A person stands 12 meters east of an intersection and watches a car driving away from the intersection to the north at 4 meters per second. At a certain instant, the car is 9 meters from the intersection. What is the rate of change of the distance between the car and the person at that instant (in meters per second)? Choose 1 answer: (A) 2.4 (B) 4sqrt10 (C) 15 (D) (20)/(3)

Question

A person stands 12 meters east of an intersection and watches a car driving away from the intersection to the north at 4 meters per second. At a certain instant, the car is 9 meters from the intersection. What is the rate of change of the distance between the car and the person at that instant (in meters per second)? Choose 1 answer: (A) 2.4 (B)  4sqrt10 (C) 15 (D)  (20)/(3)

Bytelearn · Accepted Answer

Calculate Distance: Let's call the distance between the car and the person ＂d＂. Initially, the car is 9 meters from the intersection and the person is 12 meters from the intersection. Using the Pythagorean theorem, we have: d^2 = 12^2 + 9^2 d^2 = 144 + 81 d^2 = 225 d = sqrt(225) d = 15 meters.Rate of Change: Now, we need to find the rate of change of d with respect to time, which is ＂dd/dt＂. Since the car is moving away from the intersection at 4 meters per second, we can use related rates to find dd/dt. The car's distance from the intersection is increasing, so we'll call this rate ＂dx/dt＂ = 4 m/s.Related Rates: Differentiating both sides of the Pythagorean equation with respect to time, we get: 2d*(dd/dt) = 2*12*(0) + 2*9*(dx/dt) We know that the person is not moving, so their rate is 0, and dx/dt is 4 m/s. 2d*(dd/dt) = 0 + 2*9*4 2d*(dd/dt) = 72Solve for dd/dt: Now we solve for dd/dt: dd/dt = 72 / (2d) We already found that d = 15 meters, so we substitute that in: dd/dt = 72 / (2*15) dd/dt = 72 / 30 dd/dt = 2.4 meters per second.

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