# 9.4: Speedo

Difficulty Level: At Grade Created by: CK-12

Do you know the formula D=rt\begin{align*}D=rt\end{align*}? Could you use it to solve the problem below? In this concept, we will learn how and when to use the distance-rate-time formula.

The average speed of the winner in the 30-minute bicycle race was 52 feet per second. To the nearest tenth of a mile, how many miles was the race?

### Guidance

In order to solve the problem above, use the problem solving steps along with the formula D=rt\begin{align*}D=rt\end{align*}.

• Start by describing what information is given.
• Then, identify what your job is. In these problems, your job will be to figure out either the speed, distance, or time.
• Next, make a plan for how you will solve. In these problems, you will use the given information and the formula D=rt\begin{align*}D=rt\end{align*} to solve for the missing information.
• Then, solve the problem.
• Finally, check your solution. Make sure that your solution works with D=rt\begin{align*}D=rt\end{align*}.

#### Example A

Use the \begin{align*}\textbf{D = \textit{rt}}\end{align*} formula to solve the problem. Show your steps.

The fastest moving ball in any game is the pelota in the game of Jai-Alai. In one game the pelota was clocked at 264 feet per second. At that speed, how far will the pelota travel in 0.5 seconds?

Solution:

We can use problem solving steps and the \begin{align*}\textbf{D = \textit{rt}}\end{align*} formula to solve the problem.

\begin{align*}&\mathbf{Describe:}&& \text{Speed is 264 feet per second}.\\ &&& \text{Time is 0.5 seconds}.\\ \\ &\mathbf{My \ Job:}&& \text{Figure out how far the pelota will travel}.\\ \\ &\mathbf{Plan:} && \text{Speed is another name for rate. Replace} \ r \ \text{and} \ t \ \text{with their values in the formula,} \ D = rt. \ \text{Solve}\\ &&&\text{for} \ D \ \text{in feet. } \\ \\ &\mathbf{Solve:}&& D = rt\\ &&& D = 264 ft/sec \ \times 0.5 \ sec\\ &&& D = 132 \ feet\\ &&& \text{It would travel} \ 132 \ feet.\\ \\ &\mathbf{Check:} && 264 \ ft/sec \ \text{for} \ 0.5 \ seconds \ is \ 264 \times 0.5 = 132 \ feet.\\ &&&\text{The pelota will travel} \ 132 \ feet.\end{align*}

#### Example B

Use the \begin{align*}\textbf{D = \textit{rt}}\end{align*} formula to solve the problem. Show your steps.

A sky diver in free fall sometimes reaches a speed of 200 feet per second. At that speed, how long would it take a sky diver to descend 5000 feet?

Solution:

We can use problem solving steps and the \begin{align*}\textbf{D = \textit{rt}}\end{align*} formula to solve the problem.

\begin{align*}&\mathbf{Describe:}&& \text{Speed is 200 feet per second}.\\ &&& \text{Distance is 5000 feet}.\\ \\ &\mathbf{My \ Job:}&& \text{Figure out how long it would take}.\\ \\ &\mathbf{Plan:} && \text{Speed is another name for rate. Replace} \ r \ \text{and} \ D \ \text{with their values in the formula,} \ D = rt. \ \text{Solve}\\ &&&\text{for} \ t \ \text{in seconds. } \\ \\ &\mathbf{Solve:}&& D = rt\\ &&& 5000 \ feet = 200 ft/sec \times t \ sec\\ &&& 25 \ sec = t\\ &&& t = 25 \ sec\\ &&& \text{It would take} \ 25 \ seconds.\\ \\ &\mathbf{Check:} && 200 \ ft/sec \ \text{for} \ 25 \ seconds \ is \ 200 \times 25 = 5000 \ feet.\\ &&&\text{It will take the sky diver} \ 25 \ seconds.\end{align*}

#### Example C

Use the \begin{align*}\textbf{D = \textit{rt}}\end{align*} formula to solve the problem. Show your steps.

One of the record speeds for a roller skater is 37.9 feet per second. At this speed, about how many minutes would it take to skate half a mile?

Solution:

We can use problem solving steps and the \begin{align*}\textbf{D = \textit{rt}}\end{align*} formula to solve the problem.

\begin{align*}&\mathbf{Describe:}&& \text{Speed is 37.9 feet per second}.\\ &&& \text{Distance is half a mile}.\\ \\ &\mathbf{My \ Job:}&& \text{Figure out how long it would take in minutes}.\\ \\ &\mathbf{Plan:} && \text{Speed is another name for rate. Replace} \ r \ \text{and} \ D \ \text{with their values in the formula,} \ D = rt. \ \text{Solve}\\ &&&\text{for} \ t \ \text{in seconds. Divide the number of seconds by} \ 60 \ \text{to get the number of minutes}.\\ \\ &\mathbf{Solve:}&& \text{Distance:} \ 0.5 \ miles \ \text{is} \ 0.5 \times 5280 \ feet/mile, \ \text{or} \ 2640 \ feet.\\ &&& D = rt\\ &&& 2640 = 37.9 ft/sec \times t \ sec\\ &&& 69.7 \ sec = t\\ &&& t = 69.7 \ sec\\ &&& \text{it would take about} \ 1 \ minute .\\ \\ &\mathbf{Check:} && 37.9 \ ft/sec \ \text{is} \ 37.9 \times 60, \ \text{or} \ 2274 \ ft/min\\ &&&\text{In a little over} \ 1 \ minute, \ \text{the roller skater will cover} \ 2640 \ \text{feet or half a mile.}\end{align*}

