posted by AH on .
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?
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
Thank you very much!