Question

Example 12. A motorbike accelerates uniformly
from 54 km/h to 72 km/h in 2s. Calculate (i) the
acceleration and (ii) the distance covered by the
motorbike in that time.
1
11​

Answer 1

Given:

• Initial velocity ( u ) = 54km/h
• Final velocity ( v ) = 72 km/h
• Time taken to change ( t ) = 2 s

To find:

• i) Acceleration
• ii) Distance covered at that time

Solution:

Firstly finding i) acceleration

As we know that

Acceleration = v - u/t

Acceleration = 72 - 54/2

Acceleration = 18/2

Acceleration = 9 km/h²

Now , ii) Distance covered at that time

Using the third equation of motion

S = ut + 1/2 at²

S = 54(2) + 1/2 (9)(2)²

S = 108 + 9(2)

S = 108 + 18

S = 126 km

Answers:

• Acceleration = 9 km/h²
• Distance travelled at that time = 126km

Answer 2

EXPLANATION.

• GIVEN

A motorbike accelerates uniformly from

54 km/hr to 72 kn/hr.

time = 2 seconds.

To find

1) = The acceleration

2) = The distance covered by the

motorbike in that time.

According to the question,

Initial velocity of motorbike = u

=> 54 km/hr = 54 X 5/18 = 15 m/s

Final velocity if motorbike = v

=> 72 km/hr = 72 X 5/18 = 20 m/s

Time = 2 seconds.

1 ) = To find acceleration.

Apply Newton first equation of kinematics.

=> v = u + at

=> 20 = 15 + 2a

=> 5 = 2a

=> a = 5/2 = 2.5 m/s²

Therefore,

Acceleration = 2.5 m/s²

2) = To find distance covered by the

motorbike in that time.

Apply Newton second Equation of kinematics.

=> s = ut + 1/2 at²

=> s = 15 X 2 + 1/2 X 5/2 X 2 X 2

=> s = 15 X 2 + 5

=> s = 35 m

Therefore,

Distance covered = 35 m