Question

Question #6cde5

Answer 1

The distance is `=16m`

Explanation:

The speed of the car is

`u=40kmh^-1=40*1000/3600=11.11ms^-1`

The distance is `s=4m`

We apply the equation of motion

`v^2=u^2+2as`

`0=11.11^2+2a*4`

`a=-11.11^2/8=-15.43ms^-2`

The new speed is

`u_1=80kmh^-1=80*1000/3600=22.22ms^-1`

We apply the same equation of motion

`0=22.22^2-2*15.43*s_1`

`s_1=22.22^2/30.86=16m`

We doubled the speed and the distance to stop is squared.