Question

A 1600 kg car is traveling over a hill that has a radius of curvature of 25 m. The car is slowing down as it goes over the hill. It slows down at a constant rate from a speed of 25 m/s to a speed of 10 m/s over a distance of 50 m ending at the top of the hill. The net acceleration of the car at the top of the hill is most nearly
a. 5.3 m/s/s.
b. 9.3 m/s/s.
c. 4.0 m/s/s.
d. 26 m/s/s.
e. 6.6 m/s/s.

Answer 1

Answer:

e) $a=6.6m/s^2$

Explanation:

At the top of the hill, the centripetal acceleration necessary to make the curve will be perpendicular to the tangential acceleration experimented from the reduction of speed. We calculate them both first, and calculate then the magnitude of the sum of those perpendicular vectors.

The tangential acceleration can be calculated considering the final speed at the top of the hill and the initial speed given over the distance given with the formula:

$a_t=\frac{v_f^2-v_i^2}{2d}=\frac{(10m/s)^2-(25m/s)^2}{2(50m)}=-5.25m/s^2$

The centripetal acceleration at the top of the hill will be given by the formula:

$a_c=\frac{v_f^2}{r}=\frac{(10m/s)^2}{25m}=4m/s^2$

We now calculate the magnitude of the sum of those perpendicular vectors using the Pythagoras theorem:

$a=\sqrt{(-5.25m/s^2)^2+(4m/s^2)^2}=6.6m/s^2$