Two cars are driving away from an intersection in perpendicular directions. The first car's velocity is 7 meters per second and the second car's velocity is 3 meters per second. At a certain instant, the first car is 5 meters from the intersection and the second car is 12 meters from the intersection. What is the rate of change of the distance between the cars at that instant (in meters per second)? Choose 1 answer: (A) (99)/(13) (B) 13 (C) sqrt58 (D) (71)/(13)

Question

Two cars are driving away from an intersection in perpendicular directions. The first car's velocity is 7 meters per second and the second car's velocity is 3 meters per second. At a certain instant, the first car is 5 meters from the intersection and the second car is 12 meters from the intersection. What is the rate of change of the distance between the cars at that instant (in meters per second)? Choose 1 answer: (A)  (99)/(13) (B)  13 (C)  sqrt58 (D)  (71)/(13)

Bytelearn · Accepted Answer

Identify Variables: Now we'll find the rate of change of the distance between the cars using derivatives. Let's call the distance between the cars 's', the distance of the first car from the intersection 'x', and the distance of the second car from the intersection 'y'. According to the Pythagorean theorem, s^2 = x^2 + y^2. Differentiating both sides with respect to time 't', we get: 2s(ds/dt) = 2x(dx/dt) + 2y(dy/dt) Now we plug in the velocities of the cars for dx/dt and dy/dt, and the current distance 's' to find ds/dt. 2*13(ds/dt) = 2*5*7 + 2*12*3 26(ds/dt) = 70 + 72 26(ds/dt) = 142 ds/dt = 142 / 26 ds/dt = (71)/(13) meters per second

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