Answer
$v_2 = 20.6 ~m/s$
Work Step by Step
We can use work and energy to solve this question. Let $W_f$ be the work done by friction as the car moves from point 1 to point 2. Then,
$E_2 = E_1 + W_f$
$\frac{1}{2}mv_2^2 = \frac{1}{2}mv_1^2 + mgy -0.23mgd$
$v_2^2 = v_1^2 + 2gy-2(0.23gd)$
$v_2 = \sqrt{v_1^2 + 2gy-2(0.23gd)}$
$v_2 = \sqrt{(1.30 ~m/s)^2 + (2)(9.80 ~m/s^2)(32 ~m) - (2)(0.23)(9.80 ~m/s^2)(45.0 ~m)}$
$v_2 = 20.6 ~m/s$