Question

Two cars start from rest and begin accelerating, one at 10m/s210m/s2 and the other at 12m/s212m/s2. How long will it take for the second car to be 100m100m ahead of the first car? A :

Answer 1

The time taken by the second car to be 100 m ahead of the first car is 10 seconds.

Step-by-step explanation:

Given that,

Initial speeds of both car is 0 as they starts at rest.

The acceleration of car 1,
`a_1=10\ m/s^2`

The acceleration of car 2,
`a_2=12\ m/s^2`

The distance covered by car 1 is :

`d_1=ut+(1)/(2)a_1t^2`

`d_1=(1)/(2)at^2`

`d_1=(1)/(2)* 10t^2`

`d_1=5t^2`.......(1)

The distance covered by car 2 is :

`d_2=ut+(1)/(2)a_2t^2`

`d_2=(1)/(2)a_2t^2`

`d_2=(1)/(2)* 12t^2`

`d_2=6t^2`......(2)

It is mentioned that the second car to be 100 m ahead of the first car, so,

`6t^2-5t^2=100`

`t^2=100`

t = 10 seconds

So, the time taken by the second car to be 100 m ahead of the first car is 10 seconds. Hence, this is the required solution.