NIOS Class 10 Science & Technology Chapter 9 Solution

NIOS Class 10 Science & Technology Chapter 9 Solution – Motion And Its Description

NIOS Class 10 Science & Technology Solution Chapter 9 Motion And Its Description. NIOS Class 10 Science & Technology Chapter 9 Question Answers Download PDF. NIOS Class 10 Science Notes.

NIOS Class 10 Science & Technology Chapter 9 Solution

Board NIOS
Class 10th  (Secondary)
Subject Science and Technology (212)
Topic Question Answer, Solution, Notes



(Motion And Its Description)



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

(c) distance travelled = displacement

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

Ans: – Option (c)distance travelled = displacement


2)In a circular motion the distance travelled is

(a) always > displacement

(b) always < displacements

(c) always = displacement

(d) zero when displacement is zero.

Ans: – Option (a) always>displacement.


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.

(a) Their distances travelled are same

(b) Their displacement is same

(c) The displacement of I > the displacement of II

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

Ans: – Option (b)Their displacement is same.


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

(c) displacement = 2R

(d) displacement = πR.

Ans: – Option (a) distance= displacement.


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

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

Ans: – Option (c) Zero.


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

Ans: – 



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:

Ans: –

Column I

Column II

(a)    1km/h (iii) 5/18 m/s.
(b)   18km/h (iv) 5m/s.
(c)    72km/h (i)20m/s.
(d)   36km/h (ii)10m/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.



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.

Ans: – Option (v).


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.

Ans: –


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

Ans: – We know that average speed is = (total distance/total time).

So, average speed= (1200/100) =12m/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.

Ans: – Here total time is 16h.

800km covered so the remaining distance is (1200-800) =400km.

If the train goes 400km in 6hrs then speed of (400/6 = 66.7 km/h.

The average speed= (total distance / total time)

= 1200/16= 75km/h.


8)A bird flies from a tree A to the tree B with the speed of 40 km h–1 and returnsto tree A from tree B with the speed of 60 km h–1. What is the average speedof the bird during this journey?

Ans: – We all know that average speed= (total distance/ total time).

So, average speed is =(40+ 60 /2)= 50 km/h.


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

Ans: –  From the image we seen that the path of R is more as compared to the other as it covers more distance from others.



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

Ans: – In first five seconds object is seen moves at constant speed at 2ms–1. But in 5 to15 second it is in rest and then from 15 to 20 seconds object moving with the constant speed 2 ms–1.

So, motion of the object is not uniform.


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

Ans: – Object A velocity is 4 times to the velocity of object B.


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

Ans: – 


4) A car accelerates from rest uniformly and attains a maximum velocity of2 ms–1 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.

Ans: –


5) A body moving with a constant speed of 10 ms–1 suddenly reverses its direction of motion at the 5th second and comes to rest in next 5 second. Draw a position-time graph of the motion to represent this situation.

Ans: – 



1)A ball is thrown straight upwards with an initial velocity 19.6 ms–1. It was caughtat 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 ms–2)

Ans: – We know, v^2 = u^2 + 2as;

Here, v=final velocity=0; u= initial velocity= 19.5;

So, S= (19.5)^2 / 2×9.8=19.40m.

And, v= u + at,

Or, 0= 19.5 + 9.8×t;

Or, t = 1.98 or 2.

Then total time required is (2 + 2)= 4s.


2)A brick is thrown vertically upwards with the velocity of 192.08 ms–1to thelabourer at the height of 9.8 m. What are its velocity and acceleration when itreaches the labourer?

Ans: – As we all know the acceleration is same its just be in negative when it goes in upper direction so the acceleration will be -9.8m/s^2.

As the labour will catch the ball so it will have no velocity in that.


3)A body starts its motion with a speed of 10 ms–1 and accelerates for 10 s with10 ms–2. What will be the distance covered by the body in 10 s?

Ans: – We know, S= ut + 1/2at^2;

Here, u=10 m/s, a=10 m/s^2; t=10s.

So, S= 10×10 + ½× 10×10^2.

Or, S= 600m.


4)A car starts from rest and covers a distance of 50 m in 10 s and 100 m in next10 s. What is the average speed of the car?

Ans: – The average speed is = total distance/ total time.

