Chapter 9: MOTION AND ITS DESCRIPTION

Chapter 9: Motion and Its Description Class 10 Science and Technology (212) NIOS Textbook Solutions

Chapter 9. MOTION AND ITS DESCRIPTION

INTEXT QUESTIONS 9.1

Choose the correct answer in the followings:

1. For an object moving along a straight line without changing its direction the

(a) distance travelled > displacement

(b) distance travelled < displacement

(d) distance is not zero but displacement is zero.

Answer: (c) distance travelled = displacement

[ When an object moves along a straight line without changing direction, the path length (distance) and the straight-line distance from start to finish (displacement) are the same. Displacement only differs from distance if the object changes direction or returns to its starting point. ]

2. In a circular motion the distance travelled is

(a) always > displacement

(b) always < displacements

(d) zero when displacement is zero

Answer: (a) always > displacement

[ In circular motion, the distance travelled is the actual length of the curved path, while displacement is the shortest straight-line distance between the starting and ending positions.]

3. Two persons start from position A and reach to position B by two different paths ACB and AB respectively as shown in Fig. 9.10.

Photo

(a) Their distances travelled are same

(b) Their displacement are same

(d) The distance travelled by I < distance travelled by II.

Answer:

4. In respect of the top point of the bicycle wheel of radius R moving along a straight road, which of the following holds good during half of the wheel rotation.

(a) distance = displacement

(b) distance < displacement

(d) displacement = πR

Answer: (c) displacement = 2R.
[ During half a rotation, the top point moves from top to bottom. Displacement equals the wheel's diameter (2R), while distance is πR (half circumference). So displacement = 2R.]

5. An object thrown vertically upward to the height of 20 m comes to the hands of the thrower in 10 second. The displacement of the object is

(a) 20 m (b) 40 m (c) Zero (d) 60 m

Answer: (c) Zero
[ The object returns to the thrower's hand, so its initial and final positions are the same. Displacement depends only on the starting and ending points, not the path. Therefore, displacement is zero.]

6. Draw a distance-displacement graph for an object in uniform circular motion on a track of radius 14 m.

Answer:

INTEXT QUESTIONS 9.2

1. Some of the quantities are given in column I. Their corresponding values are written in column II but not in same order. You have to match these values corresponding to the values given in column I:

Column I

Column II

(a) 1 kmh-1

(b) 18 kmh-1

(d) 36 kmh-1

(i) 20 ms-1

(ii) 10 ms-1

(iii) 5/18 ms-1

(iv) 5 ms-1

Answer: The correct matching is:

(a) 1 km/h

→ (iii) 5/18 m/s
(b) 18 km/h

→ (iv) 5 m/s
(c) 72 km/h

→ (i) 20 m/s
(d) 36 km/h

→ (ii) 10 m/s

2. A cyclist moves along the path shown in the diagram and takes 20 minutes from point A to point B. Find the distance, displacement and speed of the cyclist.

Fig. 9.17

Answer:

3. Identify the situation for which speed and average speed of the objects are equal.

(i) Freely falling ball

(ii) Second or minute needle of a clock

(iii) Motion of a ball on inclined plane

(iv) Train going from Delhi to Mumbai

(v) When object moves with uniform speed

Answer: (v) When object moves with uniform speed

[ In uniform motion, instantaneous speed remains constant, so average speed equals that constant speed at all instants.]

4. The distance-time graph of the motion of an object is given. Find the average speed and maximum speed of the object during the motion.

Fig. 9.18

Answer:

5. The distance travelled by an object at different times is given in the table below. Draw a distance-time graph and calculate the average speed of the object. State whether the motion of the object is uniform or non-uniform.

Table 9.7

Time (s)→	0	10	20	30	40	50
Distance (m)→	0	2	4	6	8	10

Answer:

6. A player completes his half of the race in 60 minutes and next half of the race in 40 minutes. If he covers a total distance of 1200 m, find his average speed.

