7.98 ft/second
You can create a right triangle to express the distance between the man and the woman. One leg of the triangle remains constant at 500 ft. (If you think about problem, the east/west separation between the man and the woman will remain constant. And that starts at 500 feet). So the other leg of the triangle needs to be calculated.
The man is walking north for 20 minutes (5 minutes before the woman starts and 15 minutes after the woman started), so his distance north is 20 * 60 * 2 = 2400 ft. The woman has been walking south for 15 minutes, so her distance south is 15 * 60 * 6 = 5400 ft. So their exact north south separation is 2400 + 5400 = 7800 feet. And finally, the north south separation between the man and woman continues to increase at the rate of 8 ft per second.
Now let's express the distance between the man and woman at time x where x is the number of seconds from where the woman has already walked for 15 minutes. That would be
sqrt((7800 + 8x)^2 + 500^2)
sqrt((7800 + 8x)^2 + 250000)
Since we're looking for rate of change, that screams "first derivative". So let's calculate the first derivative.
sqrt((7800 + 8x)^2 + 250000)
= 1/2((8x + 7800)^2 + 250000)^(1/2) * d/dx((8x + 7800)^2 + 250000)
= (d/dx((8x + 7800)^2 + d/dx(250000))/(2*sqrt((8x+7800)^2 + 250000))
= (2(8x + 7800) * d/dx(8x + 7800))/(2*sqrt((8x+7800)^2 + 250000))
= (8 * d/dx(x) + d/dx(7800))(8x + 7800)/sqrt((8x+7800)^2 + 250000)
= 8(8x + 7800)/sqrt((8x+7800)^2 + 250000)
= (64x + 62400)/sqrt((8x+7800)^2 + 250000)
Now let's calculate the value of the first derivative at x=0, so
(64x + 62400)/sqrt((8x+7800)^2 + 250000)
=(64*x + 62400)/sqrt((8*0+7800)^2 + 250000)
=62400/sqrt(7800^2 + 250000)
=62400/sqrt(60840000 + 250000)
=62400/sqrt(61090000)
=62400/7816.009
=7.983613927
So the rate at which the man and woman are moving away from each other is 7.98 ft/second.
