# 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]?

Apr 28, 2017

$a \left(t\right) = 1 - t$

$\implies v \left(t\right) = t - {t}^{2} / 2 + \alpha$

$v \left(0\right) = 0 \implies \alpha = 0$

So: $v \left(t\right) = t - {t}^{2} / 2$

$\implies x \left(t\right) = {t}^{2} / 2 - {t}^{3} / 6 + \alpha$

$x \left(0\right) = 0 \implies \alpha = 0$

$\implies x \left(t\right) = {t}^{2} / 2 - {t}^{3} / 6$

${v}_{a v e} {|}_{0}^{1} = \frac{{\int}_{0}^{1} v \left(t\right) \mathrm{dt}}{1 - 0}$

$= \frac{x \left(1\right) - x \left(0\right)}{1} = \frac{1}{3} \text{ m/s}$

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

We note:

$v \left(t\right) = t - {t}^{2} / 2 = t \left(1 - \frac{t}{2}\right)$.

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

And for $t \in \left[0 , 1\right] , v \left(t\right) \ge 0 \implies s = \left\mid v \right\mid = v$

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