Welcome to the Chapter 8 - Motion, Class 9 Science NCERT Solutions page. Here, we provide detailed question answers for Chapter 8 - Motion. The page is designed to help students gain a thorough understanding of the concepts related to natural resources, their classification, and sustainable development.
Our solutions explain each answer in a simple and comprehensive way, making it easier for students to grasp key topics Motion and excel in their exams. By going through these Motion question answers, you can strengthen your foundation and improve your performance in Class 9 Science. Whether you’re revising or preparing for tests, this chapter-wise guide will serve as an invaluable resource.
| Speed | Velocity |
|---|---|
| The distance travelled by a moving object per unit time is called its speed. It does not have any direction. | The distance travelled by moving an object in a particular direction per unit time is called its velocity. It has a unique direction. |
|
Speed is given by the relation:
|
Velocity is given by the relation:
|
| It is a scalar quantity. | It is a vector quantity. |
When the total distance travelled by the object is equal to the displacement, then its average speed would be equal to its average velocity.
The odometer of an automobile measures the distance covered by a vehicle.
An object having uniform motion then the path would be a straight line.
Time taken by the signal to reach the ground station from the spaceship
= 5 min = 5 × 60 = 300 s
Speed of the signal = 3 × 108 m/s
∴Distance = Speed × Time taken = 3 × 108 × 300 = 9 × 1010 m.
Hence, distance = 9 × 1010 m.
(i) Uniform acceleration : A body is said to be in uniform acceleration when its velocity is changing at the same rate i.e when its acceleration in unit time is the same.
example : An object in freefall is in uniform acceleration.
(ii) Non-uniform acceleration : When a body moves with unequal velocity in the equal interval of time, the body is said to be moving with non-uniform acceleration.
example : If a car covers 10 meters in the first two seconds and 15 meters in the next two seconds.
Initial speed of the bus, u = 80 km/h =
Final speed of the bus, v = 60 km/h =
Time take to decrease the speed, t = 5 s
Here, the acceleration of bus is -1.112 m/s2
Initial velocity of the train, u = 0 (since the train is initially at rest)
Final velocity of the train, v = 40 km/h = 40 x 1000 / 60 x 60 = 11.1 m/s
Time taken, t = 10 min = 10 × 60 = 600 s
Now acceleration is given by the relation :
Hence, the acceleration of the train is 0.0185 m/s2.
For uniform motion, the distance−time graph of an object is a straight line (as shown in the following figure).
For non-uniform motion, the distance−time graph of an object is a curved line (as shown in the given figure).
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 the time passes. Thus, the object is at rest.
The speed of the object is not changing with the time it means the object is in uniform motion. So , the object is moving with uniform motion.
The area occupied under the speed time graph gives you distance traveled by object.
(a) Initial speed of the bus, u = 0
Acceleration, a = 0.1 m/s2
Time taken, t = 2 minutes = 60 x 2 = 120 s
Let v be the final speed acquired by the bus.
v = u + at = 0 + 0.1 x 120 = 12 m/s
∴ The speed acquired by the bus is 12 m/s
(b) According to the equation of motion:
v2 = u2 + 2as
v2 − u2 = 2as
(12)2 - (0)2 =2(0.1)s
144 = 0.2 s
s = 144 / 0.2
s = 720 m
Distance travelled by the bus is 720 m.
Initial speed of the train, u = 90 km/h = 90 x 1000 / 60 x 60 = 25 m/s
Final speed of the train, v = 0
Acceleration = −0.5 m s-2
Using equation of motion:
v2 = u2 + 2 as
(0)2 = (25)2 + 2 (−0.5) s
0 = 625 - s
s = 625 m
Where, s is the distance covered by the train
The train will cover a distance of 625 m before it comes to rest.
Initial velocity of the trolley, u = 0
Acceleration, a = 2 cm s−2
Time, t = 3 s
According to the first equation of motion:
v = u + at
Where, v is the velocity of the trolley after 3 s from start
v = 0 + 2 × 3 = 6 cm/s
Hence, the velocity of the trolley after 3 s from start is 6 cm/s.
