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 by the function h(t)= -16t^2+20t

Answer 1

Answer:

12 feet per second.

Step-by-step explanation:

Please consider the complete 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$h(t)=-16t^2+20t$.

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?  

We will use average rate of change formula to solve our given problem.

$\text{Average rate of change}=\frac{f(b)-f(a)}{b-a}$

$\text{Average rate of change}=\frac{h(\frac{1}{2})-h(0)}{\frac{1}{2}-0}$

$\text{Average rate of change}=\frac{-16\cdot(\frac{1}{2})^2+20\cdot \frac{1}{2}-(-16\cdot(0)^2+20\cdot (0))}{\frac{1}{2}-0}$

$\text{Average rate of change}=\frac{-16\cdot\frac{1}{4}+10-(0)}{\frac{1}{2}}$

$\text{Average rate of change}=\frac{-4+10}{\frac{1}{2}}$

$\text{Average rate of change}=\frac{6}{\frac{1}{2}}$

$\text{Average rate of change}=\frac{2\cdot 6}{1}$

$\text{Average rate of change}=12$

Therefore, Spot's average rate of ascent is 12 feet per second.