Class 9 Science Chapter 7 question answer Motion

Ncert Solutions for Class 9 Science Chapter 7: Motion class 9 questions answers

Textbook	Ncert
Class	Class 9
Subject	Science
Chapter	Chapter 7
Chapter Name	Class 9 motion ncert solutions
Category	Ncert Solutions
Medium	English

Are you looking for Class 9 Science Chapter 7 question answer ? Now you can download Motion class 9 questions answers pdf from here.

In Text Questions Page No: 74

Question 1: An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Answer 1: Yes, an object can have zero displacement even if it has moved through a distance.

Example: If a person walks 5 meters forward and then 5 meters back to the starting point, the total distance is 10 meters, but the displacement is zero because the initial and final positions are the same.

Question 2: 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 2: Given:

• Side of square field = 10 m
• Time taken to go around the boundary = 40 s
• Total time = 2 min 20 s = 140 s

Find number of rounds completed

One round = 40 s
In 140 s, number of rounds = 140 ÷ 40 = 3 full rounds + 20 s

Distance covered in 20 s

Speed = Perimeter ÷ Time = (4 × 10) ÷ 40 = 40 ÷ 40 = 1 m/s
So in 20 s, distance = 1 × 20 = 20 m

Path in last 20 s

After 3 full rounds (returns to starting point), in the next 20 m he moves along the square’s sides.
He covers:

• 10 m (1 side)
• then 10 m on the next side

So he ends at the corner diagonally opposite the starting point (after 2 sides).

Displacement

Displacement = diagonal of the square =
\(\sqrt{10^2 + 10^2} = \sqrt{200} = 14.14\ \text{m (approx)}\)

14.14 m

Question 3: 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 3: Neither of the statements is true.

(a) Given statement is false because the displacement of an object which travels a certain distance and comes back to its initial position is zero.

(b) Given statement is false because the displacement of an object can be equal to, but never greater than the distance travelled.

In Text Questions Page No: 76

Question 1: Distinguish between speed and velocity.

Answer 1: Difference between Speed and Velocity:

Speed	Velocity
Scalar quantity	Vector quantity
Has only magnitude	Has both magnitude and direction
Speed = Distance / Time	Velocity = Displacement / Time
Always positive	Can be positive, negative, or zero
Does not indicate direction	Indicates direction of motion

Example:
If a car moves 60 km in 1 hour, its speed is 60 km/h.
If it moves 60 km north in 1 hour, its velocity is 60 km/h north.

Question 2: Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Answer 2: Since average speed is the total distance travelled in a time frame and velocity is the total displacement in the time frame, the magnitude of average velocity and average speed will be the same when the total distance travelled is equal to the displacement.

Question 3: What does the odometer of an automobile measure?

Answer 3: The distance covered by an automobile is recorded by the odometer of an automobile.

Question 4: What does the path of an object look like when it is in uniform motion?

Answer 4: An object has a straight-line path when it is in uniform motion.

Question 5: 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 5: Time (t) = 5 min = 5 x 60 s = 300 s,
Distance = Speed x Time
= 3 x 10⁸ ms^-1 x 300 s
= 9 x 10¹⁰ m

In Text Questions Page No: 77

Question 1: When will you say a body is in
(i) uniform acceleration? (ii) nonuniform acceleration?

Answer 1: (i) Uniform acceleration: A body is in uniform acceleration if its velocity increases or decreases by equal amounts in equal intervals of time.
Example: A freely falling object under gravity.

(ii) Nonuniform acceleration: A body is in nonuniform acceleration if its velocity changes by unequal amounts in equal intervals of time.
Example: A car moving through traffic.

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 2: Given, the initial velocity (u) = 80km/hour = 80000m/3600s= 22.22 m.s^-1

The final velocity (v) = 60km/hour = 60000m/3600s= 16.66 m.s^-1

Time frame, t = 5 seconds.

Therefore, acceleration (a) =(v-u)/t = (16.66 m.s^-1 – 22.22 m.s^-1)/5s

= -1.112 m.s^-2

Therefore, the total acceleration of the bus is -1.112m.s^-2. It can be noted that the negative sign indicates that the velocity of the bus is decreasing.

Question 3: A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h⁻¹ in 10 minutes. Find its acceleration.

Answer 3: Given:

• Initial speed, u = 0 m/s (starts from rest)
• Final speed, v = 40 km/h = \(\frac{40 \times 1000}{3600}\) = 11.11 m/s
• Time, t = 10 minutes = 600 seconds

Using the formula:
\(a = \frac{v – u}{t} = \frac{11.11 – 0}{600} = \frac{11.11}{600} \approx 0.0185 \, \text{m/s}^2\)

The acceleration of the train is 0.0185 m/s².

In Text Questions Page No: 81

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

Answer 1: When the motion is uniform, the distance time graph is a straight line with a slope.

When the motion is non uniform, the distance time graph is not a straight line.It can be any curve.

Question 2: What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

Answer 2: The distance-time graph can be plotted as follows.

