped)
Distance (height), \( s = 20 \) m
Acceleration, \( a = 10 \) m/s\(^2 \)
To find final velocity \( v \):
Using equation: \( v^2 - u^2 = 2as \)
\( v^2 - 0^2 = 2 \times 10 \times 20 \)
\( v^2 = 400 \)
\( v = \sqrt{400} \)
\( v = 20 \) m/s
To find time \( t \):
Using equation: \( v = u + at \)
\( 20 = 0 + 10 \times t \)
\( 10t = 20 \)
\( t = \frac{20}{10} \)
\( t = 2 \) s
In simple words: A ball is dropped from 20 meters high and speeds up as it falls due to gravity. We figure out how fast it will be going when it hits the ground and how long that fall will take.

Exam Tip: For objects dropped, initial velocity is zero. Acceleration due to gravity is usually taken as \( 9.8 \) m/s\(^2 \) or \( 10 \) m/s\(^2 \) as specified. Use the appropriate kinematic equations to solve for velocity and time.

Question 8. The speed-time graph for a car is shown in the figure.
(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance traveled by car during the period.
(b) Which part of the graph represents the uniform motion of the car?
Answer:(a) Distance covered \( = \) area under the speed-time graph.
Assuming the graph shows a triangular area up to 4 seconds, the distance would be calculated as: Distance \( = \frac{1}{2} \times \text{base} \times \text{height} \). Without the specific graph, the given value of \( 16.53 \) m cannot be verified but implies a complex shape or specific numerical interpretation not directly visible from the OCR.
The area under the graph up to 4 seconds should be shaded to represent the distance traveled.
(b) After 6 seconds, the car moves with uniform motion. This is represented by a horizontal line segment on the speed-time graph, indicating constant speed.
In simple words: For the first 4 seconds, we find how far the car traveled by looking at the area under its speed-time graph. The part of the graph that's a flat horizontal line (after 6 seconds) shows when the car is moving at a steady speed.

Exam Tip: The area under a speed-time graph represents the distance traveled. Uniform motion on a speed-time graph is a horizontal line, indicating constant speed and zero acceleration.

Question 9. State which of the following situations are possible and give an example for each of these:
(a) An object with a constant acceleration but with zero velocity
(b) An object moving in a certain direction with an acceleration in the perpendicular direction.
Answer:(a) Yes, a body can have constant acceleration even when its velocity is zero. For example, when a body is thrown straight up, at the highest point of its trajectory, its instantaneous velocity is zero, but it still has a constant acceleration equal to the acceleration due to gravity (downwards).
(b) Yes, an object can move in a certain direction with an acceleration in a perpendicular direction. For instance, when an airplane is flying horizontally, the acceleration due to gravity acts downwards, which is perpendicular to its horizontal motion. Another example is uniform circular motion, where velocity is tangential, and acceleration (centripetal) is directed towards the center, perpendicular to the velocity.
In simple words:(a) Yes, it's possible. Imagine throwing a ball straight up; at its very top point, it stops for a split second, but gravity is still pulling it down, so it still has acceleration. (b) Yes, it's possible. Think of an airplane flying straight; gravity pulls it down, which is a force (and thus acceleration) at a right angle to its forward movement. Or a car turning a corner, its speed might be constant but direction changes, causing perpendicular acceleration.

Exam Tip: Understanding the vector nature of velocity and acceleration is key. Acceleration can exist even with zero velocity (e.g., at the peak of a projectile's path) or be perpendicular to velocity (e.g., in circular motion or projectile motion).

Question 10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Answer: Given,
Radius, \( r = 42250 \) km
Time, \( t = 24 \) h \( = 24 \times 60 \times 60 \) s \( = 86400 \) s
The distance covered in one revolution is the circumference of the orbit: \( D = 2\pi r \)
\( D = 2 \times \frac{22}{7} \times 42250 \) km \( \approx 265571.4 \) km
Speed \( v = \frac{\text{Distance}}{\text{Time}} \)
\( v = \frac{265571.4 \text{ km}}{24 \text{ h}} \)
\( v = 11065.475 \) km/h
To convert to m/s: \( v = 11065.475 \times \frac{5}{18} \approx 3073.74 \) m/s
Or, \( v = 3.07 \) km/s (as \( 11065.475 \) km/h is approximately \( 3.07 \) km/s)
In simple words: We need to find how fast a satellite is moving. We know how big its orbit is (radius) and how long it takes to go around once. We calculate the total distance of its circular path and divide it by the time taken to find its speed.

Exam Tip: For circular orbits, the distance covered in one revolution is the circumference (\( 2\pi r \)). Remember to ensure units are consistent (e.g., convert hours to seconds or km/h to m/s) depending on the desired final unit for speed.