Problem

A potted cactus is thrown upward with a velocity of 40 feet per second. Its height in feet at time t is given by the formula h(t) = 40t – 16t². Find its velocity 2 seconds after it is released.

Solution

The velocity at time t = 2 is equal to the slope of the line tangent to the curve h(t) at t = 2, so

This simplifies to

The negative in the result means the cactus is traveling in the direction opposite that in which height is positive, or downward, at 24 feet per second.

This graph shows the height of the cactus from the time it is released to t = 2.