Question

Two cars start moving from the same point. one travels south at 60 mi/h and the other travels west at 25 mi/h. at what rate is the distance between the cars increasing four hours later?

Answer 1

Oops, I missed they we're moving perpendicularly. Gotta be more careful.  I'm gonna blame this immovable answer box popping up and hiding the question on my browser. Plus I skimmed it.

We wont worry about the signs; they won't matter.  We have car 1  (x,y)=(0,60t) and car 2 (x,y)=(25t,0).  So the squared distance is

$d^2 = (25t)^2 + 60^2 = 5^2((5t)^2 + (12t)^2) = 5^2 13^2 t^2$

recognizing the Pythagorean Triple.  Everything's positive so

$d = 65 t$

So they're moving apart at a constant 65 mph.

Answer: 65 mph

Answer 2

To answer this question you need to write the distance with parameter t (time)

At any time t the first cars distance from the starting point (origin) is 60t

The second cars distance from origin is 25t

The two cars travel in perpendicular direction.

The distance between the two cars can be found using Pythagoras theorem. Draw it if that helps you visualize.

So the distance squared is equal to (60t)^2+(25t)^2 = 65^2 * t^2

Thus D = 65t

This is a linear function so the rate of change is constant, it is 65.