A person stands 151515 meters east of an intersection and watches a car driving towards the intersection from the north at 111 meter per second. At a certain instant, the car is 888 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)

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jtellezd jtellezd · Answer 1 · 2021-01-13T03:45:15+01:00

Answer:

dL/dt = 110,98 m/s

Step-by-step explanation:

Let´s call A the point where a person is, the intersection (o) and the car from the north form a right triangle. According to Pythagoras Theorem

L² = x² + y² where L is the distance between the person and the car ( x is constant since the person does not move)

Tacking derivatives with respect to time we have:

2*L*dL/dt = 2*y*dy/dt (1)

x = 15 mts y = 888 and dy/dt = 111 m/s

L² = x² + y²

L = √ (15)² + (888)² ⇒ L = √ 225 + 788544

L = 888,13 m

2*L* dL/dt = 2* (888)*111

dL/dt = (888)*111/888,13

dL/dt =0,99*111

dL/dt = 110,98 m/s