Distance Problems Help

Example 1

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.

Example 2

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.

Example 1

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.

Example 2

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.

Example 3

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.

Example 4

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.

Example 1

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 .

Example 2

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