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/s2. Find the displacement in the next 4 sec.?
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.
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?
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?
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?
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/s2. Find the displacement of the particle in 10s?
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?
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?
$v=u+at$
$a = \frac {(80 - 20) \times 60 }{ 4} = 900 km/h$
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.
