Last updated on December 28th, 2022 at 08:11 am

This page consists of interesting **Motion Numericals of class 9 physics**. 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)## Motion Numericals Class 9 Physics – questions & answers [solved]

**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/s ^{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 **car

**.**

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.

**Summary** of this post