Answer to Question #203621 in Physics for Ahmad Badsha

Question

A car is traveling north at 20.8 m/s. After 10.2 s its velocity is 13.2 m/s in the same direction. The magnitude and direction of the car's average acceleration is:

Answer 1

By definition, the acceleration is:

$$a = \dfrac{v_f - v_i}{t}$$

where $v_f = 13.2 m/s$ is the final velocity, $v_i = 20.8m/s$ is the initial velocity, and $t = 10.2s$ is the time. Obtain:

$$a = \dfrac{13.2m/s - 20.8m/s}{10.2s} \approx -0.745\space m/s^2$$

Thus, the magnitude of the acceleration is $0.745 m/s^2$ and '-' sign denotes that the direction is the opposite to the direction of motion, i.e. south.

Answer. 0.745 m/s^2 south.