When the slope of the distance-time graph is a straight line parallel to the time axis, the object is at the same position as time passes. That means the object is at rest.

Question 3: What can you say about the motion of an object if its speedtime graph is a straight line parallel to the time axis?

Answer 3: A straight line parallel to the x-axis in a distance-time graph indicates that the position of the object does not change with time.

Therefore, the object is said to be at rest.

Question 4: What is the quantity which is measured by the area occupied below the velocity-time graph?

Answer 4: The area occupied below the velocitytime graph is a measure of the distance travelled by the body or the displacement of the body.

In Text Questions Page No: 82

Question 1: A bus starting from rest moves with a uniform acceleration of 0.1 m s^-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

Answer 1: Given:

• Initial speed, \( u = 0 \, \text{m/s} \) (starting from rest)
• Acceleration, \( a = 0.1 \, \text{m/s}^2 \)
• Time, \( t = 2 \, \text{minutes} = 120 \, \text{seconds} \)

(a) Speed acquired:

Using the formula:
v = u + at = 0 + (0.1 × 120) = 12 m/s

(b) Distance travelled:

Using the formula:
\(s = ut + \frac{1}{2} a t^2 = 0 + \frac{1}{2} \times 0.1 \times (120)^2\)
\(s = 0.05 \times 14400 = 720 \, \text{m}\)

(a) Speed acquired = 12 m/s
(b) Distance travelled = 720 m

Question 2: A train is travelling at a speed of 90 km h^–1 . Brakes are applied so as to produce a uniform acceleration of – 0.5 m s^-2. Find how far the train will go before it is brought to rest.

Answer 2: Given:

• Initial speed, \( u = 90 \, \text{km/h} = \frac{90 \times 1000}{3600} = 25 \, \text{m/s} \)
• Final speed, v = 0 m/s (train comes to rest)
• Acceleration, a = −0.5 m/s²

To find: Distance ( s )

Using the formula:
v² = u² + 2as
0 = 25² + 2(-0.5)s
0 = 625 − s ⇒ s = 625m

The train will travel 625 meters before coming to rest.

Question 3: A trolley, while going down an inclined plane, has an acceleration of 2 cm s^-2. What will be its velocity 3 s after the start?

Answer 3: Given:

• Acceleration, a = 2 cm/s^-2 = 0.02 m/s^-2
• Time, t = 3 s
• Initial velocity, u = 0 m/s (starts from rest)

Using the formula:
v = u + at = 0 + (0.02×3) = 0.06m/s

The velocity after 3 seconds is 0.06 m/s.

Question 4: A racing car has a uniform acceleration of 4 m s^-2. What distance will it cover in 10 s after start?

Answer 4: Given:

• Initial velocity, u = 0m/s (starts from rest)
• Acceleration, a = 4m/s^-2
• Time, t = 10 s

Using the formula:
\(s = ut + \frac{1}{2} a t^2 = 0 + \frac{1}{2} \times 4 \times (10)^2\) = 2 × 100 = 200m

The racing car will cover 200 meters in 10 seconds.

Question 5: A stone is thrown in a vertically upward direction with a velocity of 5 m s^-1. If the acceleration of the stone during its motion is 10 m s^–2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Answer 5: Given:

• Initial velocity, u = 5m/s (upward)
• Final velocity at highest point, v = 0m/s
• Acceleration, a = −10m/s² (since it’s acting downward)

(a) Time to reach maximum height

Using the formula:
v = u + at ⇒ 0 = 5 + (−10)t ⇒ t = \(\frac{5}{10}\) = 0.5 seconds

(b) Maximum height attained

Using the formula:
\(s = ut + \frac{1}{2} a t^2 = 5 \times 0.5 + \frac{1}{2} \times (-10) \times (0.5)^2\)
s = 2.5 – 1.25 = 1.25 m

• Height attained = 1.25 m
• Time to reach there = 0.5 s

Exercises

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 1: Given:

• Diameter of circular track = 200 m → Radius r = 100 m
• Time for one round = 40 s
• Total time = 2 min 20 s = 140 s

Step 1: Distance covered

Number of complete rounds = \( \frac{140}{40} = 3.5 \) rounds
Circumference of the track = \( 2\pi r = 2 \times 3.14 \times 100 = 628 \, \text{m} \)
Distance covered = 3.5 × 628 = 2198m

Step 2: Displacement

After 3 full rounds, the athlete is back at the starting point.
After another half round, they are at the point diametrically opposite the start.

