Question

If the velocity of a body changes from 13 m/s to 30 m/s while undergoing constant acceleration, what's the average velocity of the body?
A. 28 m/s
B. 17 m/s
C. 21.5 m/s
D. 19.5 m/s

Answer 1

Since the acceleration is constant, the average velocity is simply the average of the initial and final velocities of the body:

$v_{avg} = \frac{v_f+v_i}{2}=\frac{30 m/s+13 m/s}{2}=21.5 m/s$

We can proof that the distance covered by the body moving at constant average velocity $v_{avg}$ is equal to the distance covered by the body moving at constant acceleration a:

- body moving at constant velocity $v_{avg}$: distance is given by

$S=v_{avg}t = \frac{v_f+v_i}{2}t$

- body moving at constant acceleration $a=\frac{v_f-v_i}{t}$: distance is given by

$S=v_i t+ \frac{1}{2}at^2 = v_i t + \frac{1}{2}\frac{v_f-v_i}{t}t^2=(v_i+\frac{1}{2}(v_f-v_i))t=\frac{v_f+v_i}{2}t$

Answer 2

The answer is 21.5 m/s