# Homework Help: Newton's Laws: Distance of a car accelerating, then slamming on the breaks.

1. Mar 11, 2012

### jillime

A car which is originally at rest accelerates down a road with a net force of 32000 N acting on a 1980 kg car. At t = 12 seconds the driver slams on the brakes to avoid hitting a deer. Calculate the distance traveled by the car.
Coefficient of kinetic friction for the rubber on pavement is assumed to be .8

F⃗ net=ΣF⃗ =ma⃗ (a = F/m)
fk=μkN

v=v0+at
x=x0+v0t+(1/2)at2
v2=v20+2aΔx
a = Δv/Δt
blablahblah

Anyone who can tell me how to start this problem?
FBD shows force of friction going -X, normal force y, weight -y.

32000 = 1980a
a = 16.16 m/s2

v = 0 + 16.16x12
v = 193.92 m/s

I obviously have to do a second xvat for t=12 seconds and onward. These net force/ newtons laws problems are killing me because they involve a lot of creativity with choosing working equations, which I don't have.

2. Mar 11, 2012

### MotoPayton

We know the force from braking is equal to Fn*uk.

Force braking is (1980)(9.8)*.8

acceleration from braking is therefore 9.8*.8 in the negative x direction from F=ma

d1+d2=total distance

d1:

use d1=.5at^2

d2:

use d2=v0t+.5at^2

where a is negative for the d2 equation and I found it above