# A car exhibits a constant acceleration of $0.300 m/s^2$ parallel to the roadway. The car passes over a rise in the roadway such that the top of the rise is shaped like a circle of radius $500\; m$. At the moment the car is at the top of the rise, its velocity vector is horizontal and has a magnitude of $6.00\; m/s$. What is the direction of the total acceleration vector for the car at this instant? ￼

$\begin{array}{1 1}-13.5 ^{\large\circ}\\-23.7 ^{\large\circ}\\-26.8 ^{\large\circ}\\-11.3 ^{\large\circ}\end{array}$

Answer : $-13.5 ^{\large\circ}$
With v = 6.00 m/s and r = 500 m, we find that the radial acceleration is
$a_r=-\large\frac{v^2}{r}$