# 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)at ^{2} ……….(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.**

v

v

^{2}=u^{2 }+ 2as………………(4)=> s = (

**v**)/(2a) ……………. (5)

^{2}-u^{2}**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/s^{2}. 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/s^{2}

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/s^{2}.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/s^{2}

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)at^{2}

=> s = 40×20 + (½)x10x20^{2}

=> 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/s^{2}. 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/s^{2}

distance = s = 15 m^{}time taken t = ?

s= ut + (1/2)at^{2}

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

s= (1/2)at^{2}

=> t^{2 }=(2 s /a ) = (2×15)/1 = 30

time **t = 5.48 sec**

4. A car starts from rest and accelerates at +2.50 m/scar^{2}for a distance of 150.0 m. It then slows down with an acceleration of –1.50 m/s^{2}until the velocity is +10.0 m/s. Determine the total displacement of the.

Solution:

First part:

displacement = s = 150 m

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

v^{2} = u^{2} + 2as

=> v^{2} = 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/s^{2}

v= 10 m/s

v^{2} = u^{2} + 2as

10^{2} = 27.4^{2} + 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: v^{2} = u^{2} + 2as

a = ( v^{2} – u^{2}) /(2s) = (27.6^{2} – 21.8^{2})/(2×36.5) = 3.92 m/s^{2}

Acceleration = 3.92 m/s^{2}

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/s^{2}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/s ^{2}**

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)at^{2}

=> s = 6×12 + (1/2)(-0.5)12^{2}

=>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)at^{2}

Average speed =Total distance/Time taken

So, Average speed = (1/2)at^{2} / 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/s ^{2}**

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

The time taken by the loco to stop = 50 s

To find the distance traveled we can use this equation s = (**v ^{2}-u^{2}**)/(2a)

here v=0

so, the distance traveled s =

**-u**/(2a) = – 15

^{2}^{2}/[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)at ^{2} ]**

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.**