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?

Answer 1

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 NS-EW grid
let the woman's position after t seconds be B
Draw AB, the distance between them.
Complete a right-angled 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

Answer 2

Thank you very much!