In math, distance, rate, and time are three important concepts you can use to solve many problems if you know the formula. Distance is the length of space traveled by a moving object or the length measured between two points. It is usually denoted by *d* in math problems.

The rate is the speed at which an object or person travels. It is usually denoted by *r* in equations. Time is the measured or measurable period during which an action, process, or condition exists or continues. In distance, rate, and time problems, time is measured as the fraction in which a particular distance is traveled. Time is usually denoted by *t* in equations.

### Solving for Distance, Rate, or Time

When you are solving problems for distance, rate, and time, you will find it helpful to use diagrams or charts to organize the information and help you solve the problem. You will also apply the formula that solves distance, rate, and time, which is *distance = rate x tim*e. It is abbreviated as:

d = rt

There are many examples where you might use this formula in real life. For example, if you know the time and rate a person is traveling on a train, you can quickly calculate how far he traveled. And if you know the time and distance a passenger traveled on a plane, you could quickly figure the distance she traveled simply by reconfiguring the formula.

### Distance, Rate, and Time Example

You'll usually encounter a distance, rate, and time question as a word problem in mathematics. Once you read the problem, simply plug the numbers into the formula.

For example, suppose a train leaves Deb's house and travels at 50 mph. Two hours later, another train leaves from Deb's house on the track beside or parallel to the first train but it travels at 100 mph. How far away from Deb's house will the faster train pass the other train?

To solve the problem, remember that *d* represents the distance in miles from Deb's house and *t* represents the time that the slower train has been traveling. You may wish to draw a diagram to show what is happening. Organize the information you have in a chart format if you haven't solved these types of problems before. Remember the formula:

distance = rate x time

When identifying the parts of the word problem, distance is typically given in units of miles, meters, kilometers, or inches. Time is in units of seconds, minutes, hours, or years. Rate is distance per time, so its units could be mph, meters per second, or inches per year.

Now you can solve the system of equations:

50t = 100(t - 2) (Multiply both values inside the parentheses by 100.)

50t = 100t - 200

200 = 50t (Divide 200 by 50 to solve for t.)t = 4

Substitute *t = 4* into train No. 1

d = 50t

= 50(4)= 200

Now you can write your statement. "The faster train will pass the slower train 200 miles from Deb's house."

### Sample Problems

Try solving similar problems. Remember to use the formula that supports what you're looking for—distance, rate, or time.

d = rt (multiply)

r = d/t (divide)

t = d/r (divide)

### Practice Question 1

A train left Chicago and traveled toward Dallas. Five hours later another train left for Dallas traveling at 40 mph with a goal of catching up with the first train bound for Dallas. The second train finally caught up with the first train after traveling for three hours. How fast was the train that left first going?

Remember to use a diagram to arrange your information. Then write two equations to solve your problem. Start with the second train, since you know the time and rate it traveled:

Second traint x r = d

3 x 40 = 120 milesFirst traint x r = d8 hours x r = 120 milesDivide each side by 8 hours to solve for r.8 hours/8 hours x r = 120 miles/8 hoursr = 15 mph

### Practice Question 2

One train left the station and traveled toward its destination at 65 mph. Later, another train left the station traveling in the opposite direction of the first train at 75 mph. After the first train had traveled for 14 hours, it was 1,960 miles apart from the second train. How long did the second train travel? First, consider what you know:

First trainr = 65 mph, t = 14 hours, d = 65 x 14 milesSecond trainr = 75 mph, t = x hours, d = 75x miles

Then use the *d = rt *formula as follows:

d (of train 1) + d (of train 2) = 1,960 miles

75x + 910 = 1,960

75x = 1,050x = 14 hours (the time the second train traveled)