Initial velocity of the racing car, u = 0
Acceleration, a = 4 m/s2
Time taken, t = 10 s
Distance, s = ?
According to the equation of motion:
where s is the distance covered by the racing car,
Hence, the distance covered by the racing car after 10 s from start is 200 m.
Initially, velocity of the stone,u = 5 m/s
Final velocity, v = 0
Acceleration of the stone, a = acceleration due to gravity, g = 10 m/s2
Since, u is upward & a is downward, it is a retarded motion.
Acceleration, a = −10 m/s2
Let s be the maximum height attained by the stone in time t.
According to the first equation of motion:
v = u + at
0 = 5 + (−10) t
According to the third equation of motion:
v2 = u2 + 2 as
(0)2 = (5)2 + 2(−10) s
However , the stone attains a height of 1.25 m in 0.5 s.
Diameter of a circular track, d = 200 m
Radius of the track, r = d/2 = 100 m
Circumference = 2πr = 2π (100) = 200π m
In 40 s, the given athlete covers a distance of 200π m.
In 1 s, the given athlete covers a distance = 200π / 40 m
The athlete runs for 2 minutes 20 s = 140 s
∴Total distance covered in 140 s =
He takes 3 rounds in 40 × 3 = 120 s. Thus, after 120 s his displacement is zero.
Then, the net displacement of the athlete is in 20 s only. In this interval of time, he moves at the opposite end of the initial position. Since displacement is equal to the shortest distance between the initial and final position of the athlete, displacement of the athlete will be equal to the diameter of the circular track.
∴ Displacement of the athlete = 200 m
Distance covered by the athlete in 2 min 20 s is 2200 m and his displacement is 200 m.
Distance covered by satellite in 24 hours, s = 2πr = 2 × 3.14 × 42250 = 265330 km
Time = 24 h
Speed = 265330 / 24
= 11055.4 km/h
= 11055.4 / 3600 = 3.07 km/s
(a) From end A to end B
Distance covered A to B = 300 m
Time taken to cover that distance = 2 min 30 seconds = 2 x 60 + 30 = 150 s
Average speed = 300 / 150 = 2 m/s
Average velocity = 300 / 150 = 2 m/s
The average speed and average velocity of Joseph from A to B are the same and equal to 2 m/s.
(b) From end A to end C
Total distance covered = Distance from A to B + Distance from B to C
= 300 + 100 = 400 m
Total time taken = Time taken to travel from A to B + Time taken to travel from B to C
= 150 + 60 = 210 s
Average speed = 400 / 210 = 1.90 m/s
Average velocity = 200 / 210 = 0.95 m/s
The average speed of Joseph from A to C is 1.90 m/s and his average velocity is 0.95 m/s.
Case I: While driving to school
Average speed of Abdul’s trip = 20 km/h
Total distance = d
Let total time taken = t1
Speed = Distance / time
20 = d / t1
t1 = d / 20…(i)
Case II: While returning from school
Total distance = d
Speed = 30 km/h
Now,total time taken = t2
Speed = distance / time
30 = d / t2
t2 = d / 30 ..... (ii)
Total distance covered in the trip = d + d = 2d
Total time taken, t = Time taken to go to school + Time taken to return to school
= t1 + t2
Total time taken = d / 20 + d / 30
= 3d + 2d / 60
= 5d / 60
= d / 12
From equations (i) and (ii),
Average Speed = 120/5
= 24 km/h
Hence, the average speed for Abdul’s trip is 24 km/h.
Initial velocity, u = 0
Acceleration of the motorboat, a = 3 m/s2
Time taken, t = 8 s
According to equation of motion:
Distance covered by the motorboat, s
Hence, the boat travels a distance of 96 m.