So, average speed= (50 + 100)/20=150/20=7.5m/s.

Then the average speed of the car will be 7.5m/s.



1)In circular motion the point around which body moves

(a) always remain in rest

(b) always remain in motion

(c) may or may not be in motion

(d) remain in oscillatory motion.

Ans: – Option (a).


2)In uniform circular motion

(a) speed remain constant

(b) velocity remain constant

(c) speed and velocity both remain constant

(d) neither speed nor velocity remain constant.

Ans: – Option (a).


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

(a) always uniform circular motion

(b) always uniformly accelerated circular motion

(c) may be uniform or non-uniform circular motion

(d) variable accelerated circular motion.

Ans: – Option (b).




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) at/2 ; (b) 2a t ; (c)1/2at^2 ; (d)1/2a^2t.

Ans: –  Option (a) at/2.


2) A car starts from rest with a uniform acceleration of 4 ms–2. 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

(c) 2, 6, 10, 14 (d) 4, 16, 32, 64.

Ans: – Option (c).


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

Ans: – Yes, the velocity direction is decided the direction of the acceleration.


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

Ans: – The relation between the acceleration and distance travelled by the body is ,

S=ut + 1/2at^2.

Where, s=total distance, u=initial velocity, t=time, and a=acceleration of motion.


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.

Ans: – There is two cases where the acceleration is present and in the other acceleration is absent. In the first case the speed is constant so the according to Newton first law acceleration will not be produced so in second case the direction is changing so acceleration will produce.


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:

Ans: –




(a)    Positive Positive Ball rolling down on a slop like slide or ramp.
(b)   Positive Negative Bullet fired on water.
(c)    Positive Zero Car going on road with constant speed.
(d)   Negative Positive Ball falling down from upward.
(e)   Negative Negative Compressing a spring.
(f)     Negative Zero For going backwards with constant speed.
(g)    Zero Positive When a ball thrown upward at peak velocity is zero but acceleration is 9.8m/s^2.
(h)   Zero Negative When we hit a ball towards the wall at the time of striking velocity is zero and acceleration is negative.


7) A car travelling initially at 7 ms–1 accelerates at the rate of 8.0 ms–2 for an interval of 2.0 s. What is its velocity at the end of the 2 s?

Ans: – As we all know, V=u + at,

Where, v=final velocity, u=initial velocity=7m/s.a=acceleration=8m/s^2, t=time=2s.

So, v=7 + 8×2= 25m/s.


8)A car travelling in a straight line has a velocity of 5.0 ms–1 at some instant. After 4.0 s, its velocity is 8.0 ms–1. What is its average acceleration in this time interval?

Ans: – We know, v=u + at.

Here, v=8m/s, u=5m/s, a=? T=4s.

So, average acceleration is=(8-5)/4

= 3/4=0.75m/s^2.


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 interval0 to 5.0 s, 5.0 s to 15.0 s and 0 to 20.0 s.

Ans: –


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

(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?

Ans: –


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


Table 9.13


(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?



Ans: –


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.

Ans: –


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

Ans: – we know, s= ut + ½ at^2;

Here, a= 9.8; s=19.6; u=0;

So, 19.6 = 0 + 1/2× 9.8× t^2;

Or, t=2s.

Distance travelled in n second be, S(n) = u + a/2(2n -1).

Here, n=2;

So, S(2)= 0 + 9.8/2 × ( 4-1) = 14.7m.


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).

Ans: – Initial velocity=u, final velocity=v,

Let acceleration=a, time=t.

The arithmetic means of initial and final velocity=(v+u)/2.

We know, v=u + at.

Or, t=(v-u)/a.

Here, s=ut + 1/2at^2.

Or, s=(u-v)/a ×((u+v)/2).

Average velocity=s/t=u+ v/ 2.

So, we clearly prove it that average velocity is equal to the arithmetic mean of initial 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).

Ans: –


16)A body accelerates from rest and attains a velocity of 10 ms-1 in 5 s. What isits acceleration?

Ans: – we know that, v=u + at.

Here, v=final velocity=10m/s, u= initial velocity, a=?,

T=time= 5s.

So, 10= 0 + 5a.

Or, a=10/5=2m/s^2.

Updated: February 21, 2022 — 1:11 pm

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