Answer
Car A hit car B with a speed of $~~12.1~m/s$
Work Step by Step
We can find the deceleration of car A:
$mg~sin~\theta-mg~\mu~cos~\theta = ma$
$a = g~sin~\theta-g~\mu~cos~\theta$
$a = (9.8~m/s^2)~sin~12.0^{\circ}-(9.8~m/s^2)~(0.60)~cos~12.0^{\circ}$
$a = -3.714~m/s^2$
Note that the negative sign shows that the acceleration vector is directed up the slope.
We can find the speed of car A after sliding $d = 24.0~m$:
$v_f^2 = v_0^2+2ad$
$v_f = \sqrt{v_0^2+2ad}$
$v_f = \sqrt{(18.0~m/s)^2+(2)(-3.714~m/s^2)(24.0~m)}$
$v_f = 12.1~m/s$
Car A hit car B with a speed of $~~12.1~m/s$
