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A policeman is standing near a highway using a radar gun to catch speeders. He aims the gun at a car that has just passed his position and, when the gun is pointing at an angle of 45(degrees) to the direction of the highway, notes that the distance between the car and the gun is increasing at a rate of 100km/h. How fast is the car travelling?

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2 Answers

4 votes

Final answer:

To find the speed of the car, the 45-degree angle and the rate of distance increase of 100 km/h are used; since the angle is 45 degrees, the rates of change along both the direction of the car and the radar gun are equal, so the car's speed is 100 km/h.

Step-by-step explanation:

The question involves finding the actual speed of the car using the information that the distance between the car and the radar gun is increasing at the rate of 100 km/h when the radar gun is pointed at a 45-degree angle relative to the highway's direction. This is a classic application of trigonometry and related rates in calculus.

Let's define v as the speed of the car along the highway and dr/dt as the rate at which the distance from the police officer to the car is increasing, which is given as 100 km/h. The car's motion and the line from the car to the police officer form a right-angled triangle, with the motion of the car forming the base. We're given that the radar gun is pointed at a 45-degree angle, which means the triangle is isosceles and the car’s speed along the highway (v) is equal to dr/dt.

Therefore, the speed of the car v is also 100 km/h. The key insight here is that because the angle is 45 degrees, the rates of change along both axes are the same.

User BvdVen
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The speed obtained is the y-component of the actual speed of the car. The y-component and actual speed are related by:
Vy = Vsin(∅)
100 = Vsin(45)
V = 141.4 km/h
User Maxim Efimov
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7.8k points