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Cutting up Boards

Walk and Ride

Joan takes a walk at 4 mph and then gets a ride back at 16 mph. The entire trip took 1/2 hour. How far did Joan walk? The solution of this problem depends on the realization that three inter-related relationships are involved. First is the rate formula, time X rate =distance. Entering the pertinent data in a table with row labels walk and ride gives:
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|     |time|rate|distance|
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|walk |    |  4 |        |
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|ride |    | 16 |        |
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Certainly more information is needed! A second relationship, walk time + ride time = total time, can be used to relate the times, but requires adding a row to the table, giving:
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|     |time|rate|distance|
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|walk |    |  4 |        |
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|ride |    | 16 |        |
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|total| 0.5|    |        |
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The third relationship is more subtle. In particular it must be recognized that the distance traveled by walking is equal to the distance traveled by riding. But there are no distance entries in the table. They could be obtained, however, from the rate equation, if the times were known. The times can be determined algebracically by letting the "walk time" be the unknown 'u'. Then the time relationship allows us to solve for "ride time" which is "0.5 - u". Entering these values and calculating the distances gives:
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|     |time |rate| distance |
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|walk |  u  |  4 |   4*u    |
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|ride |0.5-u| 16 |16*(0.5-u)|
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|total| 0.5 |    |          |
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Replaceing the values 4, 16 and 0.5 by their cell names and setting the two distances equal to each other produces:
        (walk time)X(walk rate) = (ride rate)X(total time - walk time)
Solving for walk time - produces the first statement of the program:
        walk time = (ride rate X total time)/(walk rate + ride rate);
With walk time known, the distance is easily calculated:
        walk distance = walk time X walk rate;

Return to the previous page to see the table, program and result.