Solve Uniform Motion Applications

Solve Uniform Motion Applications

When planning a road trip, it often helps to know how long it will take to reach the destination, or how far to travel each day. We would use the distance, rate, and time formula, $D = rt$, which we have already seen.

In this section, we will use this formula in situations that require a little more algebra to solve than the ones we saw earlier. Generally, we will be looking at comparing two scenarios, such as two vehicles travelling at different rates or in different directions. When the speed of each vehicle is constant, we call applications like this uniform motion problems.

Our problem-solving strategy will still apply here, but we will add to the first step. The first step will include drawing a diagram that shows what is happening in the example. Drawing the diagram helps us understand what is happening so that we will write an appropriate equation. Then we will make a table to organize the information, like we did for the money applications.

Example. An express train and a local train leave Pittsburgh to travel to Washington, D.C. The express train can make the trip in 4 hours and the local train takes 5 hours for the trip. The speed of the express train is 12 miles per hour faster than the speed of the local train. Find the speed of both trains.

Both trains travel the same distance, from Pittsburgh to Washington, D.C. Let $r$ represent the speed of the local train, so $r + 12$ represents the speed of the express train. Fill in the rates and times, then multiply to get an expression for each distance:

Since both trains travel the same distance, the distance traveled by the express train equals the distance traveled by the local train:

Solve the equation:

So the speed of the local train is $48$ mph. The speed of the express train is $r + 12 = 48 + 12 = 60$ mph.

Check. The express train travels $60 \text{ mph} \cdot 4 \text{ hours} = 240$ miles. The local train travels $48 \text{ mph} \cdot 5 \text{ hours} = 240$ miles. ✓ Both distances match.

The speed of the local train is $48$ mph and the speed of the express train is $60$ mph.

In the last example, we had two trains traveling the same distance. The diagram and the chart helped us write the equation we solved. Let’s see how this works in another case, where the two distances add up to a fixed total instead of being equal to each other.

Example. Christopher and his parents live 115 miles apart. They met at a restaurant between their homes to celebrate his mother’s birthday. Christopher drove 1.5 hours while his parents drove 1 hour to get to the restaurant. Christopher’s average speed was 10 miles per hour faster than his parents’ average speed. What were the average speeds of Christopher and of his parents as they drove to the restaurant?

Let $r$ represent the average speed of the parents, so $r + 10$ represents Christopher’s average speed.

Christopher and his parents drove toward each other from opposite directions, so the distance Christopher traveled plus the distance his parents traveled must add up to the total $115$ miles between their homes:

Solve the equation:

So the parents’ speed was $40$ mph, and Christopher’s speed was $r + 10 = 40 + 10 = 50$ mph.

Check. Christopher drove $50 \text{ mph} \cdot 1.5 \text{ hours} = 75$ miles. His parents drove $40 \text{ mph} \cdot 1 \text{ hour} = 40$ miles. Together, $75 + 40 = 115$ miles. ✓

Christopher’s speed was $50$ mph, and his parents’ speed was $40$ mph.

Example. Two truck drivers leave a rest area on the interstate at the same time. One truck travels east and the other one travels west. The truck traveling west travels at 70 mph and the truck traveling east has an average speed of 60 mph. How long will they travel before they are 325 miles apart?

Both trucks travel the same amount of time, so let $t$ represent that time.

The trucks start at the same rest area and travel in opposite directions, so the distance traveled west plus the distance traveled east must add up to $325$ miles:

Solve the equation:

So it will take the trucks $2.5$ hours to be $325$ miles apart.

Check. The truck going west travels $70 \text{ mph} \cdot 2.5 \text{ hours} = 175$ miles. The truck going east travels $60 \text{ mph} \cdot 2.5 \text{ hours} = 150$ miles. Together, $175 + 150 = 325$ miles. ✓

It will take the trucks $2.5$ hours to be $325$ miles apart.

Example. When Katie Mae walks to school, it takes her 30 minutes. If she rides her bike, it takes her 15 minutes. Her speed is three miles per hour faster when she rides her bike than when she walks. What are her walking speed and her speed riding her bike?

Let $r$ represent Katie Mae’s walking speed, so $r + 3$ represents her biking speed. The speed is in miles per hour, so we need to express the times in hours, too, since one hour is 60 minutes:

The distance from Katie Mae’s home to her school is the same whether she is walking or riding her bike:

Clear the fractions by multiplying both sides by the LCD, $8$:

So Katie Mae’s walking speed is $3$ mph, and her biking speed is $r + 3 = 3 + 3 = 6$ mph.

Check. Walking, she covers $3 \text{ mph} \cdot 0.5 \text{ hour} = 1.5$ miles. Biking, she covers $6 \text{ mph} \cdot 0.25 \text{ hour} = 1.5$ miles. ✓ Either way, Katie Mae travels $1.5$ miles to school.

Katie Mae’s walking speed is $3$ mph, and her speed riding her bike is $6$ mph.

In the distance, rate, and time formula, time represents the actual amount of elapsed time (in hours, minutes, etc.). If a problem gives us starting and ending times as clock times, we must find the elapsed time in order to use the formula.

Example. Hamilton loves to travel to Las Vegas, 255 miles from his home in Orange County. On his last trip, he left his house at 2:00 pm. The first part of his trip was on congested city freeways. At 4:00 pm, the traffic cleared and he was able to drive through the desert at a speed 1.75 times as fast as when he drove in the congested area. He arrived in Las Vegas at 6:30 pm. How fast was he driving during each part of his trip?

We know the total distance is $255$ miles. We are looking for the rate of speed for each part of the trip. The rate in the desert is $1.75$ times the rate in the city. If we let $r$ represent the rate in the city, then the rate in the desert is $1.75r$.

The times here are given as clock times. Hamilton started from home at 2:00 pm and entered the desert at 4:00 pm, so he spent two hours driving the congested freeways in the city. Then he drove faster from 4:00 pm until 6:30 pm in the desert, so he drove 2.5 hours in the desert.

The sum of the distance driven in the city and the distance driven in the desert is $255$ miles:

Solve the equation:

So Hamilton drove $40$ mph in the city. His desert speed was $1.75r = 1.75(40) = 70$ mph.

Check. In the city, he drove $40 \text{ mph} \cdot 2 \text{ hours} = 80$ miles. In the desert, he drove $70 \text{ mph} \cdot 2.5 \text{ hours} = 175$ miles. Together, $80 + 175 = 255$ miles. ✓

Hamilton drove 40 mph in the city and 70 mph in the desert.

Key terms

uniform motion problem — an application in which one or more objects travel at a constant speed, so the distance, rate, and time formula $D = rt$ applies to each. elapsed time — the actual amount of time that passes during a trip; if a problem gives clock times instead, subtract to find the elapsed time before using $D = rt$.

This section is adapted from Elementary Algebra 2e, Section 3.5: Solve Uniform Motion Applications by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recast the rate/time/distance charts as markdown tables and the diagrams as prose descriptions; corrected an inconsistency in the source’s clock-time example (the worked solution used a 4:00 pm changeover, matching its diagram and 2-hour/2.5-hour split, though the prose read “4:30 pm”); omitted the Be Prepared quiz, Self Check checklist, media links, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.