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Homework Help: This is a related rate problem.

  1. Sep 6, 2006 #1
    This is a related rate problem.

    A man starts walking north at 4 ft/s from a point P. Five minutes later a woman starts walking south at 5 ft/s from a point 500 ft due east of P. At what rate are the people moving apart 15 min after the woman starts walking?

    I'm just trying to interpret it into the numbers and drawing a triangle. But I don't get a right equations for it.

    the answer is 837/ sqaure root of 8674 which is 8.99 ft/s

    Thank you so much.
  2. jcsd
  3. Sep 7, 2006 #2
    First be careful of your units in this problem, the speeds are given in ft/s but the times are given in minutes. Second, you need to determine where the two are at when the lady starts walking. You should get a nice number for the "y" value of his position. Third, you need to find out where they are after the fifteen minutes have passed. Fourth, the use the Pythagorean theorem as the relationship between them, differentiate wrt time, but be careful because one of the time derivatives is zero. Can you see why? Finally, substitute the values you found earlier, you should end up with the rate of change between the two as the only unknown. I worked it out and the answer you have is what I got.
  4. Sep 7, 2006 #3


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    I work better thinking in coordinates than triangles so what I would do is set up a coordinate system with (0,0) at P, positive y-axis north, positive x-axis east.

    Then the man's position, t seconds after he starts walking is (0, 4t).

    The woman starts at (500,0) and walks south. Since she starts walking 5 min= 300 seconds later, at t seconds after the man starts walking she has been walking for t- 300 seconds. Her position will be (500, 5(t-300)). What is the distance from (0, 4t) to (500, 5(t-300)) as a function of t? What is the rate of change (derivative) of that function? And, as xman said, the question asks about 15 minutes after the woman started walking which is 20 minutes= 1200 seconds after the man started walking: t= 1200 s.
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