Case A:
Initial speed of the car, u1 = 52 km/h = 52 x (5 / 18) = 14.44 m/s
Time taken, t1 = 5 s
Final speed = 0 m/s
Case B:
Initial speed of the car, u2 = 3 km/h = 3 x (5 / 18) = 0.833 m/s ≅ 0.83 m/s
Time taken, t2 = 10 s
Final speed = 0 m/s
Plot of the two cars on a speed−time graph is shown in the following figure:
Distance covered by each car is equal to the area under the speed−time graph.
Distance covered in case A,
Distance covered in case B,
Then, the car1 travelling with a speed of 52 km/h travels farther after brakes were applied.
(a) Object B
(b) No
(c) 5.714 km
(d) 5.143 km
(a)
From the graph, it is clear B covers more distance in less time. Therefore, B is the fastest
(b) All three objects A, B and C never come at the same point .
(c) On the distance axis:
7 small boxes = 4 km
∴1 small box = 4 / 7 km
Initially, object C is 4 blocks away from the origin.
∴Initial distance of object C from origin= 16 / 7
Distance of object C from origin when B passes A = 8 km
Distance covered by C
Hence, C has travelled a distance of 5.714 km when B passes A.
(d) Distance covered by B at the time it passes C = 9 boxes
Hence, B has travelled a distance of 5.143 km when B passes C.
Distance covered by the ball, s = 20 m
Acceleration, a = 10 m/s2
Initially, velocity, u = 0
Final velocity, v = ?
Time, t = ?
According to the third equation of motion:
v2 = u2 + 2 as
v2 = 0 + 2 (10) (20)
v = 20 m/s
According to the first equation of motion:
v = u + at
20 = 0 + 10 (t)
t = 2 s
Hence, time taken to reach at the ground is 2 seconds.
(a) Shaded area =
The distance covered by the car in first 4 seconds is 12 m.
(b) The red colour between time 6 s to 10 s represents uniform motion of the car.
(a) Possible
When an object is thrown up at maximum height, it has zero velocity, although it will have constant acceleration due to gravity, which is equal to 9.8 m/s2.
(b) Possible
When an object is moving in a circular track, its acceleration is perpendicular to its direction.
Yes. An object that has moved through a distance can have zero displacement.
Example: If an object moves from point A and reaches to the same point A.
In this case, the total distance covered by the man is 20 m + 20 m + 20 m
+ 20 m = 80 m. Hence, his displacement is zero because the shortest distance between his initial and final position is zero.
In fig. ABCD is a square field of side 10 m.
Time for one round = 40 s
Total time = 2 min 20 s
= (2 x 60 + 20)s = 140s
Number of round completed =
If farmer starts from A, it will complete 3 rounds (A → B → C → D → A) at A. In the last 0.5 round starting from A, he will finish at C.
Displacement of farmer
(a) Not true ; If an object moving from point A and reaches to the same point A, then displacement can be zero.
(b) Not true ; Its magnitude can not be greater than the distance travelled by an object. However, sometimes, it can be either equal or less the distance travelled by the object.
The displacement of a moving object in a given interval is zero i.e., the object comes back to its initial position in the given time, whereas distance travelled by the object is not zero.
For example, an athlete moving on a circular track. If he starts from A and completes one round and reaches back to point A, his displacement is zero whereas distance travelled by him is not zero but 2πr.
As the object is moving with a uniform velocity, so acceleration will be zero, a = 0.
