Question

What is the average speed of an object that is still at `t=0` and accelerates at a rate of `a(t) = 36-t^2` from `t in [2, 6]`?

Answer 1

`117.3ms^-1`

Explanation:

To get the average speed, we must sum the velocity and divide by the time traveled. We have a continuous function so we must integrate to sum the velocity. So average velocity is given by:

`v_(avg) = 1/(Delta t)int_(t_i)^(t_f)v(t)dt`

Now, we have been given the acceleration which means we must first integrate to find the formula for velocity, so:

`v(t) = inta(t)dt = int 36-t^2 dt=36t-1/3t^3+C`

Using the fact that the object is still at at `t=0` then:

`v(0) = 36(0)-1/3(0)^3+C=0 -> C=0`

So we have for the velocity:

`v(t) = 36t-1/3t^3`

Now we can use this to get the average velocity, in our case the initial time `t_i=2s` and the final time `t_f=6s` so `Delta t = 6-2=4s`

Now putting that into the expression for the average velocity:

`1/(Delta t)int_(t_i)^(t_f) v(t)dt = 1/4int_2^6 36t-1/3t^3dt`

`=1/4[18t^2-1/12t^4]_2^6`

`=1/4({18(6)^2-1/12(6)^4}-{18(2)^2-1/12(2)^4})`

`=1408/12~~117.3ms^-1`