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A police car is located 30 feet to the side of a straight road. A red car is driving along the road in the direction of the police car and is 120 feet up the road from the location of the police car. The police radar reads that the distance between the police car and the red car is decreasing at a rate of 95 feet per second. How fast is the red car actually traveling along the road

physics
Guest Mar 7, 2015

Best Answer 

 #10
avatar+889 
+5

Actually Melody, my first method is exactly equivalent to Alan's method.

The equation ds/dt = (D/s)(dD/dt) can be written as (s/D)(ds/dt) = dD/dt, which, in Alan's notation is

v.cos(theta)=95, (with or without the negative sign)

Bertie  Mar 9, 2015
 #1
avatar+92623 
0

Alan, you have divided by cos 14 and I have multiplied by cos14,

could you please explain to me where my logic is faulty?

 

 

We have here a right angled triangle with legs 30' and 120'. So they are in the ratio  1:4

For a right triangle with legs 1 and 4, the hypenuse will be  $$\sqrt{17}$$

 

 

$$\\\frac{x}{95}=\frac{4}{\sqrt{17}}\\\\
x=\frac{4}{\sqrt{17}}\times 95\\\\$$

 

$${\frac{{\mathtt{4}}}{{\sqrt{{\mathtt{17}}}}}}{\mathtt{\,\times\,}}{\mathtt{95}} = {\mathtt{92.163\: \!537\: \!513\: \!806\: \!529\: \!9}}$$

 

so maybe the car is travelling at     $$92\;feet/ sec$$

 

The distance travelled is less in the same amount of time so that also makes me think the velocity should be less than the clocked 95' /sec

I am really confused 

 

Melody  Mar 7, 2015
 #2
avatar+26716 
+5

The measured speed would be greater if the radar were measuring the speed of the car head on, so it should be that v*cos(θ) = 95, where v is the car's speed and θ is the angle between the road and the line from the initial position of the car to the police car.

 

θ = tan-1(30/120) ≈ 14°, so car speed v ≈ 95/cos(14°) = 97.9 ft/s

.

Alan  Mar 7, 2015
 #3
avatar+92623 
0

 

Thanks Alan,

I have altered my answer a little but I still do not understand what is wrong with my logic.

Can you please take a look and talk me through it?     Thank you.   

Melody  Mar 7, 2015
 #4
avatar+26716 
+5

Melody, this diagram describes the situation as I see it (note: I picture the car driving from left to right rather than bottom to top!):

 

radar speed

 

Note that the red vector is longer than the blue vector.

 

Imagine that the radar gun was a further 120 ft to the left (i.e. immediately above the initial position of the car), then it would detect zero speed.  If directly in front of the car (i.e. on the road) then it would detect the full speed of the car.  In its specified position it detects a component of the speed.  It can't detect a speed greater than that of the car!

Alan  Mar 7, 2015
 #5
avatar+1037 
+5

This is known as P*G TRIG.

 

 

The graph below may help to elaborate on Alan’s explanation.

Radar guns detect the Doppler shift in frequency to determine speed. The Doppler shift is proportional to the speed of the target’s approach or retreat from the receiver. The moving target will appear to slow and stop as the angle approaches 90, then accelerate again. At this point the cosine values are negative and the Doppler shift in frequency is now negative with respect to the broadcast frequency.

 

Basic trig functions are used to reconcile the two vector components of a radar beam. The cosine division returns the X vector component of the actual speed. This may seem counterintuitive because the target is not moving in the Y direction, but the radar beam does move in the Y direction relative to the target -- unless it is dead on, then the indicated speed equals the true speed of the target.

 

Obtaining a dead on fix requires standing in the path of the target. There are cops dumb enough to do this very thing, especially in the early days of this technology. The smarter cops stood in the retreating path. For the former, it’s pigs in a blanket; for the latter, another dozen doughnuts.

 

Teaching pigs (physics) trig is almost as annoying as teaching them to sing.

https://www.youtube.com/watch?feature=player_detailpage&v=rfmxx_72Coc#t=105

 

 

Nauseated  Mar 7, 2015
 #6
avatar+92623 
+5

Thanks Alan and Nauseated.

