#### Answer

(a) The speed of car A just before the collision was 21.6 m/s.
(b) Car A was speeding. Car A was exceeding the speed limit by 13.3 mph.

#### Work Step by Step

We can find the magnitude of deceleration as the cars were sliding.
$F_f = ma$
$ma = mg~\mu_k$
$a = g~\mu_k$
$a = (9.80~m/s^2)(0.65)$
$a = 6.37~m/s^2$
We can find the speed $v_2$ of the cars as they started sliding together.
$v_2^2 = 0 - 2ax$
$v_2 = \sqrt{-2ax} = \sqrt{-(2)(-6.37~m/s^2)(7.15~m)}$
$v_2 = 9.54~m/s$
We can use conservation of momentum to find car A's initial speed.
$m_1v_1 = m_2v_2$
$v_1 = \frac{m_2v_2}{m_1}$
$v_1 = \frac{(3400~kg)(9.54~m/s)}{1500~kg}$
$v_1 = 21.6~m/s$
The speed of car A just before the collision was 21.6 m/s.
(b) We can convert car A's initial speed to mph.
$v_1 = (21.6~m/s)(3600~s/h)(\frac{1~mi}{1609~m})$
$v_1 = 48.3~mph$
Car A was speeding. Car A was exceeding the speed limit by 13.3 mph.