Calculate the average acceleration and average speed of a particle

Question

A particle has zero velocity initially (i.e., at time $t=0$) and its acceleration at $t$ seconds is $a(t)=72t−4t^3$ meters-per-second per second. During the time interval $[5,8]$ seconds find the average acceleration and the average speed of the particle. I got the average acceleration by taking the integral and then plugging that into $f(8)-f(5)/8-5$ but I have tried working through the average speed and I can't find the answer for that one.

Answer 1

Acceleration is the derivative of the velocity, $a(t) = \frac{dv}{dt}$, so you can find an expression for the velocity by integrating and then do the same averaging procedure that you did for the acceleration.

Answer 2

Find the velocity at time $t$ by integrating $72t-4t^3$. We get $36t^2-t^4+C$. But the velocity at $t=0$ is $0$, and therefore $C=0$. So the velocity at time $t$ is given by $v(t)=36t^2-t^4$.

We are asked for the average speed from $t=5$ to $t=8$, not for the average velocity (trick question, some of us are nasty). Note that the velocity is negative when $t\gt 6$. So for the average speed, we find $$\int_5^6 (36t^2-t^4)\,dt+\int_6^8 (t^4-6t^2)\,dt,$$ and divide the result by the total elapsed time.