A man starts walking north at 2 ft/s from a point P. Five minutes later a woman starts walking south at 3 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?

The rate at which both of them are moving apart is 4.9761 ft/sec.

Step-by-step explanation:

Given:

Rate at which the woman is walking, = 3 ft/sec

Rate at which the man is walking, = 2 ft/sec

Collective rate of both, = 5 ft/sec

Woman starts walking after 5 mins so we have to consider total time traveled by man as (5+15) min  = 20 min

Now,

Distance traveled by man and woman are and ft respectively.

⇒

⇒

As we see in the diagram (attachment) that it forms a right angled triangle and we have to calculate .

Lets calculate h.

Applying Pythagoras formula.

⇒  

⇒

Now differentiating the Pythagoras formula we can calculate the rate at which both of them are moving apart.

Differentiating with respect to time.

⇒

⇒

⇒                         ...as

⇒ Plugging the values.

⇒                       ...as ft/sec

⇒  ft/sec

So the rate from which man and woman moving apart is 4.9761 ft/sec.

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