So, displacement = diameter = 200 m

• Distance covered = 2198 m
• Displacement = 200 m

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

Answer 2: Given:

• Distance from A to B = 300 m
• Time from A to B = 2 min 30 s = 150 s
• Distance from B to C = 100 m
• Time from B to C = 1 min = 60 s
• Total time from A to C = 150 + 60 = 210s

(a) From A to B

• Distance = 300 m
• Displacement = 300 m (straight line)
• Time = 150 s

Average speed = \( \frac{\text{Distance}}{\text{Time}} = \frac{300}{150} = 2 \, \text{m/s} \)
Average velocity = \( \frac{\text{Displacement}}{\text{Time}} = \frac{300}{150} = 2 \, \text{m/s} \)

(b) From A to C

• Distance = AB + BC = 300 + 100 = 400 m
• Displacement = AC = 300 − 100 = 200 m
• Time = 210 s

Average speed = \( \frac{400}{210} \approx 1.90 \, \text{m/s} \)
Average velocity = \( \frac{200}{210} \approx 0.95 \, \text{m/s} \)

(a) A to B

• Average speed = 2 m/s
• Average velocity = 2 m/s

(b) A to C

• Average speed = 1.90 m/s
• Average velocity = 0.95 m/s

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 3: Distance travelled to reach the school = distance travelled to reach home = d (say)

Time taken to reach school = t₁

Time taken to reach home = t₂

therefore, average speed while going to school = total distance travelled/ total time taken = d/t₁ = 20 kmph

Average speed while going home = total distance travelled/ total time taken = d/t₂= 30 kmph

Therefore, t₁ = d/20 and t₂ = d/30

Now, the average speed for the entire trip is given by total distance travelled/ total time taken

= (d+d)/(t₁+t₂)kmph = 2d/(d/20+d/30)kmph

= 2/[(3 + 2)/60]

= 120/5 kmh^-1 = 24 kmh^-1

Therefore, Abdul’s average speed for the entire trip is 24 kilometers per hour.

Question 4: A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s^–2 for 8.0 s. How far does the boat travel during this time?

Answer 4: Given Initial velocity of motorboat,  u = 0
Acceleration of motorboat,  a = 3.0 m s^-2
Time under consideration,  t = 8.0 s
We know that Distance, s = ut + (1/2)at²
Therefore, The distance travel by motorboat = 0 ×8 + (1/2)3.0 × 8²
= (1/2) × 3 × 8 × 8 m
= 96 m

Question 5: A driver of a car travelling at 52 km h−1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h−1 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 travelled farther after the brakes were applied?

Answer 5: As given in the figure below PR and SQ are the Speed-time graph for given two cars with initial speeds 52 kmh⁻¹ and 3 kmh⁻¹ respectively.

Distance Travelled by first car before coming to rest =Area of △ OPR
= (1/2) × OR × OP
= (1/2) × 5s × 52 kmh⁻¹
= (1/2) × 5 × (52 × 1000) / 3600) m
= (1/2) × 5 × (130 / 9) m
= 325 / 9 m
= 36.11 m

Distance Travelled by second car before coming to rest =Area of △ OSQ
= (1/2) × OQ × OS
= (1/2) × 10 s × 3 kmh⁻¹
= (1/2) × 10 × (3 × 1000) / 3600) m
= (1/2) × 10 x (5/6) m
= 5 × (5/6) m
= 25/6 m
= 4.16 m

Question 6: . 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 6: (a) The slope of B is greater than that of A and C. Hence, B is moving the fastest.

(b) No, because the three lines are not intersecting at any point.

(c) By the time B passes A, C covers a distance of about 9 km or more.

(d) By the time B passes C, it has travelled 6 km.

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 7: Given:

• Initial velocity, u = 0m/s (dropped gently)
• Acceleration due to gravity, a = 10 m/s²
• Height (displacement), s = 20m

(a) Final velocity ( v )

Using the equation:
v² = u² + 2as = 0+2×10×20 = 400 ⇒ v = \(\sqrt{400}\) = 20m/s

(b) Time to strike the ground ( t )

Using the equation:
v = u + at ⇒ 20 = 0 + 10t ⇒ t = \(\frac{20}{10}\) = 2s

• Velocity when it strikes the ground = 20 m/s
• Time taken to strike the ground = 2 seconds

Question 8: The speed-time graph for a car is shown is 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 8: (a) The shaded area which is equal to 1/2 × 4 × 6 = 12 m represents the distance travelled by the car in the first 4 s.

(b) The part of the graph in red colour between time 6 s to 10 s represents uniform motion of the car.

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

Answer 9: (a) It is possible; an object thrown up into the air has a constant acceleration due to gravity acting on it. However, when it reaches its maximum height, its velocity is zero.

(b) it is possible; acceleration implies an increase or decrease in speed, and uniform speed implies that the speed does not change over time Circular motion is an example of an object moving with acceleration but with uniform speed. An object moving in a circular path with uniform speed is still under acceleration because the velocity changes due to continuous changes in the direction of motion.

(c) It is possible; for an object accelerating in a circular trajectory, the acceleration is perpendicular to the direction followed by the object.

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 10: Radius of the circular orbit, r = 42250 km
Time taken to revolve around the earth, t= 24 h
Speed of a circular moving object, v = (2π r)/t
= [2× (22/7)×42250 × 1000] / (24 × 60 × 60)
= (2×22×42250×1000) / (7 ×24 × 60 × 60) m s^-1
= 3073.74 m s ^-1