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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Fill in the blanks
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
Crossword Puzzle
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
Answer
Across
Velocity, Vector, speed, acceleration, Odometer Down
Retardation, Scalar, Zero, Speedometer
Short Answer type
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? Answer
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? Answer
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/s2. Find the displacement in the next 4 sec.? Answer
u=5 m/s ,a=2.5m/s2, 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 /s2. Find how far the train will go before it is brought to rest. Answer
u=60 km/hr=16.66 m/s ,a=-0.5 m /s2,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? Answer
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 /s2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there? Answer
u=10 m/s , v=0 , a=-10 m /s2, 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? Answer
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? Answer
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
Long answer type
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. Answer
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. Answer
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/s2. Find the displacement of the particle in 10s? Answer
u=2 m/s ,a=0.2m/s2 ,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/s2.
(a) Find the time it will take to come to rest.
(b) Find the distance covered by it before coming to rest? Answer
u=12 m/s ,a =-0.5m/s2 ,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? Answer
$v=u+at$
$a = \frac {(80 - 20) \times 60 }{ 4} = 900 km/h$
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? Answer
Distance = $2 \pi r = 100 \times 3.14 $
Average speed = distance /time = 1.309 m/sec
Displacement =0
Average velocity=0
Summary
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.
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