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=-\frac{1\pm√(21)}2`, but only one is positive and falls in the interval
`[0,3]`. So the velocity reaches 30 cm/s when
`t=-\frac{1-√(21)}2\approx1.79`.