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?

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cryssatemp cryssatemp · Answer 1 · 2020-04-04T05:48:03+02:00

Answer: 12 ft/s

Step-by-step explanation:

We are told the following function models Spot's height at time [tex]t[/tex]:

[tex]h_{t}=-16t^{2}+20t[/tex] (1)

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

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

Where:

[tex]h_{t=\frac{1}{2}[/tex] is the height of Spot at time [tex]t=\frac{1}{2}s[/tex]

[tex]h_{t=0[/tex] is the height of Spot at time [tex]t=0 s[/tex]

[tex]t_{f}=\frac{1}{2} s[/tex] is the final time

[tex]t_{o}=0 s[/tex] is the initial time

So, firstly we need to calculate [tex]h_{t=\frac{1}{2}[/tex] and [tex]h_{t=0[/tex]:

[tex]h_{t=\frac{1}{2}}=-16(\frac{1}{2})^{2}+20(\frac{1}{2})[/tex] (3)

[tex]h_{t=\frac{1}{2}}=6 ft[/tex] (4)

[tex]h_{t=0}=0 ft[/tex] (5)

Substituting these values in (2):

[tex]V=\frac{6 ft-0 ft}{\frac{1}{2} s-0 s}[/tex] (6)

Finally:

[tex]V=12 ft/s[/tex] This is Spot's average rate of ascent