Question

The brakes applied by a car produces an acceleration of $6m/s^2$ in opposite direction of motion. If the car takes 2 sec to stop after the application of brakes. Calculate distance during this time.

Answer 1

Step 1, Given data

Acceleration $a=-6m/s^2$

Time $t=2s$

Final velocity $v=0m/s$

Let initial velocity be $u$

Step 2, Finding the distance

From the third equation of motion,

$v^2-u^2=2as$

Putting values

$0^2-u^2=2(-6)s$

$12s=u^2$-----------------------(1)

Let distance be $s$ From the first equation. of the motion we get'

$v=u+at$

So, $0=u+(-6)\left(2\right)$

So, $u=12m/s$

put the value of u in equation (1)

We get

$12\times s={12}^2⟹s=12m$

Hence the distance traveled is 12 m