Question

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

Answer 1

Given acceleration `a(t)=5-t`
OR,`(dv)/(dt)=5-t`
,`=>dv=5dt-tdt`
,`=>intdv=5intdt-inttdt+c`,where c= integration constant
`=>v=5t-t^2/2+c` ....(1)
Given at `t=0 ,v=2m/s`
So `c = 2`
and eq(1) becomes
`=>v=5t-t^2/2+2`

Now distance traversed during `t=0 to t= 3s`

`s=int_0^3vdt=int_0^3(5t-t^2/2+2)dt`
`=[5t^2/2-t^3/6+2t]_0^3=5*3^2/2-3^3/6+2*3=24m`
Hence Average speed
`=s/t=24/3=8m/s`