Answer:
Both are moving apart with the rate of 8.99 feet per sec.
Step-by-step explanation:
From the figure attached,
Man is walking north with the speed = 4 ft per second
[tex]\frac{dx}{dt}=4[/tex] feet per sec.
Woman starts walking due south with the speed = 5ft per second
[tex]\frac{dy}{dt}=5[/tex] ft per sec.
We have to find the rate of change in distance z.
From the right angle triangle given in the figure,
[tex]z^{2}=(x+y)^{2}+(500)^{2}[/tex]
We take the derivative of the given equation with respect to t,
[tex]2z.\frac{dz}{dt}=2(x+y)(\frac{dx}{dt}+\frac{dy}{dt})+0[/tex] -----(1)
Since distance = speed × time
Distance covered by woman in 15 minutes or 900 seconds = 5(900) = 450 ft
y = 4500 ft
As the man has taken 5 minutes more, so distance covered by man in 20 minutes or 1200 sec = 4×1200 = 4800 ft
x = 4800 ft
Since, z² = (500)² + (x + y)²
z² = (500)² + (4500 + 4800)²
z² = 250000 + 86490000
z = √86740000
z = 9313.43 ft
Now we plug in the values in the formula (1)
2(9313.43)[tex]\frac{dz}{dt}[/tex] = 2(4800 + 4500)(4 + 5)
18626.86[tex]\frac{dz}{dt}[/tex] = 18(9300)
[tex]\frac{dz}{dt}=\frac{167400}{18626.86}[/tex]
[tex]\frac{dz}{dt}=8.99[/tex] feet per sec.
Therefore, both the persons are moving apart by 8.99 feet per sec.
