0192837465

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Please help with the following question:

Suppose a policeman traveling at 5 m/s is firing his gun at a rate of 20 bullets per minute is chasing a bank robber who is peddling his Huffy at 50 m/s. At what rate do the bullets reach the bank robber (use 500 m/s for the speed of a bullet)?

ANSWER IS: 18 bullets/min


THANK YOU!
 

Kaustikos

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Please help with the following question:

Suppose a policeman traveling at 5 m/s is firing his gun at a rate of 20 bullets per minute is chasing a bank robber who is peddling his Huffy at 50 m/s. At what rate do the bullets reach the bank robber (use 500 m/s for the speed of a bullet)?

ANSWER IS: 18 bullets/min


THANK YOU!
I would pay twice the aamc fee to get questions like THIS.:laugh:

V +/- Vd
--------
V -/+ Vs

The source is the officer (I think this is a trick question, especially with the 500m/s of the bullet which does not matter in this case)
The detector is the robber

We can assume the officer moves at 0 m/s while the robber goes 45 m/s. The detector is moving away from the source so use the bottom sign for the Vd:
V = 330 m/s
V - 45
-------
V
285/300 = 0.87

The equation is f' = f(what I figured - [(V+/-Vs)/(V-/+Vd)]
f = 20
20 x 0.87

If it helps, try to remember the basics of this;
the observed frequency will be lower for targets moving away and higher for targets moving together. Knowing this will help eliminate answers and help you narrow down on your target, not lag behind like the cop in this example.
 

MDWannabe2009

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can anyone confirm if this is a correct way of solving this problem

Use the equation: Δf/f = -v/c

it is negative b/c they are moving away from eachother, thus the frequency will be lower.

So: Δf/20 = -45/330

Δf = -2.72

Δf = Fobserved - Femitted

-2.72 = Fobserved - 20

Fobserved = ~18

Is this a correct way to get the answer?
 
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GeorgianCMV

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I know it sounds crazy, but I find it much easier to use the long equation for this.

f' (observed frequency)= ?
f (actual frequency)= 20 bullets/min
V of source (policeman)= 5 m/s= 300 m/min
V of detector (robber)= 50 m/s = 3000 m/min
Actual speed of bullets= 500 m/s= 30,000 m/min

f'= V-Vdetector (use minus sign, bottom sign, because the robber, the detector, is moving away from cop) / V-Vsource (use minus sign, top sign, because the cop, the source, is moving towards the robber) * 20 (actual frequency)


f'= (30,000-3000)/(30,000-300), all multiplied by 20. Answer is 18.18.
 

PRodulous

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0192837465 said:
Please help with the following question:

Suppose a policeman traveling at 5 m/s is firing his gun at a rate of 20 bullets per minute is chasing a bank robber who is peddling his Huffy at 50 m/s. At what rate do the bullets reach the bank robber (use 500 m/s for the speed of a bullet)?

ANSWER IS: 18 bullets/min
The method I used was by the equation: f' = f[(V+/-Vd)/(V-/+Vs)]

f' = detected frequncy
f = source frequency
v = rate of fire (frequency) of bullet fire
vd = velocity of detector
vs = velocity of source

When choosing +/- for the numerator, consider the relative velocity of the detector. The robber (detector) moves away the police officer (source) that decreases the detected frequency. To decrease the frequency in the numerator, a minus sign must be used.

The same logic is used in the denominator. What is the velocity of the source doing to the frequency? The policeman (source) moves toward the robber (detector) that should increase frequency. To increase the detected frequency (f'), a minus sign is used to decrease the denominator.

With this said, you get the equation: f' = f[(V -Vd)/(V- Vs)]

Plug in the variables: f' = (20)[(500-50)/(500-5)] = (20)(450/495)

For the MCAT, an equation to work out like this will take a long time, so I round the denominator. It changes the answer somewhat, but you can eliminate choices based on this final approximation.

f' = (20)(450/495) = (20)(450/500) = (20)(900/1000) = (20)(0.9) = 18 bullets per minute

In the MCAT, you may see answers such as:

A. 17.8 bullets per minute
B. 18 bullets per minute
C.18.2 bullets per minute
D. 22 bullets per minute

In this case, since we increased the denominator, the answer should be slightly greater than the approximation. Hence, A can be canceled. 18 bullets per minute is merely an approximation, so it can be canceled. From our work, the bullet rate should decrease and in D., the bullet rate increases. It can be canceled. The answer is C.

In the MCAT, they expect you to approximate to simplify the work, I'm sure the answer choices won't be as complicated as this. I hope this explains your question thoroughly. I'd gladly clarify anything I missed.
 

