Distance Problems Help (page 3)
Distance, Rate, and Time
Another common word problem type is the distance problem, sometimes called the uniform rate problem. The underlying formula is d = rt (distance equals rate times time). From d = rt , we get two other relationships: r = d/t and t = d/r . These problems come in many forms: two bodies traveling in opposite directions, two bodies traveling in the same direction, two bodies traveling away from each other or toward each other at right angles. Sometimes the bodies leave at the same time, sometimes one gets a head start. Usually they are traveling at different rates, or speeds. As in all applied problems, the units of measure must be consistent throughout the problem. For instance, if your rates are given to you in miles per hour and your time is given in minutes, you should convert minutes to hours. You could convert miles per hour into miles per minute, but this would be awkward.
When the bodies move in the same direction, the rate at which the distance between them is changing is the difference between their rates.
A cyclist starts at a certain point and rides at a rate of 10 mph. Twelve minutes later, another cyclist starts from the same point in the same direction and rides at 16 mph. How long will it take for the second cyclist to catch up with the first?
When the second cyclist begins, the first has traveled miles. The “10” is the rate and the is the twelve minutes converted to hours. Because the cyclists are moving in the same direction, the rate at which the distance between them is shrinking is 16 – 10 = 6 mph. Then, the question boils down to “How long will it take for something traveling 6 mph to cover 2 miles?”
Let t represent the number of hours the second cyclist is traveling.
It will take the second cyclist of an hour or 20 minutes to catch up the first cyclist.
A car passes an intersection heading north at 40 mph. Another car passes the same intersection 15 minutes later heading north traveling at 45 mph. How long will it take for the second car to overtake the first?
In 15 minutes, the first car has traveled miles. The second car is gaining on the first at a rate of 45 – 40 = 5 mph. So the question becomes “How long will it take a body traveling 5 mph to cover 10 miles?”
Let t represent the number of hours the second car has traveled after passing the intersection.
It will take the second car two hours to overtake the first.
Find practice problems and solutions at Distance Problems Practice Problems - Set 1.
Traveling in Opposite Directions
When two bodies are moving in opposite directions, whether towards each other or away from each other, the rate at which the distance between them is changing, whether growing larger or smaller, is the sum of their individual rates.
Two cars meet at an intersection, one heading north; the other, south. If the northbound driver drives at 30 mph and the southbound driver at 40 mph, when will they be 35 miles apart?
The distance between them is growing at the rate of 30 + 40 = 70 mph. The question then becomes, “how long will it take a body moving 70 mph to travel 35 miles?”
Let t represent the number of hours the cars travel after leaving the intersection.
In half an hour, the cars will be 35 miles apart.
Katy left her house on a bicycle heading north at 8 mph. At the same time, her sister Molly headed south at 12 mph. How long will it take for them to be 24 miles apart?
The distance between them is increasing at the rate of 8 + 12 = 20 mph. The question then becomes “How long will it take a body moving 20 mph to travel 24 miles?”
Let t represent the number of hours each girl is traveling.
The girls will be 24 miles apart after hours or 1 hour 12 minutes.
Find practice problems and solutions at Distance Problems Practice Problems - Set 2.
Traveling Towards Eachother
When two bodies travel towards each other (from opposite directions) the rate at which the distance between them is shrinking is also the sum of their individual rates.
Dale left his high school at 3:45 and walked towards his brother’s school at 5 mph. His brother, Jason, left his elementary school at the same time and walked toward Dale’s high school at 3 mph. If their schools are 2 miles apart, when will they meet?
The rate at which the brothers are moving towards each other is 3 + 5 = 8 mph. Let t represent the number of hours the boys walk.
The boys will meet after an hour or 15 minutes; that is, at 4:00.
A jet airliner leaves Dallas going to Houston, flying at 400 mph. At the same time, another jet airliner leaves Houston, flying to Dallas, at the same rate. How long will it take for the two airliners to meet? (Dallas and Houston are 250 miles apart.) The distance between the jets is decreasing at the rate of 400 + 400 = 800 mph. Let t represent the number of hours they are flying.
The planes will meet after hours or minutes or 18 minutes 45 seconds.
Find practice problems and solutions at Distance Problems Practice Problems - Set 3.
Traveling at Right Angles
For distance problems in which the bodies are moving away from each other or toward each other at right angles (for example, one heading east, the other north), the Pythagorean Theorem is used. This topic will be covered in the last chapter.
Some distance problems involve the complication of the two bodies starting at different times. For these, you need to compute the head start of the first one and let t represent the time they are both moving (which is the same as the amount of time the second is moving). Subtract the head start from the distance in question then proceed as if they started at the same time.