Answer: Here, Total distance = 1200 m and Total time = 60 + 40 = 100 minutes.

We have,

Average speed =Total distanceTotal time

Average speed=1200 m100 min=12m60 s=15=0.2 m/s

7. A train has to cover a distance of 1200 km in 16 h. The first 800 km are covered by the train in 10 h. What should be the speed of the train to cover the rest of the distance? Also find the average speed of the train.

Answer: Total distance = 1200 km , Total time = 16 h , Distance covered in first part = 800 km and Time taken = 10 h

Remaining distance = 1200 − 800 = 400 km
Remaining time = 16 − 10 = 6 h

Speed needed for the remaining distance:

Speed=DistanceTime=4006=66.67 km/h

So, the train should travel at 66.67 km/h for the remaining distance.

Again,

Average speed

8. A bird flies from a tree A to the tree B with the speed of 40 km/h and returns to tree A from tree B with the speed of 60 km/h . What is the average speed of the bird during this journey?

Answer: Here, v1=40 km/h (going from A to B)

v2=60 km/h (returning from B to A)

The average speed when an object travels the same distance at two different speeds, we use the formula:

Average speed=2v1×v2v1+v2=2×40×60 40+60=4800100=48 km/h

9. Three players P, Q and R reach from point A to B in same time by following three paths shown in the Fig. 9.19. Which of the player has more speed, which has covered more distance ?

Photo Fig. 9.19

Answer:

INTEXT QUESTIONS 9.3

1. Describe the motion of an object shown in Fig. 9.28.

Photo

Answer:

2. Compare the velocity of two objects where motion is shown in Fig. 9.29.

Photo

Answer:

3. Draw the graph for the motion of object A and B on the basis of data given in Table 9.10.

Table 9.10

Time (s)	10	20	30	40	50
Position (m) for A	5	5	5	5	5
Position (m) for B	2	4	6	8	10

4. A car accelerates from rest uniformly and attains a maximum velocity of 2 m/s in 5 seconds. In next 10 seconds it slows down uniformly and comes to rest at the end of 10th second. Draw a velocity-time graph for the motion. Calculate from the graph (i) acceleration, (ii) retardation, and (iii) distance travelled.

Answer:

Graph of Photo

The V-T graph has two straight line segments:

OA — rising line (acceleration phase): 0 to 5 s, velocity rises 0 → 2 m/s

AB — falling line (retardation phase): 5 s to 15 s, velocity falls 2 → 0 m/s

(i) Acceleration (first 5 s):

a=ΔvΔt=2-05-0=25=0.4 m/s²

(ii) Retardation (next 10 s):

Retardation

=ΔvΔt=0-215-5=-210=-0.2 m/s²

Magnitude of retardation = 0.2 m/s²

(iii) The graph forms a triangle with base = 15 s and height = 2 m/s.

Area=12×base×height=12×15×2=15 m

5. A body moving with a constant speed of 10 m/s suddenly reverses its direction of motion at the 5^th second and comes to rest in next 5 second. Draw a position-time graph of the motion to represent this situation.

Answer:

Photo of graph

From 0–5 s, straight line upward (10 m/s). At 5 s, direction reverses; line bends downward with decreasing slope, becoming horizontal at 10 s (rest). This represents sudden reversal then uniform deceleration to zero velocity.

INTEXT QUESTIONS 9.4

1. A ball is thrown straight upwards with an initial velocity 19.6 m/s . It was caught at the same distance above the ground from which it was thrown:

(i) How high does the ball rise?

(ii) How long does the ball remain in air? (g = 9.8 m/s² )

Answer: Given, u=19.6 m/s (upward) , g=-9.8 m/s² and v=0 m/s

(i) We have,

v2=u2+2as

⇒02=19.62+2×-9.8×h

⇒0=19.62-1.96×h

⇒19.6 h=19.62

⇒h=19.6×19.619.6

⇒h=19.6 m

So, the ball rises to 19.6 m.

