Aball 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?

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



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



This has two solutions, , but only one is positive and falls in the interval . So the velocity reaches 30 cm/s when .

A= r2 × 3.14 11×11 ↓ ↓ 121 × 3.14 = 379.94 → 379.9