Question

A car exhibits a constant acceleration of 0.300 m/s2 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?

Answer 1

Let's first find the magnitude of the centripetal acceleration:

$$a_c=\dfrac{v^2}{R}=\dfrac{(6.0\ \dfrac{m}{s})^2}{500\ m}=0.072\ \dfrac{m}{s^2}.$$

We can find the magnitude of the total acceleration from the Pythagorean theorem:

$$a=\sqrt{a_t^2+a_c^2}=\sqrt{(0.3\ \dfrac{m}{s^2})^2+(0.072\ \dfrac{m}{s^2})^2}=0.309\ \dfrac{m}{s^2}.$$

We can find the direction of the total acceleration vector for the car at this instant from the geometry:

$$\theta=tan^{-1}(\dfrac{a_c}{a_t})=tan^{-1}(\dfrac{0.072\ \dfrac{m}{s^2}}{0.3\ \dfrac{m}{s^2}})=-13.5^{\circ}.$$