This can be solved by using derivatives and integration.
We know acceleration is the rate of velocity change and velocity is rate of change of displacement or distance. So in general notation, if a(t) is acceleraion, v(t) is velocity and s(t) is the displacement or distance at any given time t, then,
`(ds)/(dt) = v`
`(dv)/(dt) = a`
`a(t) = 6 - 2t-3t^2`
So, `(dv)/(dt) = 6 - 2t-3t^2`
By integrating this wrt t, we can find v(t),
`v(t) = int(6 - 2t-3t^2) dt`
`v(t) = 6t-t^2-t^3+c`
Where c is a constant and we have to find it using given data,
We know that the initial velocity is zero. So at `t= 0, v(t) = 0.`
`0 = c`
Therefore c is zero.
So, the velocity profile is,
`v(t) = 6t-t^2-t^3`
But we know, `v = (ds)/(dt).`
`(ds)/(dt) = 6t-t^2-t^3`
By integrating this wrt t, we can find s(t).
`s(t) = int(6t-t^2-t^3)dt`
`s(t) = 3t^2-1/3t^3-1/4t^4+d`
Where d is a constant we have to find it using given data.
Let's assume at t = 0, the distance is zero. Then,
`0 = +d`
`d = 0.`
Therefore the required function is,
`s(t) = 3t^2-1/3t^3-1/4t^4`
To find the distance travel in 13 seconds, we can find s(13).
`s(13) = 3(13^2)-1/3(13^3)-1/4(13^4)`
`s(13) = 507-723.33 - 7140.25`
`s(13) = -7356.58`
Therefore the distance travelled in 13 seconds is 7356.58 in the opposite direction.