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A man starts walking north at 4 ft/s from a point P. Five minutes later a woman starts walking south at 5 ft/s from a point 500 ft due east of P. At what rate are the people moving apart 15 minutes after the woman starts walking

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Answer:

Both are moving apart with the rate of 8.99 feet per sec.

Explanation:

From the figure attached,

Man is walking north with the speed = 4 ft per second


(dx)/(dt)=4 feet per sec.

Woman starts walking due south with the speed = 5ft per second


(dy)/(dt)=5 ft per sec.

We have to find the rate of change in distance z.

From the right angle triangle given in the figure,


z^(2)=(x+y)^(2)+(500)^(2)

We take the derivative of the given equation with respect to t,


2z.(dz)/(dt)=2(x+y)((dx)/(dt)+(dy)/(dt))+0 -----(1)

Since distance = speed × time

Distance covered by woman in 15 minutes or 900 seconds = 5(900) = 450 ft

y = 4500 ft

As the man has taken 5 minutes more, so distance covered by man in 20 minutes or 1200 sec = 4×1200 = 4800 ft

x = 4800 ft

Since, z² = (500)² + (x + y)²

z² = (500)² + (4500 + 4800)²

z² = 250000 + 86490000

z = √86740000

z = 9313.43 ft

Now we plug in the values in the formula (1)

2(9313.43)
(dz)/(dt) = 2(4800 + 4500)(4 + 5)

18626.86
(dz)/(dt) = 18(9300)


(dz)/(dt)=(167400)/(18626.86)


(dz)/(dt)=8.99 feet per sec.

Therefore, both the persons are moving apart by 8.99 feet per sec.

A man starts walking north at 4 ft/s from a point P. Five minutes later a woman starts-example-1
User Vishist Varugeese
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