**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.

**Examples**

**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.

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.

**Examples**

**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.

Find practice problems and solutions at Distance Problems Practice Problems - Set 4.

### Ask a Question

Have questions about this article or topic? Ask### Related Questions

See More Questions### Popular Articles

- Kindergarten Sight Words List
- First Grade Sight Words List
- 10 Fun Activities for Children with Autism
- Signs Your Child Might Have Asperger's Syndrome
- Theories of Learning
- A Teacher's Guide to Differentiating Instruction
- Child Development Theories
- Social Cognitive Theory
- Curriculum Definition
- Why is Play Important? Social and Emotional Development, Physical Development, Creative Development