Question

A car travelling at 108km/hr has its speed reduced to 36km/hr while travelling a distance 200m.Its acceleration and the time taken are

Answer 1

Hey there !

Solution :

We know that, 

Acceleration is defined as change in velocity in a particular period of time. It can be given as an equation. It is:

Acceleration = Change in Velocity / Time Taken

Change in Velocity = V - U

Here V represents Final Velocity, U represents Initial Velocity,

Therefore Change in velocity = 108 km/hr - 36 km/hr

=> Change in Velocity = 72 km/hr

We dont know the Time taken. But it can be found out by using this formula:

Time Taken = Total Distance / Change In velocity

=> Time Taken = 200 m / 72 km/hr

=> Time Taken = 200 m / 72 * 1000 m / 3600 sec

=> Time Taken = 200 / 72000 / 3600

=> Time Taken = 200 / 20 

=> Time Taken = 10 seconds

Therefore Acceleration = 20 m / sec / 10 sec

=> Acceleration = 2 m / s²

Therefore Acceleration = 2 m / s² and Time taken = 10 sec.

Hope it helped :-)

Answer 2

Given:

Initial speed of the car, u = 108 km/ hr

final speed of the car , v = 36km/ hr

travelling distance , S = 200m

To Find:

Find the acceleration and time taken in the journey.

Solution:

v = 36 km /hr = 10 m/s

u = 108 km/hr = 30m/s

1. acceleration:

We will apply the equation of motion,

v² = u² + 2as

putting the values,

10² = 30² + 2a(200)

100 = 900 + 400a

-400a = 900 - 100

400a = -800

a =-2m/s²

Hence there will be acceleration of -2 m/s² or we can say that there will be retardation of 2 m/s²

2. Time:

we will apply first equation of motion,

v = u + at

putting the values

10 = 30 -2t

2t = 20

t = 10 sec

Hence the time taken is 10 sec