At what rate are the people moving apart 15 min after the woman starts walking?

Question

A man starts walking north at 6 ft/s from a point P. Five minutes later, a woman starts walking south at 5 ft/s from a point 300 ft due east of P. At what rate are the people moving apart 10 min after the woman starts walking?

Answer 1

Let
n = distance man walks north
s = distance woman walks south
v = vertical distance apart
D = total distance apart
t = time in seconds

Given
dn/dt = 6 ft/sec
ds/dt = 5 ft/sec

Find dD/dt at 10 minutes after woman starts walking.
That would be t = 15*60 = 900 seconds after man starts walking.

We have

n = 6t
s = 5(t - 300) = 5t - 1500
v = n + s = 11t - 1500

dv/dt = dn/dt + ds/dt = 6 + 5 = 11 ft/sec

D² = 300² + v² = 300² + (11t - 1500)²

D² = 90,000 + 121t² - 33,000t + 2,250,000

D² = 121t² - 33,000t + 2,340,000

2D(dD/dt) = 242t - 33,000

dD/dt = (242t - 33,000) / (2D) = (121t - 16,500) / D

dD/dt = (121t - 16,500) / √(121t² - 33,000t + 2,340,000)

at t = 15 min = 15*60 = 900 seconds:

dD/dt = 92,400 / √70,650,000 = 924 / √7065 ≈ 10.99 ft/sec

Answer 2

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