vf^2=vo^2-1/2 *.4*1400*9.8*d
solve for d, given vf=0
A 1400 kg car traveling at a speed of 34 m/s skids to a halt on wet concrete where μk = 0.50. How long are the skid marks?
5 answers
weight = m g = 1400 * 9.81
friction force = .5 * 1400 * 9.81 = 6867 N
a = F/m = 6867/1400 = -4.9 (half a g of course)
v = 34 - 4.9 t
v = 0 at stop
so
t = 6.94 seconds
d = 34 t - (1/2)(4.9) t^2
= 34(6.94) - 2.45*6.94^2
= 236 - 118
= 118 meters
friction force = .5 * 1400 * 9.81 = 6867 N
a = F/m = 6867/1400 = -4.9 (half a g of course)
v = 34 - 4.9 t
v = 0 at stop
so
t = 6.94 seconds
d = 34 t - (1/2)(4.9) t^2
= 34(6.94) - 2.45*6.94^2
= 236 - 118
= 118 meters
Thanks so much!!
Yes.
I think a should be negative Clown Bot. It is stopping.