"Two cars are traveling along perpendicular roads, car A at 40mph, car B at 60mph. At noon, when car A reaches the intersection, car B is 90miles away, and moving toward it. At 1PM the distance between the cars is changing at what rate?"

Question

So lost! Please help!

Answer 1

The distance d between A and B at noon is 90

At time t thereafter,

d^2 = (90-60t)^2 + (40t)^2
At 1:00, d^2 = 30^2 + 40^2 = 50^2, so d = 50

2d d' = 2(90-60t)*(-60) + 2(40t)*(40)
100 d' = 2(30)(-60) + 2(40)(40) = -3600 + 3200 = -400
d' = -4

so, the cars are moving closer at 4 mph.

FYI, d'(1.038) = 0, so at 1:02 the distance starts to increase.