(ii) We have,

v=u+at

⇒0=19.6-9.8t

⇒9.8t=19.6

⇒t=19.69.8=2 s

Since the ball returns to the same height, total time = 2× 2 = 4 s .

2. A brick is thrown vertically upwards with the velocity of 192.08 m/s to the labourer at the height of 9.8 m. What are its velocity and acceleration when it reaches the labourer?

Answer: Given, u=192.08 m/s (upward) , v= ? , h=s=9.8 m and g=9.8 m/s²

We have, v²=u²+2as

⇒v2=192.082+2-9.8×9.8

⇒v2=36894.73-192.08

⇒v2=36702.65

⇒v=±36702.65⇒v=±191.57

⇒v=191.57 m/s

The acceleration is 9.8 m/s² downward (due to gravity).

3. A body starts its motion with a speed of 10 m/s and accelerates for 10 s with 10 m/s² . What will be the distance covered by the body in 10 s?

Answer: Given, u=10 m/s , a=10 m/s² and t=10 s

We have,

s=ut+12at2

⇒s=10×10+12×10×102

⇒s=100+5×100

⇒s=100+500=600 m

4. A car starts from rest and covers a distance of 50 m in 10 s and 100 m in next 10 s. What is the average speed of the car?

Answer: Given, Distance covered in first 10 s = 50 m and Distance covered in next 10 s = 100 m

Total distance = 50 + 100 = 150 m

Total time = 10 + 10 = 20 s

We have,

Average speed=Total distanceTotal time=15020=7.5 m/s

Therefore, the average speed of the car = 7.5 m/s

INTEXT QUESTIONS 9.5

1. In circular motion the point around which body moves

(a) always remain in rest

(b) always remain in motion

(d) remain in oscillatory motion

Answer: (a) always remain in rest

[ In circular motion, the point around which the body moves is called the center of the circle. This center is fixed and does not move (it remains at rest) for a purely circular motion about a fixed axis. ]

2. In uniform circular motion

(a) speed remain constant

(b) velocity remain constant

(d) neither speed nor velocity remain constant

Answer: (a) speed remains constant

[ In uniform circular motion, the body moves with a constant speed along a circular path. However, its velocity is not constant because the direction of motion continuously changes. ]

3. A point on a blade of a ceiling fan has

(a) always uniform circular motion

(b) always uniformly accelerated circular motion

(d) variable accelerated circular motion

Answer: (b) always uniformly accelerated circular motion
[ A point on the blade of a ceiling fan moves in a circular path. Even if the speed remains constant, the direction of velocity continuously changes, so the point experiences centripetal acceleration continuously. ]

TERMINAL EXERCISE

1. An object initially at rest moves for t seconds with a constant acceleration a. The average speed of the object during this time interval is

(a) a . t2 (b) 2a. t (c) 12a. t² (d) 12 a².t

Answer: (a) a . t2

[ Here, u=0

We have, v=u+at

⇒v=0+at ⇒v=at

Average speed =u+v2=0+at2=at2 ]

2. A car starts from rest with a uniform acceleration of 4 m/s² . The distance travelled in metres at the ends of 1s, 2s, 3s and 4s are respectively,

(a) 4, 8, 16, 32 (b) 2, 8, 18, 32

Answer: (b) 2, 8, 18, 32

[ Here, a=4 m/s²

We use the equation of motion for distance traveled from rest under uniform acceleration,

s=12at²

At t = 1 s , then

s1=12×4×1²=2×1=2 m

At t = 2 s , then

s2=12×4×2²=2×4=8 m

At t = 3 s , then

s3=12×4×3²=2×9=18 m

At t = 4 s , then

s4=12×4×4²=2×16=32 m ]

3. Does the direction of velocity decide the direction of acceleration?

Answer: No, the direction of velocity does not always decide the direction of acceleration. Acceleration depends on how velocity changes. If speed increases, acceleration is in the same direction as velocity. If speed decreases, acceleration is opposite to the direction of velocity.

