Question

At an accident scene on a level road, investigators measure a car's skid marks to be 84.9m long. The accident occurred on a road with a coefficient of friction estimated to be .34. What was the speed of the car as the driver slammed and locked the brakes?

Answer 1

1/2 x m x v^2 = Displacement x m x g(gravity) x u(coefficient of friction)
> the masses cancel out
> .5v^2 = Dgu
>plug n' chug: v^2 = (2)(84.9m)(p.81m/s^2)(.34)
>v^2 = 565.77
v = 565.77^.5
v = 23.79m/s
I figured it out. Thought i might as well share!

Answer 2

\[F_k=\mu_k~\vec n~:~F=ma\]
\[F_k=\mu_k~mg~:~F=ma\]
\[\mu_k~mg=ma\]
\[\mu_k~g=a\]
\[v_f^2=v_o^2+2a\Delta x\]
\[v_f^2=v_o^2+2\mu_k~g~\Delta x\]
seems about fair to me

Answer 3

solve for Vo

Answer 4

and since acceleration is slowing , the negative aint an issue
\[v_f^2=v_o^2+2\mu_k~g~\Delta x\]
\[v_f^2-2\mu_k~g~\Delta x=v_o^2\]
\[\sqrt{v_f^2-2\mu_k~g~\Delta x}=v_o\]
\[\sqrt{2(.34)(9.8)(84.9)}=v_o=23.79\]

Answer 5

yeah, we get to the same results

Answer 6

YAY!