# Motion, Finding stopping distance

1. Nov 16, 2012

### TheRedDevil18

I just want to confirm if what I am doing is right, please correct me because my textbook gives me a different answer.

1. The problem statement, all variables and given/known data
The mass of a minibus with the driver is 1.5*10^3 kg. The driver is moving at a speed of 30m.s when he notices that the traffic lights ahead have turned red. He applies a force of 9*10^3 N on the brake pedal.

A) How far must he have been from the traffic lights if he were able to stop in time?

2. Relevant equations

a=Fnet/m
s=d/t

3. The attempt at a solution

a=9*10^3/1.5*10^3
= 6 m.s^-2

using the equation a=v/t i got a time of 5 seconds

so
d = s*t
= 30*5
= 150 m but my textbook says it is 75 m

2. Nov 16, 2012

### doppelganger

The equation you are using for distance only works for a constant speed.

Try other equations like:

d=d0 + v0t +1/2at^2

that are suited to constant acceleration.

3. Nov 16, 2012

### TheRedDevil18

What is that 0 in the equation. Is it initial distance and final velocity?

4. Nov 16, 2012

### SammyS

Staff Emeritus
d0 is initial position.

v0 is initial velocity.

5. Nov 16, 2012

### TheRedDevil18

His initial velocity would be 30m.s but what is his initial position?

6. Nov 16, 2012

### doppelganger

If you find out how much distance the driver covers as he stops, you will know the minimum distance that he must be from the light to stop in time.

7. Nov 16, 2012

### TheRedDevil18

Ok thanks, i will try it out