## University Physics with Modern Physics (14th Edition)

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.