Question

How do you find values of t in which the speed of the particle is increasing if he position of a particle moving along a line is given by `s(t)=2t^3-24t^2+90t+7` for `t≥0`?

Answer 1

If `s(t) = 2t^3 − 24t^2 + 90t + 7`
The speed (at time t) is given by
`s'(t) = 6t^2 -48t +90`

Acceleration is given by the second derivative
`s''(t) = 12 t - 48`

Critical point for acceleration (when the speed changes from decreasing to increasing) occurs when
`s''(t) = 12t - 48 = 0`

That is when `t = 4`

This might be easier to see if we consider a table of time(t) and speed at time t:

We can see that the speed actually decreases until somewhere between `t=3` and `t=5` when it starts to increase again.