physics 4000 pound car is travelling at 40 mi/h...?

Question

a 4000 lb car is travelling at 40 mi/h when its brakes are applied and it skids to rest. The skidding tires experience a friction force about .80 times the weight of the car. How far does the car go before stopping? Take the motion to be along a straight line.

since i'm trying to find its distance. Do i use d = .5 at^2?

Answer 1

d = .5 at^2? Well you don't know the time either do you?

I suspect you need help to find the acceleration.
F = m*a -- this is Newton's 2nd
The force is 0.8*4000 lbs. What's the mass? Well
W = m*g
m = W/g = 4000 lbs / (32 ft/s^2)

Back to Newton's 2nd
F = 0.8*4000 lbs = [4000 lbs / (32 ft/s^2)]*a
a = 0.8*4000 lbs*32 ft/s^2 / 4000 lbs
a = 0.8*32 ft/s^3

You've got 4 equations of motion right? One of them should involve what you know, plus what you need. I helped you with the acceleration. What else do you know? Vo, original velocity, was 40 mph, Vf, final velocity, was 0. And then you need to find distance.

OK, how about
Vf^2 = Vo^2 + 2*a*d
(You'll need to convert 40 mph to ft/s.
60 mph = 88 ft/s)