Question

What is the average speed of an object that is still at `t=0` and accelerates at a rate of `a(t) =2t^2-2t-2` from `t in [2, 3]`?

Answer 1

`-0.5"m"/"s"`

Explanation:

Since average speed is given by the equation

`v_("av-x") = (Deltax)/(Deltat)`

We need to find the positions of the car at times `t = 2` and `t = 3`.

We can derive first the velocity equation from the acceleration, and then the position from that velocity equation, via integration, and we'll assume the initial position and velocity are `0`:

`v_x(t) = int_0^t a_xdt = 2/3t^3 - t^2 -2t`

`x(t) = int_0^t v_xdt = 1/6t^4 - 1/3t^3 - t^2`

Now we have our position equation, `x(t)`, so let's plug in the values `2` and `3` for `t` to find the position at those times:

`x(2) = 1/6(2)^4 - 1/3(2)^3 - (2)^2 = -4.0"m"`

`x(3) = 1/6(3)^4 - 1/3(3)^3 - (3)^2 = -4.5"m"`

So, the average velocity of the car on the interval `t in [2,3]` is

`v_("av-x") = ((-4.5 "m") - (-4.0 "m"))/(3"s" - 2"s") = color(red)(-0.5"m"/"s"`