NCERT Solutions - Chapter 7: Motion

Q1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Answer - Yes, an object can have zero displacement even if it has moved through a distance.

For example, if a person walks around a circular track and returns to the starting point, the distance covered is the perimeter of the circle, but the displacement is zero because the initial and final positions are the same.

Q2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Answer - Time given = 2 minutes 20 seconds = 140 seconds
Time to complete one round = 40 seconds

Number of rounds completed = 140 / 40 = 3.5

After 3 full rounds, the farmer is back to the starting point. In the next half round (0.5 round), the farmer reaches the opposite corner of the square field.

Diagonal of the square = √(10² + 10²) = √200 = 14.14 m

Therefore, the magnitude of displacement = 14.14 m

Q3. Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object.

Answer - None of the given statements are true.

Correct explanation:
(a) Displacement can be zero if the initial and final positions are the same.
(b) Displacement is always less than or equal to the distance, never greater.

Q1. Distinguish between speed and velocity.

Answer -

	Speed	Velocity
Definition	Speed is the rate of change of distance.	Velocity is the rate of change of displacement.
Nature	Scalar quantity (has only magnitude).	Vector quantity (has both magnitude and direction).
Value	Always positive.	Can be positive, negative, or zero.
Example	60 km/h	60 km/h towards north

Q2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Answer - The magnitude of average velocity is equal to the average speed when the object moves along a straight line in the same direction, i.e., when the total distance covered equals the magnitude of displacement.

Q3. What does the odometer of an automobile measure?

Answer - The odometer of an automobile measures the total distance travelled by the vehicle.

Q4. What does the path of an object look like when it is in uniform motion?

Answer - When an object is in uniform motion, its path appears as a straight line.

Q5. 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, that is, 3 × 10⁸ m s^–1.

Answer - Time taken = 5 minutes = 5 × 60 = 300 seconds
Speed of signal = 3 × 10⁸ m/s

Distance = Speed × Time = 3 × 10⁸ × 300 = 9 × 10¹⁰ m

So, the spaceship is 9 × 10¹⁰ meters away from the ground station.

Q1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?

Answer -

Type	Description
Uniform Acceleration	A body is said to be in uniform acceleration when it changes its velocity by equal amounts in equal intervals of time.
Non-uniform Acceleration	A body is said to be in non-uniform acceleration when it changes its velocity by unequal amounts in equal intervals of time.

Q2. A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.

Answer - Initial speed (u) = 80 km/h = 22.22 m/s
Final speed (v) = 60 km/h = 16.67 m/s
Time (t) = 5 s

Acceleration (a) = (v - u)/t = (16.67 - 22.22)/5 = -1.11 m/s²

The acceleration of the bus is -1.11 m/s² (negative sign indicates retardation).

Q3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km/h in 10 minutes. Find its acceleration.

Answer - Initial speed (u) = 0 (since it starts from rest)
Final speed (v) = 40 km/h = 11.11 m/s
Time (t) = 10 minutes = 600 s

Acceleration (a) = (v - u)/t = (11.11 - 0)/600 = 0.0185 m/s²

The acceleration of the train is approximately 0.0185 m/s².

Q1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

Answer -

Type of Motion	Nature of Distance-Time Graph
Uniform Motion	Straight line inclined to the time axis
Non-uniform Motion	Curved line

Q2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

Answer - If the distance-time graph is a straight line parallel to the time axis, it means that the object is at rest. There is no change in distance with time.

Q3. 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 - If the speed-time graph is a straight line parallel to the time axis, it indicates that the object is moving with uniform speed (no acceleration or deceleration).

Q4. What is the quantity which is measured by the area occupied below the velocity-time graph?

Answer - The area under the velocity-time graph gives the displacement of the object.

Q1. A bus starting from rest moves with a uniform acceleration of 0.1 m/s² for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

Answer - (a) Initial velocity (u) = 0 m/s
Acceleration (a) = 0.1 m/s²
Time (t) = 2 minutes = 120 seconds

Speed acquired (v) = u + at = 0 + (0.1 × 120) = 12 m/s

(b) Distance travelled (s) = ut + (1/2)at² = 0 + (1/2 × 0.1 × 120²) = (0.05 × 14400) = 720 m

Q2. A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of –0.5 m/s². Find how far the train will go before it is brought to rest.

Answer - Initial speed (u) = 90 km/h = 25 m/s
Final speed (v) = 0 m/s
Acceleration (a) = –0.5 m/s²

Using v² = u² + 2as
0 = (25)² + 2 × (–0.5) × s
0 = 625 – s
s = 625 / 1 = 625 m

Q3. A trolley, while going down an inclined plane, has an acceleration of 2 cm/s². What will be its velocity 3 s after the start?

Answer - Initial velocity (u) = 0 cm/s
Acceleration (a) = 2 cm/s²
Time (t) = 3 s

Velocity (v) = u + at = 0 + 2 × 3 = 6 cm/s

Q4. A racing car has a uniform acceleration of 4 m/s². What distance will it cover in 10 s after start?

