**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.

**Examples**

**Example 1 **

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.

**Example 2**

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.

**Examples**

**Example 1 **

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.

**Example 2**

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.

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