161927

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this is the best equation to use because it works for all situations
The method I used was by the equation: f' = f[(V+/-Vd)/(V-/+Vs)]

f' = detected frequncy
f = source frequency
v = rate of fire (frequency) of bullet fire
vd = velocity of detector
vs = velocity of source

When choosing +/- for the numerator, consider the relative velocity of the detector. The robber (detector) moves away the police officer (source) that decreases the detected frequency. To decrease the frequency in the numerator, a minus sign must be used.

The same logic is used in the denominator. What is the velocity of the source doing to the frequency? The policeman (source) moves toward the robber (detector) that should increase frequency. To increase the detected frequency (f'), a minus sign is used to decrease the denominator.

With this said, you get the equation: f' = f[(V -Vd)/(V- Vs)]

Plug in the variables: f' = (20)[(500-50)/(500-5)] = (20)(450/495)

For the MCAT, an equation to work out like this will take a long time, so I round the denominator. It changes the answer somewhat, but you can eliminate choices based on this final approximation.

f' = (20)(450/495) = (20)(450/500) = (20)(900/1000) = (20)(0.9) = 18 bullets per minute

In the MCAT, you may see answers such as:

A. 17.8 bullets per minute
B. 18 bullets per minute
C.18.2 bullets per minute
D. 22 bullets per minute

In this case, since we increased the denominator, the answer should be slightly greater than the approximation. Hence, A can be canceled. 18 bullets per minute is merely an approximation, so it can be canceled. From our work, the bullet rate should decrease and in D., the bullet rate increases. It can be canceled. The answer is C.

In the MCAT, they expect you to approximate to simplify the work, I'm sure the answer choices won't be as complicated as this. I hope this explains your question thoroughly. I'd gladly clarify anything I missed.
 

tncekm

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this is the best equation to use because it works for all situations
That equation takes too much time, and the MCAT will never give answers so close that the doppler approximation won't work.

Δf/f₀ = v/c and Δλ/λ₀ = v/c are the best ones to know, even though they're not as precise, because they're so simple to use.
 

DrMattOglesby

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50m/s on a huffy is F***ing unreal! (thats approx half a minute to travel a mile! the best of the best can do a mile in a time trial in 2.5 min or so)
that robber should take up triathlons or the tour de france and get on a real bike! or maybe that was a world class cyclist who got kicked out for doping and was just trying to make ends meet...?
sorry for not helping out...im just procastinating studying for a piano midterm :rolleyes:
 

tncekm

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50m/s on a huffy is F***ing unreal! (thats approx half a minute to travel a mile! the best of the best can do a mile in a time trial in 2.5 min or so)
that robber should take up triathlons or the tour de france and get on a real bike! or maybe that was a world class cyclist who got kicked out for doping and was just trying to make ends meet...?
sorry for not helping out...im just procastinating studying for a piano midterm :rolleyes:
You mean, you can't ride 1.86 miles/minute? I don't know how you can call yourself an athlete! :laugh:
 

DrMattOglesby

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You mean, you can't ride 1.86 miles/minute? I don't know how you can call yourself an athlete! :laugh:
what the heck, really?
okay: here is what i just mind-numbered (apparently without too much thought though!):

50 meters/sec
*compared this value to 1500 meters (which is just under a mile)

and figured if he was to complete this distance at his velocity...he would take 30 seconds. Because 1500m/(50m/s) = 30s
sooo--i STILL arrive at just over half a minute to complete a mile.
I train at around 1mi/3min (for a 56mi race gives me a 2hr 48min pace).
So ... after looking at this again, i STILL think this dude is hauling butt.
now, if it was 1.86 miles/HOUR ... then thats a different story...but not per minute.

EDIT: i just calculated 1.86 mi/min =
111 mph
 

tncekm

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I was just kidding :laugh:
 

DrMattOglesby

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tncekm

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Haha... its okay, I'm brain dead today as well :p

So, how is the studying going?
 

DrMattOglesby

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Haha... its okay, I'm brain dead today as well :p

So, how is the studying going?
the what now???
yeah...since my ironman event...ive been kinda zoned out and havent gotten as much study time in. I have since put in a fair amount but nothing near what i was doing beforehand.
Gotta get back on the proverbial horse and giddy up!
 

reiternick

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why did you guys use the speed of sound instead of the speed of the bullets?
 

TheBoondocks

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why did you guys use the speed of sound instead of the speed of the bullets?
They didn't. They used 500m/s which is the speed of the bullets.
 
Feb 6, 2014
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Question regarding the ratio of the formula

2 objects moving towards each other
Frequency will be higher
So this mean we add source or observers speed to v of wave
But does this mean we subtract lambda since lambda is inversely proportional to F, if trying to find new lambda?

2 objects moving away from each other
Frequency will be lower
So this mean we subtract source or observers speed from v of wave
But does this mean we add lambda since lambda is inversely proportional to F? If trying to find new lambda?
 
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