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# Rate-Time-Distance Problems

If you've driven down a highway at 60 miles per hour and determined that it will take about 2 hours to reach a destination about 120 miles away, you've calculated a rate-time-distance problem. In actuality, it's pretty common to make general estimates like this every day. Of course, it's easier to make calculations when you have access to the formula for rate-time-distance problems.

## Formula to solve for rate, time, and distance

The formula for rate, time, and distance is pretty straightforward:

$RT=D$

$\mathrm{Rate}\left(r\right)\times \mathrm{Time}\left(t\right)=\mathrm{Distance}\left(d\right)$

## Solving for distance

Let's revisit the scenario where you are traveling down a highway at 60 miles per hour for a period of 2 hours. In this scenario, the rate equals 60 miles per hour and the time equals 2 hours. Let's say you're looking for the distance.

We are given the following: rate is $\frac{60\mathrm{miles}}{1\mathrm{hour}}$ and time is 2 hours.

If you want to know how far you'll be from your starting point (distance) after driving at 60 miles per hour for 2 hours, here's what to do:

Set up an equation.

$d=\left(\frac{60\mathrm{miles}}{1\mathrm{hour}}\right)\times \left(2\mathrm{hours}\right)$

The hours cancel each other out, so you can eliminate them.

$d=60\mathrm{miles}\times 2$

Multiply:

$d=120\mathrm{miles}$

You will be a distance of 120 miles from your starting point along the road.

## Solving for rate

You can use the same rate-time-distance formula to determine the rate of speed or motion. Suppose you want to know your speed in feet per second when walking on a park trail, covering a distance of 180 feet in 2 minutes.

In this case, the rate is unknown, the time is 2 minutes, and the distance is 180 feet.

Set up an equation.

$d=r\times 2\mathrm{minutes}=180\mathrm{feet}$

Divide both sides by 2 minutes.

$r=\frac{180\mathrm{feet}}{2\mathrm{minutes}}$

Reduce the fraction.

$r=\frac{90\mathrm{feet}}{\mathrm{minute}}$

To convert the rate from feet per minute to feet per second, multiply by the conversion factor $\frac{1\mathrm{minute}}{60\mathrm{seconds}}$ .

$r=\left(\frac{90\mathrm{feet}}{1\mathrm{minute}}\right)\times \left(\frac{1\mathrm{minute}}{60\mathrm{seconds}}\right)=\frac{1.5\mathrm{feet}}{\mathrm{second}}$

This means you're walking at a rate of 1.5 feet per second.

Note: When discussing rate, you could refer to x feet per y seconds, x miles per y hours, or even x miles per y seconds.

## Solving for time

You can also use the rate-time-distance formula to determine time. Let's say that you want to skateboard 3 miles at a rate of 6 miles per hour. How long do you think it will take?

In this case, the time is unknown, the rate is 6 miles per hour, and the distance is 3 miles.

Set up an equation.

$\frac{6\mathrm{miles}}{\mathrm{hour}}\times t=3\mathrm{miles}$

Divide both sides by 6 miles per hour.

$t=\frac{3\mathrm{miles}}{6\mathrm{miles}\mathrm{per}\mathrm{hour}}$

Reduce the fraction.

$t=\frac{1}{2}\mathrm{hour}$

To convert the time from hours to minutes, multiply by the conversion factor 60 minutes per hour.

$t=\left(\frac{1}{2}\mathrm{hour}\right)\times \left(\frac{60\mathrm{minutes}}{\mathrm{hour}}\right)=30\mathrm{minutes}$

It will take you 30 minutes to skateboard 3 miles at 6 miles per hour.

## Rate-time-distance table for traveling in opposite directions

Let's briefly go over how to solve for time when moving in opposite directions. Suppose you have two cars leaving the same place but moving in opposite directions. Car #1 is traveling at 45 miles per hour, and car #2 is traveling at 50 miles per hour. In how many hours will they be 190 miles apart?

To start, you can set up a rate-time-distance table:

r | t | d | |

car #1 | |||

car #2 |

Next, fill in the table with the given information. The r for car #1 is 45 mph, the r for car #2 is 50 mph, and t is time when they are 190 miles apart.

r | t | d | |

car #1 | 45 | t | |

car #2 | 50 | t |

Now, use the formula d=rt to fill in the remaining values:

r | t | d | |

car #1 | 45 | t | 45t |

car #2 | 50 | t | 50t |

With the total distance being 190, this is our equation:

$45t+50t=190$

Combine like terms.

$95t=190$

Divide both sides by 95.

$t=2$

The cars will be 190 miles apart in 2 hours.

## Practice questions on rate, time, and distance

a. If you want to walk 4.5 miles at a rate of 3 miles per hour, how long will it take in minutes?

$t=\frac{d}{r}$

$t=\frac{4.5\mathrm{miles}}{\frac{3\mathrm{miles}}{\mathrm{hour}}}$

$t=1.5\mathrm{hours}$

$t=1.5\mathrm{hours}\times \frac{60\mathrm{minutes}}{\mathrm{hour}}=90\mathrm{minutes}$

b. If a meteor travels at a rate of 40 miles per second for 45 seconds, how many miles will the meteor travel?

$d=r\times t$

$d=\frac{40\mathrm{miles}}{\mathrm{second}}\times 45\mathrm{seconds}=1800\mathrm{miles}$

c. Suppose you plan to ride your scooter 10 miles and want to get to your destination in 40 minutes. How fast in miles per hour will you have to travel?

$40\mathrm{minutes}=\frac{40}{60}\mathrm{hours}=\frac{2}{3}\mathrm{hours}$

$r=\frac{d}{t}$

$r=\frac{10\mathrm{miles}}{\left(\frac{2}{3}\right)\mathrm{hours}}=\frac{15\mathrm{miles}}{\mathrm{hour}}$

d. If a bus is traveling at 35 miles per hour in one direction and a car is traveling at 40 miles per hour, in how many hours will they be 225 miles apart?

$\mathrm{Relative\; speed}=35+40=\frac{75\mathrm{miles}}{\mathrm{hour}}$

$t=\frac{d}{r}$

$t=\frac{225\mathrm{miles}}{\frac{75\mathrm{miles}}{\mathrm{hour}}}=3\mathrm{hours}$

e. Suppose you are driving 72 miles per hour over a period of 2.25 hours, how many miles will you travel?

$d=r\times t$

$d=\frac{72\mathrm{miles}}{\mathrm{hour}}\times 2.25\mathrm{hours}=162\mathrm{miles}$

f. Let's say you can sprint 13 miles per hour and want to run 0.06 miles. How many seconds will it take?

$t=\frac{d}{r}$

$t=\frac{0.06\mathrm{miles}}{\frac{13\mathrm{miles}}{\mathrm{hour}}}$

$t=\frac{0.06\mathrm{miles}}{\frac{13\mathrm{miles}}{3600\mathrm{seconds}}}=16.62\mathrm{seconds}$

## Topics related to the Rate-Time-Distance Problems

## Flashcards covering the Rate-Time-Distance Problems

## Practice tests covering the Rate-Time-Distance Problems

College Algebra Diagnostic Tests

## Get help learning about rate-time-distance problems

Whether your student is being asked to calculate distance, rate, or time – or they need to calculate travel in opposite directions – rate-time-distance problems can be challenging. Working alongside a tutor could make a major difference in your student's comprehension. Tutoring can also be helpful if they're studying for an upcoming exam or would like help with rate-time-distance homework assignments. During sessions, a qualified tutor can answer your student's questions and address their concerns. Contact the Educational Directors at Varsity Tutors to learn all about the perks of tutoring.

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