Answer - Initial velocity (u) = 0 m/s
Acceleration (a) = 4 m/s²
Time (t) = 10 s

Distance (s) = ut + (1/2)at² = 0 + 0.5 × 4 × 100 = 200 m

Q5. A stone is thrown in a vertically upward direction with a velocity of 5 m/s. If the acceleration of the stone during its motion is 10 m/s² in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Answer - Initial velocity (u) = 5 m/s
Final velocity (v) = 0 (at maximum height)
Acceleration (a) = –10 m/s²

Using v² = u² + 2as
0 = (5)² + 2 × (–10) × s
0 = 25 – 20s
s = 25 / 20 = 1.25 m

Time to reach maximum height (t) = (v – u)/a = (0 – 5)/–10 = 0.5 s

Q1. 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 = 2 minutes 20 seconds = 140 s
Time for one round = 40 s
Number of rounds = 140 / 40 = 3.5

Distance covered = 3.5 × Circumference
Diameter = 200 m → Radius = 100 m
Circumference = 2πr = 2 × 3.14 × 100 = 628 m
Distance = 3.5 × 628 = 2198 m

Displacement = diameter (as athlete ends halfway around the track)
Displacement = 200 m

Q2. 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 speeds and velocities in jogging (a) from A to B and (b) from A to C?

Answer - (a) From A to B:
Distance = 300 m
Time = 2 min 30 s = 150 s
Average speed = 300 / 150 = 2 m/s
Displacement = 300 m → Average velocity = 2 m/s

(b) From A to C:
Distance = 300 + 100 = 400 m
Time = 150 + 60 = 210 s
Average speed = 400 / 210 ≈ 1.90 m/s
Displacement = 200 m
Average velocity = 200 / 210 ≈ 0.95 m/s

Q3. Abdul, while driving to school, computes the average speed for his trip to be 20 km/h. On his return trip along the same route, there is less traffic and the average speed is 30 km/h. What is the average speed for Abdul’s trip?

Answer - Let one-way distance = x km

Time for onward trip = x / 20 hours
Time for return trip = x / 30 hours
Total distance = 2x
Total time = x/20 + x/30 = (3x + 2x)/60 = 5x/60 = x/12 hours

Average speed = Total distance / Total time = 2x / (x/12) = 24 km/h

Q4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m/s² for 8.0 s. How far does the boat travel during this time?

Answer - Initial velocity (u) = 0
Acceleration (a) = 3.0 m/s²
Time (t) = 8 s

Distance = ut + ½at² = 0 + 0.5 × 3 × 64 = 96 m

Q5. A driver of a car travelling at 52 km/h applies the brakes.
(a) Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?

Answer - (a) The area under the speed-time graph represents the distance travelled. Shade the area below the sloping line showing deceleration (braking phase).

(b) The horizontal part of the graph before the slope (constant speed) represents the uniform motion of the car.

Q6. Fig 7.10 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 travelling the fastest?
(b) Are all three ever at the same point on the road?
(c) How far has C travelled when B passes A?
(d) How far has B travelled by the time it passes C?

Answer - (a) The steepest graph represents the fastest object. Thus, B is travelling the fastest.
(b) The three lines do not intersect at the same point. So, they are never at the same point on the road.
(c) From the graph, note the distance covered by C at the time when B and A intersect. Read distance value for C at that time (as per diagram).
(d) Similarly, at the point where B and C intersect, read the distance value for B. That is how far B has travelled.

Q7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m/s², with what velocity will it strike the ground? After what time will it strike the ground?

Answer - Initial velocity, u = 0
Acceleration, a = 10 m/s²
Distance, s = 20 m

Use: v² = u² + 2as = 0 + 2 × 10 × 20 = 400 ⇒ v = √400 = 20 m/s

Use: v = u + at ⇒ 20 = 0 + 10t ⇒ t = 2 seconds

Q8. The speed-time graph for a car is shown in Fig. 7.11.
(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?

Answer - (a) The area under the speed-time graph (0 to 4 s) gives the distance travelled. From the graph, use triangle area: ½ × base × height = ½ × 4 × speed.
Calculate accordingly and shade the triangle under the curve.

(b) The horizontal line part of the graph where speed is constant represents uniform motion.

Q9. 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 with an acceleration but with uniform speed
(c) an object moving in a certain direction with an acceleration in the perpendicular direction

Answer - (a) Yes. A ball thrown upwards has zero velocity at the highest point, but constant acceleration due to gravity (–10 m/s²).
(b) No. If speed is uniform, acceleration must be zero. So this is not possible.
(c) Yes. A satellite in circular orbit is moving tangentially while being accelerated towards the center of the Earth. Hence, direction changes but speed remains constant.

Q10. 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 - Radius = 42250 km = 4.225 × 10⁷ m
Time = 24 h = 86400 s
Circumference = 2πr = 2 × 3.14 × 4.225 × 10⁷ = 2.6545 × 10⁸ m

Speed = Distance / Time = 2.6545 × 10⁸ / 86400 ≈ 3072 m/s