Math CentralQuandaries & Queries


Question from Tom, a student:

If you start at a stoplight and your acceleration is 16t - t^2, how far have you gone after 8 seconds?

Hi Tom,

I assume that this is a question from a calculus class and that the units of acceleration are feet per second squared. I can get you started.

The acceleration is given by $a(t) = 16t - t^2$ feet per second. You know that acceleration is the rate of change of velocity so if the velocity is given by $v(t)$ then $a(t) = v^\prime(t)$ and hence $v(t)$ is an antiderivative of $a(t).$ The antiderivative of $a(t) = 16t - t^2$ is $v(t) = 8 t^2 - \frac13 t^3 + C$ where $C$ is a constant. Since you are at a stoplight at time $t = 0$ seconds the velocity at time $t = 0$ is zero. Thus $0 = v(0) = 8 0^2 - \frac13 0^3 + C = C$ and hence $C = 0$ and

\[v(t) = 8 t^2 - \frac13 t^3.\]

You also know that velocity is the rate of change of displacement so what is the displacement function $s(t)?$


About Math Central


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
Quandaries & Queries page Home page University of Regina PIMS