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April 20, 2014

April 20, 2014

Posted by **AH** on Friday, November 2, 2012 at 1:07pm.

- Math -
**Reiny**, Friday, November 2, 2012 at 5:16pmLet 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

- Math -
**AH**, Saturday, November 3, 2012 at 12:50amThank you very much!

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