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In this page we have Important Questions and answers on Motion for Class 9 physics . Hope you like them and do not forget to like , social share
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Question 1
An airplane accelerates down a runway at 3.20 m/s
^{2} for 32.8 s until is finally lifts off the ground. Determine the distance travelled before take off.
Answer
Given,
the initial velocity = 0 m /s,
acceleration = 3.20 m/s^{2},
t = 32.8 s
This is quite a basic question of 2nd equation of motion
$s=ut + \frac {1}{2}a t^2$
s = 1/2 at^{2}
s=(.5)(3.20)(32.8)^{2}= 1721.344m
Question 2
A Jeep starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the Jeep.
Answer
Use 2nd equation of motion
$s=ut + \frac {1}{2}a t^2$
110 = [0.5a × (5.21)^{2}2]
a = 110 / (5.21^{2}2 × 0.5)
a = 8.10m/s^{2}
Question 3
John is riding the Giant Drop at Canada. If John free falls for 2.6 seconds, what will be his final velocity and how far will he fall?
Answer
Using 1st equation
v=u+at
v=gt where g =acceleration due to gravity=9.8 m/s^{2}
v=9.8X 2.6=25.48 m/s
Using 2nd equation
$s=ut + \frac {1}{2}a t^2$
$s=\frac {1}{2}g t^2$
s=33.124m
Question 4
A racing car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance travelled.
Answer
$a =\frac {(vu)}{t}$
$a=\frac {(46.118.5)}{2.47}$
a=11.47m/s^{2}
Question 5
A feather is dropped on the planet other than earth which has very low acceleration due to gravity from a height of 1.40 meters. The acceleration of gravity on the other planet is 1.67 m/s
^{2}. Determine the time of feather to fall to the surface of the other planet
Answer
Here u=0, a=1.67m/s^{2} , s=1.40 , t=?
Using 2nd equation
$s=ut + \frac {1}{2}a t^2$
t= 1.29 s
Question 6
Rocketpowered sleds are used to test the human response to acceleration. If a rocket powered sled is accelerated to a speed of 444 m/s in 1.8 seconds, then what is the acceleration and what is the distance that the sled travels?
Answer
Here u=0, v=444m/s ,t=1.8
$a= \frac {vu}{t} = \frac {444}{1.8}=246.66 m/s^2$
Now
$s=ut + \frac {1}{2}a t^2$
s=399.6 m
Question 7
Honda Activita accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike.
Answer
Here v=7.10 m/s , s=35.4 m ,u=0 ,a=?
Now,
$v^2 =u^2 + 2as$
a=.712 m/s^{2}
Question 8
An Aeronautics engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate is likely to be 3 m/s
^{2}. The take off speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the runway?
Answer
Here v=65 m/s , s=? ,u=0 ,a=3m/s^{2}
Now,
$v^2 =u^2 + 2as$
distance= 704m
Question 9
A BMW car travelling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance of the car (assume uniform acceleration)
Answer
u = 22.4 m/s,v = 0, t = 2.55 s
using First equation of motion
v = u+at
0 = 22.4+aX2.55
a = 8.784 m/s^{2}
Now using third equation of motion
$v^2=u^2 +2as$
0=(22.4)^{2} 2X8.784Xs
s=28.56m
Question 10
A kangaroo is capable of jumping to a height of 2.62 m. Determine the take off speed of the kangaroo.
Answer
Here v=0 , s=2.62m ,u=? ,a=9.8 m/s^{2}
Now,
$v^2 =u^2 + 2as$
u= 7.17m/s
Question 11
If Rahul has a vertical leap of 1.29 m, then what is his take off speed and his hang time (total time to move upwards to the peak and then return to the ground)?
Answer
Here v=0 , s=1.29m ,u=? ,a=9.8 m/s^{2}
Now,
$v^2 =u^2 + 2as$
u= 5.03m/s.
Now time going Up,
v=u+at
or
t=u/a = 5.03/9.8 =.513 s
