the full question is below?

Question

a man starts walkin north at 4m/s from point P 5 minutes later, a woman starts walking south at 5m/s from a point 500m due east of p. at what rate are the people moving apart 15 minutes after the woman starts walking

Answer 1

The man has been walking north for 20 minutes, putting him 4800m from his starting point, and the woman has been walking south for 15 minutes, putting her 4500m from her starting point. On the N/S axis, they are now 9300m apart, but there is still the 500m difference on the E/W axis. The actual distance between them is the hypotenuse of that right triangle: approximately 9313.43m

So, since they are moving on the N/S axis a total of 9m/s from each other, you multiply 9m/s by the ratio (9300/9313.43) to get the vector of the speed they are moving in the direction of the distance between them, which is approximately 8.987m/s.

Answer 2

Let the point P be (0,0).

At time t, in seconds, the man's x-coordinate is 0, and his y-coordinate is 4t.

The woman's x-coordinate is 500, and her y-coordinate is 0, for t < 300, and -5t, for t >= 300 (because she starts walking 5 min = 300 seconds later).

The distance between them is
sqrt((xm-xw)sq + (ym-yw)sq) (sorry, for some reason my up-arrow key isn't working) where (xm,ym) are the man's coordinates, and (xw,yw) are the woman's.

Assuming t >= 300, this distance is

sqrt(500 + 9t*t)

To see how rapidly this distance is increasing, differentiate with respect to t, and you get

(1/2)(1/sqrt(500+9t*t))*18t.

Plug in t=20 min=1200 seconds to find the rate at which the distance is increasing 15 minutes after the woman starts walking.