Police physics?

Question

Police find skid marks 28 m long on a highway showing where a car made an emergency stop. Assuming that the acceleration was -10 m/s^2 how fast was the car going?
Please show work.

Was the car exceding the 8o km/h speed limit?

Was the car exceding the 80 km/h speed limit?

Answer 1

V^2 = Vo^2 + 2ax

0 = Vo^2 + 2ax

Vo = sqrt(-2ax)

Vo = sqrt(2*10*28)

Vo = 23.7 m/s

Answer 2

Sonuds like a homework problem :) anyways not gonna solve it that would require too much time and effort, how about pointing you in the right direction ^.^ first draw a diagram label the knowns acc. initial and distance final. when u say assume accel is -10m/s^2 im assuming it is constant until the car stops also i assume no reverse action going on. will with all this great info sonuds like a one dimensional question. so onto the knowns we know the final velocity is zero, we know the acc is -10m/s^2 we know the change in distance is 28m and assume the acc is constant we need to know initial velocity but dont know time. now onto the equation to use we dont know time, pulling up a wiki search reveals that there is an equation that doesnt use time. so here it goes (Velocity final)^2= (Velocity initial)^2 +2*a(Change in Distance) plug in the knowns (0)^2=x^2 + -20m/s^2(28m) or
x=squareRoot(560m^2/s^2) which equals 23.66m/s initial velocity change that to km/h and you obtian an answer of 85.191 km/h. Hmmm so i hope this helps. hope you figure it out someday.

Answer 3

How could the police arrived from the assumption that the car is deccelerating at -10 m/s^2? I am just curious? then I will answer your question.

Answer 4

s=.5*a*t^2 s=28m a=10m/sec^2 solve for t. v = at.