A man starts walking north at 3 ft/s from a point P. Five minutes later a woman starts walking south at 4 ft/s from a point 500

Question

3 years ago

14

ft due east of P. At what rate are the people moving apart 15 min after the woman starts walking?

1 answer:

mrs_skeptik [129] · Answer 1 · 2022-07-02T12:50:54+03:00

Answer:

ds/dt = 6.98 ft/s

Explanation:

Given:

- The speed of man due north Vm = 3 ft/s

- The speed of woman due south Vw = 4 ft/s

- Woman starts walking 5 mins later than man

Find:

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

Solution:

- The total time for which the man is walking due north from P, is Tm:

Tm = 5 + 15 = 20 mins

- The total distance traveled by man in Tm mins is:

Dm = Tm*Vm

Dm = 20*60*3

Dm = 3,600 ft

- The total time for which the woman is walking due south from 500 ft due east from P, is Tw:

Tw = 15 = 15 mins

- The total distance traveled by man in Tw mins is:

Dw = Tw*Vw

Dw = 15*60*4

Dw = 3,600 ft

- The displacement between man and woman at any instance is (s) which can be related by pythagoras theorem as follows:

s^2 = (dm + dw)^2 + 500^2

Where, dm : Distance travelled by man at any time Tm

dw : Distance travelled by woman at any time Tw

- Differentiate s with respect to t:

2s*ds/dt = 2*(dm + dw)*(Vm + Vw)

s*ds/dt = (dm + dw)*(Vm + Vw)

ds/dt = [ (dm + dw)*(Vm + Vw) ] / s

- Evaluate the rate of separation of man and woman ds/dt by evaluating at instance Tm = 20 mins and Tw = 15 mins. We have:

ds/dt = [ (Dm + Dw)*(Vm + Vw) ] / sqrt ( (Dm + Dw)^2 + 500^2 )

- Plug in the values:

ds/dt = [ (3600 + 3600)*(3 + 4) ] / [sqrt ( (3600 + 3600)^2 + 500^2 )]

ds/dt = 6.98 ft/s