Two cars start moving from the same point. One travels south at 48 mi/h and the other travels west at 20 mi/h. At what rate is the distance between the cars increasing three hours later?

Answer 1

Let x(t) be distance of first car from starting point at time t

Let y(t) be distance of second car from starting point at time t

Let s(t) be distance between the two cars at time t. We need to evaluate ds/dt at time t = 3.

We know x²(t) + y²(t) = s²(t). Therefore 2x * dx/dt + 2y * dy/dt = 2s * ds/dt. The 2s cancel.

At t=3 hours, x = 144 = 12² miles. and y = 60 miles. We will need the distance between the cars at t=3:

s(3) = √x(3)² + y(3)² = √144² + 60² = √24336 = 156.

Now fill in x * dx/dt + y * dy/dt = s * ds/dt at time t = 3 and solve for ds/dt:

144 * 48 + 60 * 20 = 156 * ds/dt

ds/dt = 8112/156 = 52 miles/hr