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"Distance" Word Problems: More Examples (page 2 of 2)

  • Two cyclists start at the same time from opposite ends of a course that is 45 miles long. One cyclist is riding at 14 mph and the second cyclist is riding at 16 mph. How long after they begin will they meet?
    •   d r t
      slow guy d 14 t
      fast guy 45 – d 16 t
      total 45 --- ---


    Why is t the same for both cyclists? Because I am measuring from the time they both started to the time they meet somewhere in the middle. And how did I get "d" and "45 – d" for the distances? Because once I'd assigned the slow guy as having covered d miles, that left 45 – d miles for the fast guy to cover: the two guys together covered the whole 45 miles.

    Using "d = rt", I get d = 14t from the first row, and 45 – d = 16t from the second row. Since these distances add up to 45, I will add the distance expressions and set equal to the given total:

      45 = 14t + 16t

    Solve for t.

  • A boat travels for three hours with a current of 3 mph and then returns the same distance against the current in four hours. What is the boat's speed in calm water? How far did the boat travel one way?
    •   d r t
      downstream d b + 3 3
      upstream d b – 3 4
      total 2d --- 7

    (It may turn out that I won't need the "total" row.)

    I have used "b" to indicate the boat's speed. Why are the rates "b + 3" and "b – 3"? Because I actually have two speeds combined into one on each trip. The boat has a certain speed (the "speed in calm water" that I'm looking for; this is the speed that registers on the speedometer), and the water has a certain speed (this is the "current"). When the boat is going with the current, the water's speed is added to the boat's speed. This makes sense, if you think about it: even if you cut the engine, the boat would still be moving, because the water would be carrying it downstream. When the boat is going against the current, the water's speed is subtracted from the boat's speed. This makes sense, too: if the water is going fast enough, the boat will still be going downstream (a "negative" speed, because the boat would be going backwards at this point), because the water is more powerful than the boat. (Think of a boat in a cartoon heading toward a waterfall. The guy paddles like crazy, but he still goes over the edge.)

        d r t
      downstream d = 3(b + 3) b + 3 3
      upstream d = 4(b – 3) b – 3 4
      total 2d --- 7

    Using "d = rt", the first row (of the revised table above) gives me:

      d = 3(b + 3)

    The second row gives me:

      d = 4(b – 3)

    Since these distances are the same, I will set them equal:

      3(b + 3) = 4(b – 3)

    Solve for b. Then back-solve for d.

In this case, I didn't need the "total" row.

  • With the wind, an airplane travels 1120 miles in seven hours. Against the wind, it takes eight hours. Find the rate of the plane in still air and the velocity of the wind.
    •   d r t
      tailwind 1120 p + w 7
      headwind 1120 p – w 8
      total 2240 --- 15

    (I probably won't need the "total" row.) Just as with the last problem, I am really dealing with two rates together: the plane's speedometer reading, and the wind speed. When the plane turns around, the wind is no longer pushing the plane to go faster, but is instead pushing against the plane to slow it down.   Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved

    The first row gives me:

      1120 = 7(p + w)

    The second row gives me:

      1120 = 8(pw)

    The temptation is to just set these equal, like I did with the last problem, but that just gives me:

      7(p + w) = 8(pw)

    ...which doesn't help much. I need to get rid of one of the variables.

    I'll take that first equation:

      1120 = 7(p + w)

    ...and divide through by 7:

      160 = p + w

    Then, subtracting w from either side, I get that p = 160 – w. I'll substitute "160 – w" for "p" in the second equation:

      1120 = 8([160 – w]w)

      1120 = 8(160 – 2w)

    ...and solve for w. Then I'll back-solve to find p.

  • A spike is hammered into a train rail. You are standing at the other end of the rail. You hear the sound of the hammer strike both through the air and through the rail itself. These sounds arrive at your point six seconds apart. You know that sound travels through air at 1100 feet per second and through steel at 16,500 feet per second. How far away is that spike? (Round to one decimal place.)
    •   d r t
      air d = 1100t 1100 t
      steel d = 16,500(t – 6) 16,500 t – 6
      total --- --- 6

    However long the sound took to travel through the air, it took six seconds less to propagate through the steel. (Since the speed through the steel is faster, then that travel-time must be shorter.) I multiply the rate by the time to get the values for the distance column. (Once again, I didn't need the "total" row.)

    Since the distances are the same, I set the distance expressions equal to get:

      1100t = 16,500(t – 6)

    Solve for the time t, and then back-solve for the distance d by plugging t into either expression for the distance d.

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Cite this article as:

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


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