Question

Question #0996a

Answer 1

`-80"ft"/"s"`

Explanation:

We're asked to find the average rate of change (of position) with respect to the time interval `[2,3]`. This rate of change is the average velocity of the silver dollar, which is negative, as it is falling downward (assuming you take the positive `y`-axis as the "upward" direction).

The average velocity of the silver dollar is the rate of change of position with respect to time, given by the equation

`v_(av-y) = (s_2-s_1)/(3"s"-2"s") = (Deltay)/(Deltat)`

where

• `s` is the position/height (in `"ft"`) at times `t = 2"s"` (`s_1`) and `t = 3"s"` (`s_2`) (the change in `y`-position `Deltay` is always the final position minus the initial position, never the reverse), and

• `t` is the time, in `"s"`.

To find the positions at times `t = 2"s"` and `t = 3"s"`, we simply plug in the values `2` and `3` in for the variable `t` in the equation:

`t = 2`

`-16(2)^2 + 555 = color(red)(491"m"`

`t = 3`

`-16(3)^2 + 555 = color(green)(411"m"`

So, plugging in these values into the average velocity equation, we have

`v_(av-y) = (color(green)(411"ft")-color(red)(491"ft"))/(3"s"-2"s") = color(blue)(-80"ft"/"s"`