Solved Physics Class 9 Motion Numericals Examples

In this post, you will find solved examples of numericals from chapter 8 of Motion given in the NCERT science book of class 9^th.

You must learn all formulas for solving numericals.

Example -1 – An object travels 16m in 4 s and then another 16 m in 2 s. What is the average speed of the object?

Given values 

Distances=  16 m+ 16m =32 m

Time = 4s +2 s= 6s

Average speed =?

Solution 

`\text (Average    speed)` 

`=  \text (Total    distance    travelled)/(\text (total    time    taken)) `

`=  32/6`

 

 = 5.33 m/s

 So the average speed of the object is 5.33 m/s

Example -2- The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of the trip. If the trip took 8 h. calculate the average speed of the car in km/h and m/s.

Given values

Distance(s) = 2400 – 2000 =400 km

Time (t)= 8h

Average speed(v) =?

Solution 

`v_(av)=  s/t` 

`=  400/8` 

= 50  km / h

Now we will find the average speed in m/s 

1 km = 1000 m

1 h= 60 × 60 =3600 s

So,

 `(50×1000)/3600`

  `50000/3600` =13.9 m/s

Example -3-Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha

Given values 

Distance (s) =180 m

Displacement(s) =0 

Time(t) = 1 minute = 60 s

Average speed (v) =?

Average velocity =?

Solution 

`\text (Average speed) =s/t` 

`=180/60`  =3 m/s 

The average speed of Usha is 3 m/s.

`\text (Average velocity)` 

`=(displacement )/(time )`

 

`=0/60`  =0 m/s 

So, the  average velocity of Usha is 0 m/s

Example -4 - Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m/s in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m/s in the next 5 s. calculate the acceleration of the bicycle in both cases.

Given values

For the 1^st situation

Initial velocity (u) = 0

Final Velocity (v)= 6m/s

Time (t)= 30 s

Acceleration (a)=?

For 2^nd situation

Initial velocity (u) =  6 m/s

Final Velocity (v)=  4 m/s

Time (t)= 5 s

Acceleration (a)=?

Solution 

 For the 1st situation 

`a=(v-u)/t` 

`a=(6 -0 )/30` 

`a=(6  )/30 =0.2` 

`a=0.2 m/s^2`

For 2nd situation 

`a=(v-u)/t` 

`a=(4 -6 )/5` 

`a=(-2 )/5=-0.4`  

`a=-0.4 m/s^2`

Example -5- A train starting from rest attains a velocity of 72km/h in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and (ii) the distance traveled by the train for attaining this velocity.

Given values

Initial velocity (u) =0

Final velocity (v) =72 km/h = `72000/3600`  = 20 m/s

Time (t) =5 minute = 5 ×60 = 300 s

Acceleration(a) =?

Distance(s) =?

Solution 

`a=(v-u)/t` 

`a=(20-0)/300` 

`a=1/15` 

The acceleration of the train is 1/15`m/s^2`

`s=  v^2/(2a)` 

`s=  20^2/(2×1/15)` 

`s=  (400 ×15)/2` 

`s=  6000/( 2)` =3000 

S= 3000 m =3 km

The distance covered by train is 3 km

Example – 6 – A car accelerates uniformly from 18 km/h to 36 km /h in 5 s. calculate (i) the acceleration and (ii) the distance covered by the car in that time.

Given values

Initial velocity (u) = 18 km/h = 5 m/s

Final velocity (v) = 36 km/h = 10 m/s

Time (t) = 5 s

Acceleration (a) =?

Distance (s)= ?

Solution 

`a=(v-u)/t` 

`a=(10-5 )/5`

 

`a=1 m/s^2` 

The acceleration of the car is 1 `m/s^2`

`s=ut +  1/2  at^2` 

`s=5× 5 +  1/2  ×1×5^2` 

`s=25+12.5`  

 `s=37.5 m`

  

Example – 7 – The brakes applied to a car produce an acceleration of 6 m/s^2 in the opposite direction to the motion. If the car takes 2 seconds to stop after the application of brakes. Calculate the distance it travels during this time.

Given values

Acceleration (a) =  - 6 m/s^2

Time (t) = 2 s

Final velocity (v)= 0

Distance (s)= ?

Solution 

`v=u+at` 

`0=u+(-6)×2` 

`u=12 m/s` 

`s=ut +  1/2  at^2` 

`s=12× 2 +  1/2  ×(-6)×2^2` 

`s=24-12`  

`s=12 m` 

so, these numericals for class 9 Motion have been taken from the NCERT science book.