Total hang time = 2 * .513 =1.03s
Question 12
A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of the riffle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume a uniform acceleration).
Answer
Here v=521 m/s , s=.840m ,u=0 ,a=?
Now,
$v^2 =u^2 + 2as$
$a = 162 \times 10^3$ m/s^{2}
Question 13
A baseball is popped straight up into the air and has a hang time of 6.25 s. Determine the height to which the ball rises before it reaches its peak. (Hint: the time to rise to the peak is one half the total hangtime.)
Answer
Time to rise = 6.25/2= 3.125 s
s=? ,a =9.8 m/s^{2}
Now,
$s= \frac {1}{2} at^2$
s= 47.9m
Question 14
The observation deck of the tall skyscraper 370 m above the street. Determine the time required for a penny to free fall from the deck to the street below.
Answer
s=370m ,a =9.8 m/s^{2}
Now,
$s= \frac {1}{2} at^2$
t= 8.69s
Question 15
A bullet is moving at a speed of 367 m/s when it embeds into a lump of moist clay. The bullet penetrates for a distance of 0.0621 m. Determine the acceleration of the bullet while moving into the clay. (assume a uniform acceleration.)
Answer
u=367 m/s ,a =? , s=.0621 m,v=0
Now,
$v^2 =u^2 + 2as$
$a = 1.08 \times 10^6$ m/s^{2} (Negative sign shows deceleration)
Question 16
A coin is dropped into a deep well and is heard to hit the water 3.41 s after being dropped. Determine the depth of the well.
Answer
depth of well= 57.0 m
Question 17
It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar skidded to a stop with a constant acceleration of 3.90 m/s
^{2} , determine the speed of the Jaguar before it began to skid.
Answer
speed= 47.6 m/s
Question 18
A plane has a take off speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the acceleration of the plane and the time required to reach this speed.
Answer
acceleration = 2.86 m/s^{2}
time= 30.8s
Question 19
A dragster accelerates to a speed of 112 m/s over a distance of 398 m. Determine the acceleration (Assume uniform) of the dragster.
Answer
acceleration = 15.8 m/s^{2}
Question 20
With what speed in miles/ h (1 m/s= 2.23 mi/hr) must an object be thrown to reach a height of 91.5 m (equivalent to one football field)? Assume negligible air resistance.
Answer
v=0 (at maximum height),u=?,h=91.5m ,g=9.8m/s^{2}
From 3rd equation of motion
$v^2u^2=2gh$
Substituting and solving, we get
$u=\sqrt (2gh)$
=42.34m/s
=94.43 miles/hr
Question 21
A Truck starting from rest moves with a uniform acceleration of 0.2 m /s
^{2} for 2 minutes. Find
(a) the speed acquired
(b) the distance travelled.
Answer
a=0.2 m /s^{2}, t=2 min=120 s, u=0
$v=u+at$
v=24 m/s
$s=ut+ \frac {1}{2}at^2$
s=1440 m
Question 22
Match the column
Column A

Column B

Speed

Scalar

Distance

Vector

Displacement


Acceleration


Velocity


Answer
Scalar : Speed ,distance
Vector: Velocity,Acceleration ,displacement
Question 23
Match the column
Column A

Column B

22 km/h

.266 m/s

100 Km/h

6.11 m/s

10 m/min

27.77 m/s

2 km/s

83.33 m/s

5 km/min

2000m/s

Answer
22 km/h=6.11 m/s
100 Km/h=27.77 m/s
10 m/min= .266 m/s
2 km/s=2000 m/s
5 km/min=83.33 m/s
Question 24
Study the speed time graph
Find following based on the graph
a) Which paths have constant speed
b) when is the maximum speed reached
c) when is the acceleration happened
d) when is the deceleration happened
Answer
a) AB and DE
b) C
c) OA and BC
d) CD
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