- #1

Lucaburra

- 3

- 0

## 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...?