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
