Calculate the average acceleration and average speed of a particle 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.
 A: 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.
A: 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. 
A: The integral of acceleration is velocity, and we are given a point defined as $v = 0$ when $t = 0$. We can then integrate acceleration and solve for c to get the equation of velocity.
$$v(t)=\int{a(t)dt}$$ $$v(t)=\int{72t - 4t^4}$$ $$v(t) = 36t^2 - t^4 + c$$ $$0 = 0 + c$$ $$v(t) = 36t^2 - t^4$$
Then, to find the average speed on the interval $[5, 8]$, find the velocity values for $t=5$ and $t=8$ and average those. 
$$v(5) = 275$$
$$v(8) = -1,729$$
The speed is just the absolute value of velocity, so the two speeds are $275$ and $1,729$.
Now, just average the two speeds. 
$$avg = \frac{275 + 1,729} {2}$$
$$avg = 1,002 m/s$$
To find the average acceleration, simply follow the same procedure.
$$a(5) = -140$$
$$a(8) = -1472$$
$$ avg = \frac{-140 - 1472}{2}$$
$$ avg = -806 m/s^s$$
