Stopping distance and time of a vehicle

  • Thread starter Lucaburra
  • Start date
  • #1
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...?
 

Answers and Replies

  • #2
1,198
5
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
 
  • #3
1,065
10
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.
 
  • #4
1,198
5
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.
 

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