4. Establish the relation between acceleration and distance travelled by the body.

Answer: We know the equations of motion,

v=u+at

⇒v-u=at

⇒at=v-u

⇒t=v-ua

And the average velocity,

s=v+u2 t

⇒s=v+u2v-ua=v2-u22a

⇒s=v2-u22a

⇒2as=v2-u2

⇒v²=u²+2as

5. Explain whether or not the following particles have acceleration:

(i) a particle moving in a straight line with constant speed, and

(ii) a particle moving on a curve with constant speed.

Answer: (i) No, the particle has no acceleration because its speed and direction remain constant. Since there is no change in velocity, acceleration is zero.

(ii) Yes, the particle has acceleration because its direction of motion keeps changing. Since velocity changes due to change in direction, acceleration is present.

6. Consider the following combination of signs for velocity and acceleration of an object with respect to a one dimensional motion along x-axis and give example from real life situation for each case:

Table 9.11

Velocity

Acceleration

Example

(a) Positive

(b) Positive

(d) Negative

(e) Negative

(f) Negative

(g) Zero

(h) Zero

Positive

Negative

Zero

Positive

Negative

Zero

Positive

Negative

Ball rolling down on a slope like slide or ramp

Answer:

Based on one-dimensional motion along the x-axis, here is the completed table with real-life examples for each case:

Velocity	Acceleration	Example
(a) Positive	Positive	A car moving forward (positive direction) and speeding up (e.g., pressing the accelerator while driving straight).
(b) Positive	Negative	A car moving forward but braking to slow down (e.g., applying brakes while moving forward).
(c) Positive	Zero	A car moving forward at a constant speed on a straight, flat road (no acceleration).
(d) Negative	Positive	A car moving backward (reversing) but slowing down (e.g., applying brakes while reversing).
(e) Negative	Negative	A car moving backward and speeding up in reverse (e.g., pressing the accelerator while reversing).
(f) Negative	Zero	A car moving backward at a constant speed (steady reverse motion on a straight path).
(g) Zero	Positive	A ball dropped from rest just after release (velocity is zero at the top, but acceleration due to gravity is downward/positive).
(h) Zero	Negative	A ball thrown upward at its highest point (velocity is zero momentarily, but acceleration due to gravity is downward/negative).

7. A car travelling initially at 7 m/s accelerates at the rate of 8.0 m/s² for an interval of 2.0 s. What is its velocity at the end of the 2 s?

Answer: Here, u=7 m/s , a=8.0 m/s² , t=2.0 s and v=?

We have, v=u+at

⇒v=7+8.0×2.0=7+16

⇒v=23 m/s

Therefore, the velocity at the end of 2 seconds is 23 m/s.

8. A car travelling in a straight line has a velocity of 5.0 m/s at some instant. After 4.0 s, its velocity is 8.0 m/s . What is its average acceleration in this time interval?

Answer: Here, u=5.0 m/s , a= ? , t=4.0 s and v=8.0 m/s

We have, v=u+at

⇒8.0=5.0+a×4.0=5+4a

⇒8-5=4a

⇒4a=3

⇒a=34=0.75 m/s²

Therefore, the average acceleration of the car is 0.75 m/s².

9. The velocity-time graph for an object moving along a straight line has shown in Fig. 3.32. Find the average acceleration of this object during the time interval 0 to 5.0 s, 5.0 s to 15.0 s and 0 to 20.0 s.

Photo

Answer:

10. The velocity of an automobile changes over a period of 8 s as shown in the table given below:

Table 9.12

Time (s)	0.0	1.0	2.0	3.0	4.0	5.0	6.0	7.0	8.0
Velocity (m/s)	0.0	4.0	8.0	12.0	16.0	20.0	20.0	20.0	20.0

(i) Plot the velocity-time graph of motion.

(ii) Determine the distance the car travels during the first 2 s.

