Answer
Just before the collision, the SUV was traveling with a speed of 12.0 m/s and the sedan was traveling with a speed of 21.0 m/s.
Work Step by Step
We can find the distance $d$ that the cars slid after the collision.
$d = \sqrt{(5.39~m)^2+(6.43~m)^2}$
$d = 8.39~m$
We can find the magnitude of deceleration as the cars were sliding.
$F_f = ma$
$ma = mg~\mu_k$
$a = g~\mu_k$
$a = (9.80~m/s^2)(0.75)$
$a = 7.35~m/s^2$
We can find the speed $v_2$ of the cars as they started sliding together.
$v_2^2 = 0 - 2ad$
$v_2 = \sqrt{-2ad} = \sqrt{-(2)(-7.35~m/s^2)(8.39~m)}$
$v_2 = 11.1~m/s$
We can find the angle $\theta$ south of west.
$tan(\theta) = \frac{6.43}{5.39}$
$\theta = arctan(\frac{6.43}{5.39})$
$\theta = 50.0^{\circ}$
We can use conservation of momentum to find the SUV's initial speed.
$m_{SUV}v_x = m_2v_2~cos(\theta)$
$v_x = \frac{m_2v_2~cos(\theta)}{m_{SUV}}$
$v_x = \frac{(3700~kg)(11.1~m/s)~cos(50.0^{\circ})}{2200~kg}$
$v_x = 12.0~m/s$
We can use conservation of momentum to find the sedan's initial speed.
$m_{sedan}v_y = m_2v_2~sin(\theta)$
$v_y = \frac{m_2v_2~sin(\theta)}{m_{sedan}}$
$v_y = \frac{(3700~kg)(11.1~m/s)~sin(50.0^{\circ})}{1500~kg}$
$v_y = 21.0~m/s$
Just before the collision, the SUV was traveling with a speed of 12.0 m/s and the sedan was traveling with a speed of 21.0 m/s.