A car driving eastbound passes through an intersection at 6:00 at the rate of 30 mph. Another car driving westbound passes through the same intersection ten minutes later at the rate of 35 mph. When will the cars be 18 miles apart?
The eastbound driver has a 10-minute head start. In 10 minutes ( hours), that driver has traveled miles. So when the westbound driver passes the intersection, there is already 5 miles between them, so the question is now “How long will it take for there to be 18 – 5 = 13 miles between two bodies moving away from each other at the rate of 30 + 35 = 65 mph?”
Let t represent the number of hours after the second car has passed the intersection.
In of an hour or minutes, an additional 13 miles is between them. So 12 minutes after the second car passes the intersection, there will be a total of 18 miles between the cars. That is, at 6:22 the cars will be 18 miles apart.
Two employees ride their bikes to work. At 10:00 one leaves work and rides southward home at 9 mph. At 10:05 the other leaves work and rides home northward at 8 mph. When will they be 5 miles apart?
The first employee has ridden miles by the time the second employee has left. So we now need to see how long, after 10:05, it takes for an additional miles to be between them. Let t represent the number of hours after 10:05. When both employees are riding, the distance between them is increasing at the rate of 9 + 8 = 17 mph.
After hour, or 15 minutes, they will be an additional miles apart. That is, at 10:20, the employees will be 5 miles apart.
Two boys are 1250 meters apart when one begins walking toward the other. If one walks at a rate of 2 meters per second and the other, who starts walking toward the first boy four minutes later, walks at the rate of 1.5 meters per second, how long will it take for them to meet?
The boy with the head start has walked for 4(60) = 240 seconds. (Because the rate is given in meters per second, all times will be converted to seconds.) So, he has traveled 240(2) = 480 meters. At the time the other boy begins walking, there remains 1250 – 480 = 770 meters to cover. When the second boy begins to walk, they are moving toward one another at the rate of 2 + 1.5 = 3.5 meters per second. The question becomes “How long will it take a body moving 3.5 meters per second to travel 770 meters?”
Let t represent the number of seconds the second boy walks.
The boys will meet 220 seconds, or 3 minutes 40 seconds, after the second boy starts walking.
A plane leaves City A towards City B at 9:10, flying at 200 mph. Another plane leaves City B towards City A at 9:19, flying at 180 mph. If the cities are 790 miles apart, when will the planes pass each other?
In 9 minutes the first plane has flown miles, so when the second plane takes off, there are 790 – 30 = 760 miles between them. The planes are traveling towards each other at 200 + 180 = 380 mph. Let t represent the number of hours the second plane flies.
Two hours after the second plane has left the planes will pass each other; that is, at 11:19 the planes will pass each other.
Find practice problems and solutions at Distance Problems Practice Problems - Set 4.
There are some distance/rate problems for which there are three unknowns. You must reduce the number of unknowns to one. The clues on how to do so are given in the problem.
A semi-truck traveled from City A to City B at 50 mph. On the return trip, it averaged only 45 mph and took 15 minutes longer. How far is it from City A to City B?
There are three unknowns—the distance between City A and City B, the time spent traveling from City A to City B, and the time spent traveling from City B to City A. We must eliminate two of these unknowns. Let t represent the number of hours spent on the trip from City A to City B. We know that it took 15 minutes longer traveling from City B to City A (the return trip), so represents the number of hours traveling from City B to City A. We also know that the distance from City A to City B is the same as from City B to City A. Let d represent the distance between the two cities. We now have the following two equations.
But if the distance between them is the same, then 50 t = Distance from City A to City B is equal to the distance from City B to City A = 45 . Therefore,
We now know the time, but the problem asked for the distance. The distance from City A to City B is given by d = 50 t , so . The cities are miles apart.
Another approach to this problem would be to let t represent the number of hours the semi spent traveling from City B to City A. Then would represent the number of hours the semi spent traveling from City A to City B. The equation to solve would be .
Kaye rode her bike to the library. The return trip took 5 minutes less. If she rode to the library at the rate of 10 mph and home from the library at the rate of 12 mph, how far is her house from the library?
Again there are three unknowns—the distance between Kaye’s house and the library, the time spent riding to the library and the time spent riding home. Let t represent the number of hours spent riding to the library. She spent 5 minutes less riding home, so represents the number of hours spent riding home. Let d represent the distance between Kaye’s house and the library.
The trip to the library is given by d = 10 t , and the trip home is given by . As these distances are equal, we have that .
The distance from home to the library is
Find practice problems and solutions at Distance Problems Practice Problems - Set 5.
In the above examples and practice problems, the number for t was substituted in the first distance equation to get d. It does not matter which equation you used to find d, you should get the same value. If you do not, then you have made an error somewhere.
More practice problems for this concept can be found at: Algebra Word Problems Practice Test.
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