What is the average speed of an object that is moving at `18 m/s` at `t=0` and accelerates at a rate of `a(t) =t-8` on `t in [1,4]`?

We are given the following information:

• `|->abs(vecv)=18" m"//"s at " t=0`

• `|->a(t)=t-8`

• `|->t in [1,4]`

We can begin by recognizing that acceleration is the derivative of velocity, or equivalently, velocity is the antiderivative/integral of acceleration.

We can take the integral of the provided equation for acceleration to find the equation for velocity at time `t`.

`v(t)=int(t-8)dt`

`=>v(t)=1/2t^2-8t+C`

We can use that `v=18` when `t=0` to find the constant `C`:

`18=1/2(0)^2-8(0)+C`

`=>C=18`

Therefore, we have: `v(t)=1/2v^2-8t+18`

Average velocity is given by:

`v_"avg"=(Deltas)/(Deltat)`

We can take the integral of `v(t), t in [1,4]` to find the position `Deltas`.

`Deltas=int_1^4(1/2t^2-8t+18)dt`

`=>=1/6t^3-4t^2+18t|_1^4`

`=>1/6(4)^3-4(4)^2+18(4)-1/6(1)^3+4(1)^2-18(1)`

`=>64/6-64+72-1/6+4-18`

`=>9/2 " m"`

Substituting this into our equation for average velocity:

`v_"avg"=(9/2" m")/((4-1)"s")`

`=>=(9/2" m")/(3" s")`

`=>color(darkblue)(3/2" m"//"s")`

`=>=1.5" m"//"s"`

Speed is the magnitude of velocity, and so the answer remains the same.