# 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?

Feb 17, 2015

If s(t) = 2t^3 − 24t^2 + 90t + 7
The speed (at time t) is given by
$s ' \left(t\right) = 6 {t}^{2} - 48 t + 90$

Acceleration is given by the second derivative
$s ' ' \left(t\right) = 12 t - 48$

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

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.