Answer:
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,[tex]\frac{d(w)}{dt}[/tex] = 3 ft/sec
Rate at which the man is walking,[tex]\frac{d(m)}{dt}[/tex] = 2 ft/sec
Collective rate of both, [tex]\frac{d(m+w)}{dt}[/tex] = 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 [tex]m[/tex] and [tex]w[/tex] ft respectively.
⇒ [tex]m=2\ ft/sec=2\times \frac{60}{min} \times 20\ min =2400\ ft[/tex]
⇒ [tex]w=3\ ft/sec = 3\times \frac{60}{min} \times 15\ min =2700\ ft[/tex]
As we see in the diagram (attachment) that it forms a right angled triangle and we have to calculate [tex]\frac{dh}{dt}[/tex] .
Lets calculate h.
Applying Pythagoras formula.
⇒ [tex]h^2=(m+w)^2+500^2[/tex]
⇒ [tex]h=\sqrt{(2400+2700)^2+500^2} = 5124.45[/tex]
Now differentiating the Pythagoras formula we can calculate the rate at which both of them are moving apart.
Differentiating with respect to time.
⇒ [tex]h^2=(m+w)^2+500^2[/tex]
⇒ [tex]2h\frac{d(h)}{dt}=2(m+w)\frac{d(m+w)}{dt} + \frac{d(500)}{dt}[/tex]
⇒ [tex]\frac{d(h)}{dt} =\frac{2(m+w)\frac{d(m+w)}{dt} }{2h}[/tex] ...as [tex]\frac{d(500)}{dt}= 0[/tex]
⇒ Plugging the values.
⇒ [tex]\frac{d(h)}{dt} =\frac{2(2400+2700)(5)}{2\times 5124.45}[/tex] ...as [tex]\frac{d(m+w)}{dt} = 5[/tex] ft/sec
⇒ [tex]\frac{d(h)}{dt} =4.9761[/tex] ft/sec
So the rate from which man and woman moving apart is 4.9761 ft/sec.
