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

Answer:

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

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 :

$d_1=ut+\dfrac{1}{2}a_1t^2$

$d_1=\dfrac{1}{2}at^2$

$d_1=\dfrac{1}{2}\times 10t^2$

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

The distance covered by car 2 is :

$d_2=ut+\dfrac{1}{2}a_2t^2$

$d_2=\dfrac{1}{2}a_2t^2$

$d_2=\dfrac{1}{2}\times 12t^2$

$d_2=6t^2$......(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.