Then put a = 0
1st equation of motion is , v = u + at
v = u + 0 x t
v = u
2nd equation of motion is , s = ut + (½)at^2
s = ut + (½)0 t^2
s = ut
3rd equation of motion is , v^2 = u^2 + 2as
v^2 = u^2 + 2 x 0 x s
v^2 = u^2
For the initial 50 second, velocity is 2 m/s. After that, velocity drops of 0, as shown by the vertical line in the graph. For the next 50 second, velocity is taken in negative because displacement is becoming 0. So, the velocity-time graph will look like
The velocity attained in 8 seconds is given by
V = u+at
v = 0+5*8
v = 40m/s
The distance travelled in the first 8 sec = S1 = ut + 1/2 at2 = 0 + 1/2 x 5 x 64 = 160 m
The distance travelled in the last 4 sec with const. speed = S2 = vt = 40 x 4 = 160 m
So the total distance travelled by car in 12 sec = 160 + 160 = 320m
Speed from A to B = 30 Km/hr
Let the distance from A to B be D
Time taken to travel from A to B , T1 = ( Distance travelled / Speed )
T1 = D / 30
Speed from B to A = 20 km/hr
Time taken to travel from B to A, T2 = ( Distance travelled / Speed )
Total time taken, T = T1 + T2
D/30 + D/20 = D/12
Total distance from A to B and from B to A = 2D
Average speed = Total distance travelled / Total time taken = 2D/D/12
= 24 km/hr
(i) Since velocity is not changing, acceleration is equal to 0.
(ii) Reading the graph, velocity = 20 m/s
(iii) distance traveled by the cyclist in 15 s
= Area under v-t graph during that time interval
= 20 m/s x 15 s = 300 m
Due to retardation (acceleration due to gravity), the velocity of stone decreases and becomes zero at the highest point. Now the direction of velocity changes and increases due to acceleration due to gravity and reaches to the same velocity at ground.
Here, PQ corresponds to upward motion and QR corresponds to downward motion of stone.
Initially, difference in heights of two objects
= 150 m – 100 m = 50 m
Distance travelled by first object in 2 s = h1
= 0 + ½ (2)2 = 2 g
Distance travelled by another object in 2 s = h2
= o + ½ (2)2 = 2 g
After 2 s, height at which the first object will be h1 = 150 - 2g
After 2 s, height at which the second object will be h2 = 100 - 2g
Thus, after 2 s, difference in height = 150 - 2g - (100 - 2g) = 50 m = initial difference in height.
Thus, the difference in height does not vary with time.
For first 2 s motion of object,
u = 0, t = 2 s, s = 20 m.
For first 2 s motion of object,
u = 0, t = 2 s, s = 20 m
2a = 20
a = 10 m/s^2
Next 2 s motion of object
u = 0, s = (20 + 160) = 180 m
t = 2 + 4 = 6 sec., a = ?
Again using, s = ut + ½ at2
180 = 0 + ½ x a x (6)^2
180 = 18a
a = 10 m/s^2
This shows that the object is moving with constant acceleration.
Total motion,
u = 0, t = 7 sec, a = 10 m/s^2, v = ?
v = u + at
v = 0 + 10 x 7
v = 70 m/s
Using following data, draw time - displacement graph for a moving object:
| Time (s) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 |
| Displacement (m) | 0 | 2 | 4 | 4 | 4 | 6 | 4 | 2 | 0 |
Use this graph to find average velocity for first 4 s, for next 4 s and for last 6 s.
From the given data, displacement-time graph is shown as
Given initial velocity of electron, u = 5 x 10 ^4 m/s
Acceleration, a = 10^4 m/s^2
(i) Using first equation of motion,
Final velocity, v = 2u
v = u + at
2u = u + at
t = u/a = 5 s
(ii) Distance travelled in this time interval be s using 2nd equation of motion
s = ut + ½ at^2
s = 5 x 10^4 x 5 + ½ x 10^4 x 5^2
s = 37.5 x 10^4 m
Let the object be moving with initial velocity u m/s and uniform acceleration a m/s^2
Using the equation of motion, s = ut + at^2
Distance travelled in 5 sec, s = u x 5 + ½ a x 5^2
s = 5u + 25/2 a
Similarly, distance travelled in 4 sec, s’ = 4u + 16/2 a
s’ = 4u + 8a
Distance travelled in the interval between 4th and 5th seconds = s - s’
= 5u + 25/2 a - (4u + 8a) = (u + 9/2 a) m
At the highest point, v = 0
For the stone thrown with velocity u1,
v^2 - u^2 = 2gh
( Assume upward acceleration is –g and downward acceleration to be +g ).