Chapter 10, Problem 15CT

a.

To determine

To graph: The function using a graphing calculator.

Answer to Problem 15CT

Domain: $h\ge 0$

Range: $v\ge 0$

Explanation of Solution

Given Information: The velocity of roller coaster at bottom of the hill is given by $v=\sqrt{19.6h}$ .

Graph:

  

Domain: $h\ge 0$

Range: $v\ge 0$

Thus, domain is $h\ge 0$ and range is $v\ge 0$ .

b.

To determine

To find: The height of the hill for the velocity to be $28m/s$ at the bottom of the hill.

Answer to Problem 15CT

Height of the hill should be $40m$ .

Explanation of Solution

Given Information: The velocity of roller coaster at bottom of the hill is given by $v=\sqrt{19.6h}$ .

Putting $v=28$ in $v=\sqrt{19.6h}$ ,

  $28=\sqrt{19.6h}$

Squaring both sides,

  $\begin{array}{l}19.6h=784\\ h=40\end{array}$

Thus, height of the hill should be $40m$ .

c.

To determine

To find: The behavior of rate of change of velocity as height increases.

Answer to Problem 15CT

Rate of change of velocity decreases.

Explanation of Solution

Given Information: The velocity of roller coaster at bottom of the hill is given by $v=\sqrt{19.6h}$ .

Calculation:

Calculating the average rate of change of velocity over three intervals,

For $h=0toh=20$ ,

  $\begin{array}{l}\frac{v(20)-v(0)}{20-0}\\ =\frac{\sqrt{19.6\times 20}-\sqrt{19.6\times 0}}{20}\\ =\frac{19.8}{20}\\ =0.99\end{array}$

For $h=20toh=40$ ,

  $\begin{array}{l}\frac{v(40)-v(20)}{40-20}\\ =\frac{\sqrt{19.6\times 40}-\sqrt{19.6\times 20}}{20}\\ =\frac{28-19.8}{20}\\ =0.41\end{array}$

For $h=40toh=60$ ,

  $\begin{array}{l}\frac{v(60)-v(40)}{60-40}\\ =\frac{\sqrt{19.6\times 60}-\sqrt{19.6\times 40}}{20}\\ =\frac{34.3-28}{20}\\ =0.315\end{array}$

Above calculations clearly imply that rate of change of velocity is decreasing as the height of hill increases.

Thus, rate of change of velocity is decreasing.