# Stopping distance and time of a vehicle

## Homework Statement

The coefficient of friction is .48. The brakes are applied to a 1.05 x10^3 car travelling at 95km/hr. How far does the car move before it stops ? How long does it take for the car to stop?

## Homework Equations

Fg = Fn, Fa= u x Fn, a=Fa/m, d=Vf^2-Vi^2/2a, t=d/v

## The Attempt at a Solution

Fg = Fn =-9.81m/s^2 x 1050 kg = 10300 N
Fa = u x Fn = .48 x 10300 N = 4944.2 N
a = F/m = 4944.2 N / 1050 Kg = 4.71 m/s
d = (Vf^2 - Vi^2)/2a =
95 km/hr converts to 26.4 m/s
so if Vf = 0 then -26.4 ^2 = 697.0 m/s^2
697.0 m/s^2 / 2 x 4.71 m/s = 74.0 m
then t = d/v so time is 74m/ 26.4 m/s = 2.80 secs

Im just not convinced I have done this correct...?

You have computed the distance correctly but your time is wrong. You should divide by the average velocity or use:

dist = V0*t + .5*a*t^2, where a<0

t=v/a
ma=mgμ

t=v/gμ
Thus time taken for a body to stop depends only on initial velocity, coefficient of friction.
Mass is not in the equation.

Going back to my post, the average velocity is half the initial velocity which doubles your answer for the time. The time also is the root of:

dist = V0*t + .5*a*t^2, where a<0 or -4.71 m/sec^2 which you computed above.