At what time are they closest?

Two men are at opposite corners of a square block which is 500 feet on a side. They start to walk at the same time; one man walking east at the rate of 4 feet per second, and the other walks west at the rate of 4 feet per second. At what time are they closest?

Short answer: they will be closest when they are directly across from each other, i.e., at the middle. Thus, the required time is 250/4 = 62.5 seconds.

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Longer answer: superimpose a coordinate system such that the starting place of the man walking east is at the origin, (0, 0). The starting place of the second man is (500, 500).

After t seconds, the two men will walk 4t ft. Thus, after t seconds, the man walking east will be at (4t, 0) and the man walking west will be at (500 - 4t, 500).

By the Distance Formula, the distance between them is:

D = √[(500 - 4t - 4t)^2 + (500 - 0)^2] = √(64t^2 - 8000t + 500000).

Since D is minimized when 64t^2 - 8000t + 50000 is minimized and the minimum value of ax^2 + bx + c occurs at -b/(2a), the minimum distance occurs when:

t = -(-8000)/[2(64)] = 62.5;

in other words, the minimum distance occurs after 62.5 seconds.

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I hope this helps!

#1

62.5secs later

#2