#### Concept Problem Revisited

The average speed of the winner in the 30-minute bicycle race was 52 feet per second. To the nearest tenth of a mile, how many miles was the race?

We can use problem solving steps and the \begin{align*}\textbf{D = \textit{rt}}\end{align*} formula to solve the problem.

\begin{align*}&\mathbf{Describe:}&& \text{Speed is 52 feet per second}.\\ &&& \text{Time is 30 minutes}.\\ \\ &\mathbf{My \ Job:}&& \text{Figure out the length of the race in miles}.\\ \\ &\mathbf{Plan:} && \text{Speed is another name for rate. Since time is in minutes, figure out the rate}\\ &&& \text{in minutes. Replace} \ t \ \text{and} \ r \ \text{with their values in the formula,} \ D = rt. \ \text{Solve}\\ &&&\text{for} \ D \ \text{in feet. Divide the number of feet by} \ 5280 \ \text{to get the number of}\\ &&&\text{miles}.\\ \\ &\mathbf{Solve:}&& \text{Rate:} \ 52 \ feet/second \ \text{is} \ 52 \times 60 \ seconds/minute, \ \text{or} \ 3120 \ feet/minute.\\ &&& D = rt\\ &&& D = 3120 \ ft/min \times 30 \ min, \ \text{or} \ 93,600 \ ft\\ &&& D = \frac{93,600}{5280 \ ft/mile}\\ &&& D = 17.7 \ miles\\ &&& \text{The race was} \ 17.7 \ miles \ \text{long.}\\ \\ &\mathbf{Check:} && 3120 \ ft/min \ \text{is} \ \frac{3120}{5280}, \ \text{or} \ 0.59 \ miles/min\\ &&&\text{In} \ 30 \ minutes, \ \text{the biker will cover} \ 30 \times 0.59 \ miles, \ \text{or} \ 17.7 \ miles.\end{align*}

### Vocabulary

Rate is another word for the speed of an object. Rate tells us the distance that an object is traveling in a given amount of time. The formula \begin{align*}D=rt\end{align*} shows the relationship between Distance, rate, and time.

### Guided Practice

Use the \begin{align*}\textbf{D= \textit{rt}}\end{align*} formula to solve these problems. Show your steps.

1. One of the fastest swimmers swam 100 yards at the rate of 6.65 feet per second. How long did it take the swimmer to swim the 100 yards?

2. A hard hit table tennis ball can travel the length of the 9 foot table in 0.075 second. At that speed, how far would the table tennis ball travel in one second?

3. In 0.3 second, a well hit volley ball can travel 59 feet, the length of a volley ball court. How far could a hard hit volleyball travel in one second?

\begin{align*}D&= rt\\ 100 \ yards &= 300 \ ft\\ 300 \ ft &= 6.65 \ ft/sec \times \textbf{\textit{t}}\\ \frac{300 \ ft}{6.65 \ ft/sec} &= \textbf{\textit{t}}\\ 45.11 \ sec &= t\end{align*}

2. 120 ft/sec.

\begin{align*}D&= rt\\ 9 \ ft &= \textbf{\textit{r}} \times 0.075 \ sec\\ \frac{9 \ ft}{0.075 \ sec} & = \textbf{\textit{r}}\\ 120 \ ft/sec &= \textbf{\textit{r}}\end{align*}

3. 196.7 ft/sec

\begin{align*}D&= rt\\ 59 \ ft &= \textbf{\textit{r}} \times 0.3 \ sec\\ \frac{59 \ ft}{0.3 \ sec} &= \textbf{\textit{r}}\\ 196.7 \ ft/sec &= \textbf{\textit{r}}\end{align*}

### Practice

Use the \begin{align*}\textbf{D= \textit{rt}}\end{align*} formula to solve these problems. Show your steps.

1. A swimmer swims 150 yards at the rate of 2.8 feet per second. How long did it take the swimmer to swim the 150 yards?
2. A baseball travels 80 mph. At that speed, how far would it travel in three seconds? Determine your answer in feet.
3. In 3.3 seconds, a volleyball travels 100 feet. At that rate, how far did it travel in 1 second?
4. A plane travels at 550mph. How long does it take to go 3300 miles?
5. A biker rides 25 miles in 200 minutes. What is his speed in miles per hour?

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Date Created:
Jan 18, 2013