Question

What is the average speed, on `t in [0,5]`, of an object that is moving at `12 m/s` at `t=0` and accelerates at a rate of `a(t) =5t^2-3t` on `t in [0,4]`?

Answer 1

Given:
Initial velocity of the object `v0=12m/s` at `t=0`
To determine the average speed for`t(0,5)`
The acceleration is given by
`a(t)=5t^2-3t`
We know that
`a(t)=(dv)/dt`
Separating variables,
`dv=a(t)dt`
Substituting for `a(t)`
`dv=(5t^2-3t)dt`
Integrating both sides
`intdv=int(5t^2-3t)dt`
`v=5/3t^3-3/2t^2+v0`

At `t=0, v0=12m/s`
Substituting
`12=5/3(0)^3-3/2(0)^2+v0`
`12=0-0+v0`
`v0=12`
Thus, the expression for the velocity given by
`v(t)=5/3t^3-3/2t^2+12`
Velocity at 4 sec is
`v(4)=5/3(4)^3-3/2(4)^2+12`
`v(4)=106.667-37.5+12`
`v(4)=81.167`m/s
Average velocity for the interval is
Distance travelled is obtained by integrating the expression of velocity
`v=(ds)/dt`
Separating variables,
`ds=v(t)dt`
Substituting for `v(t)`
`dv=5/3t^3-3/2t^2+12dt`

Integrating both sides
`intdv=int(5/3t^3-3/2t^2+12)dt`
`s(t)=5/12t^4-1/2t^3+12t+s0`
Assuming the object starts from rest, s0=0
`s(t)=5/12t^4-1/2t^3+12t+s0`
At t-4s, the position is
`s(4)=5/12(4)^4-1/2(4)^3+12t+0`
`s(4)=122.667`

Total distance travelled in the interval of 4 seconds is `122.667`m
Total time elapsed is 4 seconds

average speed = distance /time
`=122.667/4=30.667 m/s`