In this page we have *Worksheet on Motion for Class 9 physics* .
(a) Fill in the blanks

(b) crossword puzzle

(c) Short questions

(d) Long answer questions

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(b) crossword puzzle

(c) Short questions

(d) Long answer questions

Hope you like them and do not forget to like , social share and comment at the end of the page.

- A car starts at rest and travel a distance 20 m in 1 sec .The car average speed is ___________
- Motion of earth around sun is an example of __________ motion
- Tractor moving with 18 km/h is _______ then car moving with 1500 m/min
- The motion of a free falling body is an example of __________ motion
- The curved speed time graph represent ________ accelerated motion
- Distance is a _______quantity while displacement is a _____________ quantity

1) 20 m/s

2) Uniform circular motion

3) Lesser

4) Uniform acceleration motion

5) Not uniform

6) Scalar , vector

*Across*

2. Physical quantity obtained by dividing displacement with time taken

3. A quantity having both magnitude and direction

4. This remains constant in uniform circular motion

6. It is the slope of speed -time graph

8. This measure the distance travelled by the car

*Down*

1. This is the other name for negative acceleration

4. The speed is said to be a

5. This is the acceleration of the body with uniform velocity

7. This measure the speed of the car

*Across*

Velocity, Vector, speed, acceleration, Odometer

*Down*
Retardation, Scalar, Zero, Speedometer

**Question 1**

A particle is moving up an inclined plane. Its velocity changes from 15m/s to 10m/s in two seconds. What is its acceleration?

u=15 m/s , v=10m/s ,t=2 sec , a=?

$a = \frac {v-u}{t} =-2.5 m/s^2$

**Question 2**

The velocity changes from 45m/s to 60m/s in Three seconds. What is its acceleration?

u=45 m/s , v=60m/s ,t=3 sec , a=?

$a = \frac {v-u}{t} =5 m/s^2$

**Question 3**

A body covered a distance of z metre along a semicircular path. Calculate the magnitude of displacement of the body, and the ratio of distance to displacement?

Let r be the radius of semicircular path

$z= \pi r$

or

$r= \frac {z}{ \pi}$

Displacement = Diameter = 2r = $\frac {2z}{ \pi}$

Now ratio of distance to displacement

$= \frac {z}{\frac {2z}{ \pi}} = \frac {\pi}{2}$

**Question 4**

A particle moving with an initial velocity of 5m/s is subjected to a uniform acceleration of 2.5m/s^{2}. Find the displacement in the next 4 sec.?

u=5 m/s ,a=2.5m/s^{2}, t=4 s ,s=?

$s= ut + \frac {1}{2}at^2$

s=40 m

**Question 5**

A train is travelling at a speed of 60 km/ h. 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.

u=60 km/hr=16.66 m/s ,a=-0.5 m /s^{2},v=0,s=?

$v^2 = u^2 + 2as$

s=277.55m

**Question 6**

A Truck covers 30km at a uniform speed of 30km/hr. what should be its speed for the next 90km if the average speed for the entire journey is 60km/h?

Total distance travelled = 30 +90 =120 km

Average speed=60 km/hr

Hence total time taken = 120/60 =2 hr

Now when travelling with 30 km/hr for 30 km. Time taken = 1 hr

So time left = 2-1 =1 hr.

So Truck has to taken 90 km in 1 hour in order to have average speed of 60 km/hr

Hence the speed should be = 90/1= 90 km/hr for the next 90 km

**Question 7**

A stone is thrown in a vertically upward direction with a velocity of 10 m/s. 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?

u=10 m/s , v=0 , a=-10 m /s^{2}, s=? , t=?

$v^2 = u^2 + 2as$

s=5 m

$v = u + at$

or t = 1 sec

**Question 8**

A person goes to market, makes purchases and comes back at a constant slower speed. Draw displacement- time and velocity time graphs of the person?

Displacement Time graph is given as below

velocity Time graph is given as below

**Question 9**

John runs for 10 min. at a uniform speed 9km/h. At what speed should he run for the next 20 min. so that the average speed comes 12km/hr?

Total time = 10 +20 =30 min

Average speed=12 km/hr

Hence total distance = 6 km

Now when travelling with 9 km/hr for 10 min. Distance = 1.5 Km

So Distance left = 6-1.5 =4.5 km.

So John has to taken 4.5 km in 20 min in order to have average speed of 12 km/hr

Hence the speed should be = (4.5/20 )*60= 13.5 km/hr for the next 20 min

**Question 10**

A particle was at rest from 1 a.m. It moved at a uniform speed 50km/hr from 1.30 a.m. to 2:00 a.m. Find the average speed between

(a) 1.00 a.m. and 2.00 a.m.

(b) 1.15 a.m. and 2.00 a.m.

(c) 1.30 a.m. and 2.00 a.m.

Distance travelled between 1.30 a.m. to 2:00 a.m = 50 /2 =25 km

a. average speed between 1.00 a.m. and 2.00 a.m = Distance /time = 25/1=25 km/hr

b. average speed between 1.15 a.m. and 2.00 a.m = (25/45)*60 =33.22 km/hr

c. average speed between 1.30 a.m. and 2.00 a.m = (25/30) *60 =50km/hr

**Question 11**

An object moves along a circular path of diameter 14cm with constant speed. If it takes 2 min. to move from a point on the path to the diametrically opposite point. Find

(a) The distance covered by the object

(b) The speed

(c) The displacement

(d) average velocity.

Distance = $\pi r =\frac {22}{7} \times 7 = 22 cm$

Speed = Distance/time = 22 /120 =.183 cm/sec

Displacement = diameter= 14 cm

Average velocity = Displacement /time = 14/120 = .116 cm/sec

**Question 12**

A particle with a velocity of 2m/s a t=0 moves along a straight line with a constant acceleration of 0.2m/s^{2}. Find the displacement of the particle in 10s?

u=2 m/s ,a=0.2m/s^{2} ,t=10 s ,s=?

$s = ut + \frac {1}{2}at^2$

s=30 m

**Question 13**

A particle is pushed along a horizontal surface in such a way that it starts with a velocity of 12m/s. Its velocity decreases at a uniform rate of 0.5m/s^{2}.

(a) Find the time it will take to come to rest.

(b) Find the distance covered by it before coming to rest?

u=12 m/s ,a =-0.5m/s^{2} ,v=0

$v=u+ at$

t=24 sec

$s = ut + \frac {1}{2}at^2$

s=432 m

**Question 14**

A train accelerated from 20km/hr to 80km/hr in 4 minutes. How much distance does it cover in this period? Assume that the tracks are straight?

$v=u+at$

$a = \frac {(80 - 20) \times 60 }{ 4} = 900 km/h$

Now

$v^2 = u^2 + 2 a s$

s = 3.33 km

**Question 15**

A cyclist moving on a circular track of radius 50m completes one revolution in 4 minutes. What is his

(a) average speed

(b) average velocity in one full revolution?

Distance = $2 \pi r = 100 \times 3.14 $

Average speed = distance /time = 1.309 m/sec

Displacement =0

Average velocity=0

This Worksheet on Motion for Class 9 physics with answers is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.

**Notes****Assignments**

Class 9 Maths Class 9 Science