AP Calc Problem?

Question

2 cars start at the same place and at the same time. One car travels west at a constant velocity of 50 miles pre hour and a second car travels south at a constant velocity of 60 miles per hour. Approximately how fast is the distance between them changing one-half hour later.

possible answers are 72, 74, 76, 78, 0r 80 miles per hour, i got 78

Answer 1

The cars are following two legs of a right triangle, and the distance between them is the hypotenuse. If the time elapsed is t, we can find the distance between them at any given time:

c^2 = a^2 + b^2

--or--

distance^2 = west^2 + south^2
distance^2 = (50t)^2 + (60t)^2
distance^2 = 2500t^2 + 3600t^2
distance^2 = 6100t^2
distance = 10tsqrt(61)

The problem is a related rate problem, so we need to take a derivative of this with respect to time t:

d(distance)/dt = 10sqrt(61)
d(distance)/dt = 78

So it turns out we really don't care about the "one-half hour later" part - the distance is always changing by 78 miles per hour.