Question

Ian tosses a bone up in the air for his dog, Spot. The height, h, in feet, that Spot is above the ground at the time t seconds after she jumps for the bone can be represented h(t)=-16²+20t.

What is Spot's average rate of ascent, in feet per second, from the time she jumps into the air to the time she catches the bone at t=1/2?

Answer 1

Answer: 12 ft/s

Explanation:

We are told the following function models Spot's height at time
`t`:

`h_(t)=-16t^(2)+20t` (1)

And we are asked to find Spot's average rate of ascent, this means its velocity
`V`, which is calculated by:

`V=\frac{h_{t=(1)/(2)}-h_(t=0)}{t_(f)-t_(o)}` (2)

Where:

`h_{t=(1)/(2)` is the height of Spot at time
`t=(1)/(2)s`

`h_{t=0` is the height of Spot at time
`t=0 s`

`t_(f)=(1)/(2) s` is the final time

`t_(o)=0 s` is the initial time

So, firstly we need to calculate
`h_{t=(1)/(2)` and
`h_{t=0`:

`h_{t=(1)/(2)}=-16((1)/(2))^(2)+20((1)/(2))` (3)

`h_{t=(1)/(2)}=6 ft` (4)

`h_(t=0)=0 ft` (5)

Substituting these values in (2):

`V=(6 ft-0 ft)/((1)/(2) s-0 s)` (6)

Finally:

`V=12 ft/s` This is Spot's average rate of ascent