I now get that the radar will have to clock the oncoming car at a smaller speed then what the car is really travelling at.

Each of your pics gave me different info and they both helped a lot .

Alan I can see that your pic gives me the desired answer but  I can't quite fathom why the pic is right.  

I am still contemplating the actual maths of it.  Is it possible for you to show me mathematically, with the units, why your answer is correct?       Why is the triangle drawn like that? Why is the right angle at the police car ?

Is that to seperate the x and y components of the velocity relative to the police car?  Maybe there is a glimmer of understanding coming.

 

Maybe I just need to think about it for a bit ?       (´・_・`)   

-----------------------------------------------------------------------

Thanks Nauseated for that most educational clip you sent me too.

Here is a return one for you.  :)

https://www.youtube.com/watch?annotation_id=annotation_485354&feature=iv&src_vid=rfmxx_72Coc&v=GKNAM8z4Spo

-----------------------------------------------------------------------

Melody  Mar 8, 2015
 #7
avatar+26716 
+5

As Nauseated indicated, the police radar detects the component of the true speed of the car as seen from the direction of the police car. I simply scaled the velocity vectors so the blue one (measured speed, 95 ft/s) occupied the whole distance from car to police car.  I could have drawn both vectors a different length, but they would still have had the same relative length.  The important relationship is: true speed*cos(angle) = measured speed.

.

Alan  Mar 8, 2015
 #8
avatar+889 
+5

Hi Melody,

Like you  I'm struggling with the physics of this problem, I don't fully understand Alan's diagram, how the radar gun measures the component of the cars speed at right angles to the direction of the radar beam, and how that relates to the diagram. The logic as to why the speed of the car should be greater than that shown by the radar gun is nice though.

 

When I first looked at the problem, I tried two different methods of solution. In both cases I put the right angle at the road as you did, so the line of the radar beam is down the hypotenuse of the triangle.

 

First method.

Call the length of the hypotenuse D and the distance that the car is down the road s, then D^2 = s^2 + 30^2, and differentiating that, 2D(dD/dt) = 2s(ds/dt) so that ds/dt = (D/s)(dD/dt).

Initially we have s = 120, D^2 = 120^2 + 30^2, so D = 123.69 and dD/dt = -95, so ds/dt = 123.69.(-95)/120 = -97.92.

 

Second method.

Consider the movement of the car over 100th of a second. The hypotenuse decreases by 0.95 ft, and will now have a length approximately 123.69 - 0.95 = 122.74, and that translates to a distance of the car on the road as s^2 = 122.74^2 - 30^2 = 14165.885, so s = 119.02. That seems to imply that the car has moved a distance of 120 - 119.02 = 0.9795 ft which in turns implies a speed of 97.95 fps.

If instead of using 100th of a second, a limiting process were used, I would expect the second result to agree with the first, (though they are pretty close anyway).

Bertie  Mar 8, 2015
 #9
avatar+92623 
0

Thanks Bertie.  

That really did help.  I especially like your first method but I can see that both would work fine.  :)

Alan, if I can remember yours I can see it is easier but i am not good at remembering things that I do not understand.  

Melody  Mar 9, 2015
 #10
avatar+889 
+5
Best Answer

Actually Melody, my first method is exactly equivalent to Alan's method.

The equation ds/dt = (D/s)(dD/dt) can be written as (s/D)(ds/dt) = dD/dt, which, in Alan's notation is

v.cos(theta)=95, (with or without the negative sign)

Bertie  Mar 9, 2015
 #11
avatar+92623 
0

Thanks Bertie,

Yes, so it is!

I am still slightly more likely to remember the derivation than i am to remember the formula.

I'm funny like that, I hardy remember any formulas.

[ I'll probably have to see it many more times before I really remember either.  :)  ]

 

I am very grateful to all three of you for your help, and I am very pleased that i actually do understand. 

Melody  Mar 9, 2015

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