(iii) What distance does the car travel during the first 4 s?

(iv) What distance does the car travel during the entire 8 s?

(v) Find the slope of the line between t = 5.0 s and t = 7.0 s. What does the slope indicate?

(vi) Find the slope of the line between t = 0 s to t = 4 s. What does this slope represent?

11. The position-time data of a car is given in the table given below:

Table 9.13

Time (s)	0	5	10	15	20	25	30	35	40	45
Position (m)	0	100	200	200	200	150	112.5	75	37.5	0

(i) Plot the position-time graph of the car.

(ii) Calculate average velocity of the car during first 10 seconds.

(iii) Calculate the average velocity between t = 10 s to t = 20 s.

(iv) Calculate the average velocity between t = 20 s and t = 25 s. What can you say about the direction of the motion of car?

12. An object is dropped from the height of 19.6 m. Draw the displacement-time graph for time when object reach the ground. Also find velocity of the object when it touches the ground.

Answer: Here, u=0 m/s (dropped from rest) , h=19.6 m , a=g=9.8 m/s²

We have,

s=ut+12gt2

⇒19.6=0×t+12×9.8×t2

⇒19.6=4.9×t2

⇒t2=19.64.9

⇒t=19.64.9=4

⇒t=2 s

Again, v=u+at

⇒v=0+9.8×2

⇒v=19.6 m/s

Therefore, the Velocity at ground = 19.6 m/s (downward).

Step 3: Displacement-Time Table

Time (s)	s=12×9.8×t²	Displacement (m)
0	4.9×0	0
0.5	4.9×0.25	1.225
1.0	4.9×1	4.9
1.5	4.9×2.25	11.025
2.0	4.9×4	19.6

Displacement-Time Graph

Graph of photo

Graph description:

X-axis: Time (t) in seconds (0 to 2 s)

Y-axis: Displacement (s) in meters (0 to 19.6 m)

The curve starts at origin (0,0), slopes gently at first, then becomes steeper as time increases (parabolic shape), ending at point (2, 19.6).

13. An object is dropped from the height of 19.6 m. Find the distance travelled by object in last second of its journey.

Answer: Given, h=19.6 m , u=0 , a=g=9.8 m/s²

We have,

h=ut+12gt2=12gt²

⇒19.6=12×9.8×t2

⇒4.9t2=19.6

⇒t2=19.64.9

⇒t=19.64.9=4=2s

So, total time of journey = 2 seconds.

For distance in 2 s:

s2=12×9.8×2²=9.8×2=19.6 m

For distance in 1 s:

s1=12×9.8×1²=9.8×1=9.8 m

Therefore, the distance during last second =19.6-4.9=14.7 m

So, the object travels 14.7 m in the last second of its fall.

14. Show that for a uniformly accelerated motion starting from velocity u and acquiring velocity v has average velocity equal to arithmetic mean of the initial (u) and final velocity (v).

Answer: Let, u = Initial velocity , u = Final velocity = v and Time taken = t

The distance travelled in time t ,

s=u+v2 t

Average velocity =st

=u+v2 tt=u+v2

Therefore, the average velocity is equal to the arithmetic mean of the initial velocity and final velocity.

15. Find the distance, average speed, displacement, average velocity and acceleration of the object whose motion is shown in the graph (Fig. 9.33).

Photo

16. A body accelerates from rest and attains a velocity of 10 m/s in 5 s. What is its acceleration?

Answer: Given, u=0 m/s (starts from rest) , v=10 m/s , t=5 s

Using the formula:

a=v-ut

⇒a=10-05=105=2 m/s²

Therefore, the acceleration of the body = 2 m/s².

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Chapter 9: MOTION AND ITS DESCRIPTION

Chapter 9: Motion and Its Description Class 10 Science and Technology (212) NIOS Textbook Solutions

Chapter 9. MOTION AND ITS DESCRIPTION

TERMINAL EXERCISE

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