### Home > INT2 > Chapter 9 > Lesson 9.3.1 > Problem9-78

9-78.

The table at right shows the height of an acrobat as she practices a trick on a trampoline. She reaches a maximum height of $16$ feet above the trampoline.

1. What is the average velocity of the acrobat between $0.25$ seconds and $0.5$ seconds? Include units with your answer.

Divide the difference of the heights by the difference of the times.

$\frac{12\;\text{ft}-7\;\text{ft}}{0.5\;\text{s}-0.25\;\text{s}}=\frac{5\;\text{ft}}{0.25\;\text{s}}$

$20$ ft/s

2. What is the average velocity of the acrobat between $0.5$ seconds and $1$ second?

Use the same techniques as in part (a).

3. What appears to happen to the velocity of the acrobat as she approaches the top of her jump? Does this make sense? Explain.

Her velocity decreases near the top of her jump. Think of the time components as $x$ and the height components as $y$. Notice that the slope of this parabola is flatter near the vertex and steeper the further away you get from the vertex.

Time (s)

Height (ft)

$0$

$0$

$0.25$

$7$

$0.5$

$12$

$0.75$

$15$

$1.0$

$16$

$1.25$

$15$

$1.5$

$12$

$1.75$

$7$

$2.0$

$0$