Rate is the distance between the cars increasing two hours later is 52 miles/hr
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 = 2.
We know x2(t) + y2(t) = s2(t). Therefore 2x * dx/dt + 2y * dy/dt = 2s * ds/dt. The 2s cancel.
At t=2 hours, x = 96 miles. and y = 40 miles.
We will need the distance between the cars at t=2:
s(2) = √x(2)2 + y(2)2 = √96 2 + 40 2 = √10816 = 104.
Now fill in x * dx/dt + y * dy/dt = s * ds/dt at time t = 3 and solve for ds/dt:
96 * 48 + 40 * 20 = 104 * ds/dt
ds/dt = 5408/104 = 52 miles/hr
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