Question

What is the average speed of an object that is still at `t=0` and accelerates at a rate of `a(t) =t-3` from `t in [3, 6]`?

Answer 1

`v_a=-3 " unit/s"`

Explanation:

`v(t)=int(t-3)*d t" "v(t)=t^2/2-3t`

`x(t)=int v(t)*d t" "x(t)=int (t^2/2-3t)*d t`

`x(t)=1/2*t^3/3-3t^2/2" "x(t)=t^3/6-(3t^2)/2`

`"for position t=3 -> x(3)=-9 unit"`

`"for position t=6 ->x(6)=-18 unit"`

`v_a=tan alpha`

`v_a=(-18+9)/(6-3)`

`v_a=-9/3`

`v_a=-3 " unit/s"`

Answer 2

`"Average speed"=-3 " units"`
`-ve` sign indicates that speed is opposite to the defined positive direction of displacement.

Explanation:

Given acceleration `a(t)=t-3`. We know that `a=x''`, therefore, to find distance moved during the period of interest `x=int int a(t).dt.dt` over the period of interest
(First integral will give us the velocity in terms of time.) Inserting given values.

`x=int int(t-3).dt.dt`
`=int_3^6(t^2/2-3t+C).dt` .....(1), where `C` is constant of integration.
We need to evaluate `C` from initial conditions
Velocity `v(t)=t^2/2-3t+C`
Now `v(0)=0^2/2-3xx0+C=0`, initial condition
or `C=0`
Inserting in (1)

`x=int_3^6(t^2/2-3t).dt`
`x=|t^3/6-3 t^2/2|_3^6`, constant of integration ignored due to definite integral

`=(6^3/6-3xx 6^2/2)-(3^3/6-3 xx3^2/2)`
`=36-54-9/2+27/2`
`=-9`,
We know that `"Average speed"=("Total Displacement")/"Time Taken"`
`"Average speed"=-9/(6-3)`
`"Average speed"=-3`