Physics Help?

Question

Here is the problem.

A man is driving a car, hits a pedestrian, and then applies the brakes. The skid mark left on the road is 280 ft long, and the coefficiant of kinetic friction is .30. How fast was the man going before applying the brakes.

I know the answer to this problem is 50 mph according to the back of the book, but i dont know how to get to the answer

Answer 1

This is a problem in kinematics where we are concerned with movement and not the forces involved in the movement.

In your book the coeficient of friction is defined as:

u = problem acceleration/acelleration of gravity

.30 = a/32 therefor the car decellerates at 9.6 feet per second^2

from kinematics velocity squared is as follows: v^2 = 2a x s where a = 9.6 and s = 280

v^2 = 2(9.6) x 280
= 5376
extract the root v = 73 feet/sec, changed to mph = 49.77 mph

Answer 2

we ned to know the mass of the car to use the coefficient of k friction.

newton's 2nd law is force = mass X acceleration

Force of friction = mu F normal
draw an FBD showing all forces but especially the force of friction
Note that gravity = the magnitude of the normal force (to keep the car from sinking into the earth)
set signs with your direction (i usually go with friction facing left and negative)
use F = ma to find the acceleration (it should be negative since he applied the brakes and isn't accelerating away from the crime scene... unlike me)

kinematics equation: v(final)^2 = v(initial)^2 +2(accel)Delta distance

therefore 0 mph (since he's at rest) = v (initial)^2 + 2 (accel) (280 ft) and solve for v(initial)

Answer 3

if uk = 0.3 then a = g*uk = -9.6 ft/s/s
g = 32 ft/s/s

-at = v0
x(t) = 280 = 0.5 a t^2 + v0 t + 0 = -0.5 a t^2 = 4.8 t^2

t = 7.64 s
v0 = 73.32 ft/s * 1 mi/5280ft * 3600s/hr = 50 mi/hr

Answer 4

Crap, I hate when I misread the word Physics as Psychics. I am so sorry I can't help you.

Answer 5

it is all physics man. all physics