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"Distance" Word Problems (page 1 of 2)

"Distance" word problems, often also called "uniform rate" problems, involve something travelling at some fixed and steady ("uniform") pace ("rate" or "speed"), or else moving at some average speed. Whenever you read a problem that involves "how fast", "how far", or "for how long", you should think of the distance equation, d = rt, where d stands for distance, r stands for the (constant or average) rate of speed, and t stands for time.

Warning: Make sure that the units for time and distance agree with the units for the rate. For instance, if they give you a rate of feet per second, then your time must be in seconds and your distance must be in feet. Sometimes they try to trick you by using the wrong units, and you have to catch this and convert to the correct units.

  • A 555-mile, 5-hour plane trip was flown at two speeds. For the first part of the trip, the average speed was 105 mph. Then the tailwind picked up, and the remainder of the trip was flown at an average speed of 115 mph. For how long did the plane fly at each speed?

    First I'll set up a grid:   Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved

        d r t
      first part d 105 t
      second part 555 – d 115 5 – t
      total 555 --- 5

    Using "d = rt", the first row gives me d = 105t and the second row gives me:

      555 – d = 115(5 – t)

    Since the two distances add up to 555, I'll add the two distance expressions, and set their sum equal to the given total:


      555 = 105t + 115(5 – t)

    Then I'll solve:

      555 = 105t + 575 – 115t
      555 = 575 – 10t
      –20 = –10t
      2 = t

    According to my grid, "t" stands for the time spent on the first part of the trip, so my answer is "The plane flew for two hours at 105 mph and three hours at 115 mph."

You can add distances and you can add times, but you cannot add rates. Think about it: If you drive 20 mph on one street, and 40 mph on another street, does that mean you averaged 60 mph?

As you can see, the actual math involved is often quite simple. It's the set-up that's the hard part. So what follows are some more examples, but with just the set-up displayed.

  • An executive drove from home at an average speed of 30 mph to an airport where a helicopter was waiting. The executive boarded the helicopter and flew to the corporate offices at an average speed of 60 mph. The entire distance was 150 miles; the entire trip took three hours. Find the distance from the airport to the corporate offices.
    •   d r t
      driving d 30 t
      flying 150 – d 60 3 – t
      total 150 --- 3

    The first row gives me the equation d = 30t. Since the first part of his trip accounted for d miles of the total 150-mile distance and t hours of the total 3-hour time, I am left with 150 – d miles and 3 – t hours for the second part. The second row gives me the equation:

      150 – d = 60(3 – t)

    Adding the two "distance" expressions and setting their sum equal to the given total distance, I get:

      150 = 30t + 60(3 – t)

    Solve for t; interpret the value; state the final answer.

  • A car and a bus set out at 2 p.m. from the same point, headed in the same direction. The average speed of the car is 30 mph slower than twice the speed of the bus. In two hours, the car is 20 miles ahead of the bus. Find the rate of the car.
    •   d r t
      car d + 20 2r – 30 2
      bus d r 2
      total --- --- ---

    (As it turns out, I won't need the "total" row this time.) The first row gives me:

      d + 20 = 2(2r – 30)

    This is not terribly helpful. The second row gives me:

      d = 2r

    Use the second equation to simplify the first equation by substituting "2r" in for "d", and then solve for "r". Interpret this value within the context of the exercise, and state the final answer.

  • A passenger train leaves the train depot 2 hours after a freight train left the same depot. The freight train is traveling 20 mph slower than the passenger train. Find the rate of each train, if the passenger train overtakes the freight train in three hours.
    •   d r t
      passenger train d r 3
      freight train d r – 20 3 + 2 = 5
      total --- --- ---

    (As it turns out, I won't need the "total" row this time.) Why is the distance just "d" for both trains? Partly, that's because the problem doesn't say how far the trains actually went. But mostly it's because they went the same distance as far as I'm concerned, because I'm only counting from the depot to wherever they met. After that meet, I don't care what happens. And how did I get those times? I know that the passenger train drove for three hours to catch up to the freight train; that's how I got the "3". But note that the freight train had a two-hour head start. That means that the freight train was going for five hours.

        d r t
      passenger train d = 3r r 3
      freight train d = 5(r – 20) r – 20 3 + 2 = 5
      total --- --- ---

    Now that I have this information, I can try to find my equation. Using the fact that d = rt, the first row gives me d = 3r (note the revised table above). The second row gives me:

      d = 5(r – 20)

    Since the distances are equal, I will set the equations equal:

      3r = 5(r – 20)

    Solve for r; interpret the value within the context of the exercise, and state the final answer.

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Cite this article as:

Stapel, Elizabeth. "'Distance' Word Problems." Purplemath. Available from Accessed


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