Stopping distance help!

I'm having troubles with this question:

A car is travelling at a constant velocity of

15ms^-1

. The driver sees a child stepping onto the road,

50m

ahead. The driver takes

0.50s

to react before applying the brakes. The brakes decelerate the car at

6.0ms^-2

. Calculate how far the car stops from where the child stepped onto the road.

Working out :
Ok so

v=15ms^-1

Distance from child =

50m

So the time it would take to reach the girl is

\frac{distance}{speed}

which is

3.3 seconds

.
As the reaction time is

0.5s

I did

3.3-0.5=2.8

.
Then I just did

6 x 2.8 = 16.8m?

. No idea if I am doing the right thing but any help would be appreciated :smile:

Reply 1

Stonebridge

13

The total stopping distance is made up of
a - the distance the car travels during the reaction time
b - the distance during deceleration

a- constant velocity for 0.5s at 15m/s
b- use a suitable SUVAT equation. you have initial u and final v velocity as well as deceleration a. - find s

Reply 2

Super199

OP

20

Original post by Stonebridge

The total stopping distance is made up of
a - the distance the car travels during the reaction time
b - the distance during deceleration

a- constant velocity for 0.5s at 15m/s
b- use a suitable SUVAT equation. you have initial u and final v velocity as well as deceleration a. - find s