A man starts walking north at 5 ft/s from a point P. Five minutes later a woman starts walking south at 4 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?
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2 answers

Let the time passes since the woman started walking be t seconds
So the an has walked for t+300 seconds, let his position be A on a NSEW grid
let the woman's position after t seconds be B
Draw AB, the distance between them.
Complete a rightangled triangle by extending AP downwards to a point C, so that PC is the woman's distance and CB= 500
AP = 5(t+300) = 5t + 1500
PC = 4t
AC = 9t+1500
AB^2 = (9t+1500)^2 + 500^2
when t = 15 min = 900 sec
2AB d(AB)/dt = 2(9t+1500)(9) + 0
d(AB)/dt = 9(9t+1500)/AB
so when t=15 min = 900sec
AB^2 = 92160000 + 250000
AB = 9613.01 ft
d(AB)/dt = 9(9600)/9613.01
= 8.99 ft/sec 👍
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Thank you very much!
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