Question

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?

Answer 1

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

Explanation:

Given:

Rate at which the woman is walking,
`(d(w))/(dt)` = 3 ft/sec

Rate at which the man is walking,
`(d(m))/(dt)` = 2 ft/sec

Collective rate of both,
`(d(m+w))/(dt)` = 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
`m` and
`w` ft respectively.

⇒
`m=2\ ft/sec=2* (60)/(min) * 20\ min =2400\ ft`

⇒
`w=3\ ft/sec = 3* (60)/(min) * 15\ min =2700\ ft`

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

Lets calculate h.

Applying Pythagoras formula.

⇒
`h^2=(m+w)^2+500^2`

⇒
`h=√((2400+2700)^2+500^2) = 5124.45`

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

Differentiating with respect to time.

⇒
`h^2=(m+w)^2+500^2`

⇒
`2h(d(h))/(dt)=2(m+w)(d(m+w))/(dt) + (d(500))/(dt)`

⇒
`(d(h))/(dt) =(2(m+w)(d(m+w))/(dt) )/(2h)` ...as
`(d(500))/(dt)= 0`

⇒ Plugging the values.

⇒
`(d(h))/(dt) =(2(2400+2700)(5))/(2* 5124.45)` ...as
`(d(m+w))/(dt) = 5` ft/sec

⇒
`(d(h))/(dt) =4.9761` ft/sec

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