Multiple choice

A police cruiser, approaching a right-angled intersection from the north, is chasing a speeding car that has turned the corner and is now moving straight east. When the cruiser is $0.6$ km north of the intersection and the car is $0.8$ km to the east, the police determine with radar that the distance of the car is increasing at $20$ km/h. Suppose that the cruiser is moving at $60$ km/h at the instant of measurement. The speed of the car is (in km/h)

  1. $70$
  2. $80$
  3. $75$
  4. $60$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let cruiser position be (0, y) and car be (x, 0). Distance D = sqrt(x^2 + y^2). D^2 = x^2 + y^2. 2D * dD/dt = 2x * dx/dt + 2y * dy/dt. D = sqrt(0.6^2 + 0.8^2) = 1.0. dD/dt = 20. dx/dt = v_car. dy/dt = -60 (cruiser moving towards intersection). 2 * 1 * 20 = 2 * 0.8 * v_car + 2 * 0.6 * (-60). 40 = 1.6 * v_car - 72. 112 = 1.6 * v_car. v_car = 112 / 1.6 = 70.

AI explanation

Using the Pythagorean theorem, the distance between them is 1 km. By Pythagoras applied to relative speeds, the equation is 0.8 multiplied by car speed plus 0.6 multiplied by 60 equals 1 multiplied by 20. Solving gives car speed equals 70 km/h.

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