Question

Prove that the average velocity of a particle calculated for the time interval from t1 to t2 really is the average value of the particle’s instantaneous velocity v(t) over the same time interval. Hint: you may not assume constant acceleration, so you will need to do an integral.

Answer 1

The average velocity of an object is defined as the total displacement covered by the particle divided by the total time taken in covering that displacement.

Let the position of the particle at any time t be
`\vec r(t)`.

Position of the particle at time
`t_2` =
`r(t_2)`.

Position of the particle at time
`t_1` =
`r(t_1)`.

Total displacement covered by the particle in time from
`t_2` to
`t_1` =
`\vec r(t_2)-\vec r(t_1)`

Therefore, the average velocity is given by

`\vec v_(av) = (\vec r(t_2)-\vec r(t_1))/(t_2-t_1)\ \ ...............\ (1).`

Now, the instantaneous velocity is defined as

`\vec v=(\text d\vec r)/(\text dt).`

The average value of instantaneous velocity in the same time interval is given by

`\vec v_(av) = (\int\limits^(t_2)_(t_1)\vec v\text dt)/(\int\limits^(t_2)_(t_1) \text dt)\\=(\int\limits^(t_2)_(t_1)(\text d\vec r)/(\text dt)\text dt)/(\int\limits^(t_2)_(t_1)\text dt)\\=(\vec r(t_2)-\vec r(t_1))/(t_2-t_1)\ \ ...............\ (2).`

Thus, from (1) and (2), it is clear that the average velocity of a particle calculated for the time interval from t1 to t2 really is the average value of the particle’s instantaneous velocity
`\vec v(t)` over the same time interval.