 # Numericals of Motion Class 9 – solved

Last updated on July 26th, 2023 at 01:05 pm

This page consists of interesting Numericals of Motion Class 9. These Numericals & questions are based on equations of motion (suvat equations). The solved answers to the Physics Numericals for class 9 motion are provided with step-by-step explanations.

Equations or formulas required to solve the motion numerical

This is the equation of the final velocity v.
v=u + at ……………….. (1)
Now see the equation of the displacement s.
s= ut + (1/2)at2 ……….(2)
Another equation of displacement without using acceleration a.
displacement = average velocity * time = [(u+v)/2]. t ………..(3)
This is another equation of the final velocity v. Please note that this equation of final velocity doesn’t have any mention of time t.
v2=u2 + 2as………………(4)

=> s = (v2-u2)/(2a) ……………. (5)
Average velocity = (u+v)/2 ……. (6) for uniform acceleration.
Again, Average velocity = displacement/time ……… (7) for uniform acceleration.

## Numericals of Motion Class 9 | solved numerical problems in physics class 9 motion

1) A boat has an acceleration of 4 m/s2. What would the final velocity of the boat be after 10 seconds if the initial velocity of the speedboat is 2 m/s?

Solution:

a = 4 m/s2
u = 2 m/s

t = 10 s

final velocity = v

v = u + at = 2 + 4×10 = 42 m/s

2) A vehicle has an acceleration of 10 m/s2. What would the final velocity of the vehicle be after 20 seconds if the initial velocity of the vehicle is 40 m/s? What is the displacement of the vehicle during the 20 s time interval?

Solution:

a = 10 m/s2
u = 40 m/s

t = 20 s

final velocity = v

v = u + at = 40 + 10×20 = 240 m/s
final velocity = 240 m/s

Displacement during the 20 s time interval = s

s = ut + (1/2)at2

=> s = 40×20 + (½)x10x202

=> s= 800 + 2000 = 2800 m

Displacement during the 10 s time interval = 2800 m

3) A child on a  flat board starts from rest and accelerates down a snow-covered hill at 1 m/s2. How long does it take the child to reach the bottom of the hill if it is 15.0 m away?

Solution: initial velocity u = 0
acceleration a = 1 m/s2
distance = s = 15 m
time taken t = ?

s= ut + (1/2)at2

As here, u = 0, so we can write:

s= (1/2)at2

=> t2 =(2 s /a ) = (2×15)/1 = 30
time t = 5.48 sec

4. A car starts from rest and accelerates at +2.50 m/s2 for a distance of 150.0 m. It then slows down with an acceleration of –1.50 m/s2 until the velocity is +10.0 m/s. Determine the total displacement of the car.

Solution:

First part:
displacement = s = 150 m

We need to find out the final velocity(v) of this accelerating part.

v2 = u2 + 2as
=> v2 = 0 + 2×2.5×150 = 750
=> v = 27.4 m/s

This velocity becomes the initial velocity of the second part of the journey.

Second part:

u = 27.4 m/s

a = -1.5 m/s2

v= 10 m/s

v2 = u2 + 2as
102 = 27.42 + 2.(-1.5).s
=> 100 = 750 – 3s

s= 650/3 = 216.7 m

Second part displacement = 216.7 m

Considering a straight-line journey without a change of direction,
total displacement = (150 + 216.7) m = 366.7 m

5. A car accelerates uniformly from a velocity of 21.8 m/s to a velocity of 27.6 m/s. The car travels 36.5 m during this acceleration.

a) What was the acceleration of the car?

b) Determine the time interval over which this acceleration occurred.

Solution:

a)

final velocity v = 27.6 m/s

initial velocity u = 21.8 m/s

distance traveled s=36.5 m

We will use this equation: v2 = u2 + 2as

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a = ( v2 – u2) /(2s) = (27.62 – 21.82)/(2×36.5) = 3.92 m/s2

Acceleration = 3.92 m/s2

b) time interval over which this acceleration occurred = t

v = u + at

=> t = (v-u)/a = (27.6-21.8) / 3.92 = 1.48 seconds.

6) A bus is moving with the initial velocity of ‘u’ m/s. After applying the brake, its retardation is 0.5 m/s2 and it stopped after 12s. Find the initial velocity (u) and distance traveled by the bus after applying the brake.

Solution:

retardation a = –0.5 m/s2
final velocity v= 0
t = 12 s
u = ?

Using the formula: v=u + at
0 = u + (-0.5)(12)
=> 0 =u -6
=>u =6 m/s

Distance traveled:
s= ut + (1/2)at2
=> s = 6×12 + (1/2)(-0.5)122
=>s = 72 – (1/2)x72
=> s = 36 m

7 ) A car travels from rest with a constant acceleration ‘a’ for ‘t’ seconds. What is the average speed of the car for its journey if the car moves along a straight road?

Solution

The car starts from rest, so u = 0
The distance covered in time t:
s= (1/2)at2

Average speed =Total distance/Time taken
So, Average speed = (1/2)at2 / t =at/2

8) At a distance L= 400m away from the signal light, brakes are applied to a locomotive moving with a velocity, u = 54 km/h. Determine the time taken by the loco to stop. Also, determine the position of the rest of the locomotive relative to the signal light after 1 min of the application of the brakes if its acceleration a = – 0.3 m/s2

Solution

Since the locomotive moves with a constant deceleration after the application of brakes, say it comes to rest in t sec.
We know,
v = u + at
Here u = 54 km/h = 54 x 5/18 =15 m/s
v = 0 at time t
given a = -0.3 m/s2
From v = u+at we get t =(v-u)/a =(0-15)/(-0.3) = (-15)/(-0.3) =50 s

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The time taken by the loco to stop = 50 s

To find the distance traveled we can use this equation s = (v2-u2)/(2a)
here v=0
so, the distance traveled s = -u2/(2a) = – 152/[2x(-0.3)]= 375 m

[note: As we have found out the time of travel we could use this equation as well to determine the distance: s= ut + (1/2)at2 ]

Thus in 1 minute after the application of brakes, the locomotive will be at a distance = 400 – 375 = 25 m from the signal light.

Summary of this post

We covered Equations or formulas required to solve the motion numerical problems and also solved a set of motion Numericals from class 9 physics.

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