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Example 1. Police are 30 feet from the side of the road. Their radar sees your car approaching at 80 feet per second when your car is 50 feet away from the radar gun. The speed limit is 65 miles per hour (which translates to 95 feet per second). Are you speeding?

First, draw a diagram of the setup (as in Fig. 1):

Fig1

Figure 1: Illustration of example 1: triangle with the police, the car, the road, D and x labelled. Next, give the variables names. The dD At D = 50, x = 40. (We know this because it’s a 3-4-5 right triangle.) In addition, D' =d(D)/ dt =−80. D' is negative because the car is moving in the −x direction.

My question is where does the relation d(D)/ dt =−80 come from ? Since v and t are normally related to X and not D, I was expacting x=-80t.

Thank you

TiyebM
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1 Answers1

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$\dfrac{dx}{dt}$ (actually its magnitude) is your actual speed (along the direction you are moving).

Magnitude of $\dfrac{dD}{dt}$ is the speed observed by the radar gun. It works on the Doppler effect principle, so it measures speeds going towards or away from it, so along the line joining the police and the car.

You can use the relation $D^2 = x^2 + 900$ and differentiate it to find the value of $\dfrac{dx}{dt}$

ab123
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  • Thank you, I understand the difference between the two speeds, but since the movement is on the road, the 80f/s speed is belong to the actual speed and so to the equation dx/dt, dD/dt should be calculated using the other pythagorean equation? – TiyebM Sep 15 '19 at 12:26
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    @TiyebM No, I think you misunderstood. $\dfrac{dD}{dt}$ is the speed along direction of the radar gun and that direction is along the line joining the police and the car. Since I said that radar gun only notices speeds going toward or away from it, it measures the rate of change of distance "along the line it sees" which means that it measures $\dfrac{dD}{dt}$. $\implies \dfrac{dD}{dt} = -80 f/s$. – ab123 Sep 16 '19 at 05:06