# Forces, find net with velocity and distance

1. Oct 2, 2013

### yogoes123

1. The problem statement, all variables and given/known data

Driver brakes 1420kg car moving at 64.8km/h [W]. Car slows down and comes to stop after 729m [w]

find net force

2. Relevant equations

a=v/t f=ma

3. The attempt at a solution

t=40.5 a= 0.44

f=1420(0.44) f= 624.8

what am i doing wrong?

Last edited: Oct 2, 2013
2. Oct 2, 2013

### Staff: Mentor

Hi yogoes123. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

Your first mistake is probably in how you worked out that 40.5 secs, as you haven't shown the equation you used.

Last edited by a moderator: May 6, 2017
3. Oct 2, 2013

### yogoes123

I used v=d/t swapped it around and solved for t=d/v and that gave me 40.5s, is that wrong?

4. Oct 2, 2013

### Staff: Mentor

That equation relates time to the distance travelled at a constant speed.

5. Oct 2, 2013

### yogoes123

So it would be correct, no? I cant figure out what i did wrong.

6. Oct 2, 2013

### Staff: Mentor

You [somehow] worked out a figure of 624.8N

How to explain that halving?

7. Oct 2, 2013

### yogoes123

The TB answer is 316, my answer is 624.8, sorry I didnt make that clear.

8. Oct 2, 2013

### Staff: Mentor

As I emphasised "at a constant speed". At no time here is the vehicle travelling at a fixed speed, the vehicle is constantly slowing.

To begin over....

you need equations relating distance, initial velocity, final velocity (= 0 here) etc.

So, write down some suitable candidates.

9. Oct 2, 2013

### yogoes123

I thought it starts to slow down after the 729m. Are you saying that is it slowing down within 729m?

10. Oct 2, 2013

### Staff: Mentor

At the 729m mark, it is motionless.

Where speed is constant it has acceleration = 0 (for straight line travel).

11. Oct 2, 2013

### yogoes123

Ok I think I got this, I used Vf^2=Vi^2+2ad

Then I did a= 18^2/2/729 and i got 0.2222222 then I used F=ma and got 316, thanks so much for clearing that up.

12. Oct 2, 2013

Good.