Question

What is the average speed of an object that is still at `t=0` and accelerates at a rate of `a(t) =1-t` from `t in [0, 1]`?

Answer 1

`a(t) = 1- t`

`implies v(t) = t- t^2/2 + alpha`

`v(0) = 0 implies alpha = 0`

So: `v(t) = t- t^2/2`

`implies x(t) = t^2/2- t^3/6 + alpha`

`x(0) = 0 implies alpha = 0`

`implies x(t) = t^2/2- t^3/6`

`v_(ave)|_0^1 = (int_0^1 v(t) dt)/(1 - 0)`

`= (x(1) - x(0))/(1) = 1/3 " m/s"`

Now here's where you have to be careful. That's the average velocity (`v`), not speed (`s`). And `s(t) = abs (v(t))`. So negative velocities count as positive speeds. We need therefore to be sure of the direction of travel of the object.

We note:

`v(t) = t- t^2/2 = t(1 - t/2)`.

So the particle stops at `t = 0, 2`

And for `t in [0,1], v(t) ge 0 implies s = abs v = v`

So, in this case, the average speed is also the average velocity.