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