Question

A ball rolls down an inclined plane such that the distance (in centimeters) that it rolls in t seconds is given by s(t) = 2t^3 + 3t^2 + 4 for 0<=t<=3 At what time is the velocity 30cm/s?

Answer 1

The ball's velocity will be represented by the derivative of its distance function:

$s'(t)=6t^2+6t$

Now find the times $t$ for which this is equal to 30, i.e. solve

$6t^2+6t=30\implies t^2+t-5=0$

This has two solutions, $t=-\dfrac{1\pm\sqrt{21}}2$, but only one is positive and falls in the interval $[0,3]$. So the velocity reaches 30 cm/s when $t=-\dfrac{1-\sqrt{21}}2\approx1.79$.