Question

An object is dropped from a height of 150 feet. Its heights s at time, t is given by the

equation s(t) =-16t^2+150 ,where s is measured in feet and t is measured in seconds.

Find the average rate of change of the height over the interval [1, 2.5).

O 1) -56 ft/sec

O2) 77 ft/sec

03) 56 ft/sec

O4) -35 ft/sec

Answer 1

The average rate of change of the height over the interval [1, 2.5] is - 56 feet per second.

Explanation:

Let
`s(t) = -16\cdot t^(2) + 150`. Geometrically speaking, average rate of change over a given interval (
`\bar v`), measured in feet per second, is determined by definition of secant line, which is defined:

`\bar v = (s(t_(2))-s(t_(1)) )/(t_(2)-t_(1))` (1)

Where:

`s(t_(1))`,
`s(t_(2))` - Initial and final position of the object, measured in feet.

`t_(1)`,
`t_(2)` - Initial and final times, measured in seconds.

If we know that
`t_(1) = 1\,s` and
`t_(2) = 2.5\,s`, then the average rate of change over the interval
`[1,2.5]`:

`s(1) = -16\cdot (1)^(2)+150`

`s(1) = 134`

`s(2.5) = -16\cdot (2.5)^(2)+150`

`s(2.5) = 50`

`\bar v = (50\,ft-134\,ft)/(2.5\,s-1\,s)`

`\bar v = -56\,(ft)/(s)`

The average rate of change of the height over the interval [1, 2.5] is - 56 feet per second.