Question

A) which hall reaches a greater maximum height?B) which ball is traveling at a faster average rate on the interval [0,2]?Explain your answers to part (a) and part (b).

Answer 1

Solution

Given the question in the image, the following are the solution steps to answer the question.

STEP 1: State the maximum height for the ball 1

The maximum height of the ball 1 is shown below:

STEP 2: State the maximum height for the ball 2

Given the function,

$-16t^2+40t+4$

The graph is as seen below and the maximum point is indicated:

Maximum height is 29

Hence, the ball that reaches the greater maximum height is ball 2

STEP 3: Answer Part B

Average rate is calculated using the formula:

$\frac{f(b)-f(a)}{b-a}$

STEP 4: Find the average rate for ball 1

$\begin{gathered} a=0,b=2 \\ f(a)=2.5,f(b)=8 \\ rate=\frac{8-2.5}{2-0}=\frac{5.5}{2}=2.75 \end{gathered}$

STEP 5: Find the average rate for ball 2

$\begin{gathered} a=0,f(a)=4 \\ b=2,f(b)=20 \\ rate=\frac{20-4}{2-0}=\frac{16}{2}=8 \end{gathered}$

Since the average rate of change for ball 2 is more than that of ball 1, therefore ball 2 was travelling at a faster average rate on the given interval.