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Can anyone help me with the answer: A man is walking 3mph next to a trolley line. He notices that for every 40 trolley cars that pass him going his direction, 60 cars pass him going the opposite direction. How fast are the trolley cars traveling?

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True. The answer is 15 mph for identical lengths and identical speeds, but the context in which the question was asked, one can't derive a correct answer without further information.

The context is a math homework problem. Thus, in many cases, we can assume such things as identical lengths and speeds unless specifically stated. Most math textbooks I've seen don't put in trick questions or ones that need more information to solve.

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"There are no time values giving outside of the man walking at 3mph. At what intervals did each car pass him in minutes."

Time doesn't matter. Here's how I figure it:

Distance = velocity * time

The distance is the length of the trolley's cars (L) multiplied by how many passed the man (40 or 60).

Velocity is v +- 3, depending on which trolley it is.

Time is simply t.

Thus, the two equations are:

40L = (v-3)t

60L = (v+3)t

If x=y and a=b, x/a=y/b, so:

(40L) / (60L) = [(v-3)t] / [(v+3)t]

Simplify to find:

40/60 = (v-3) / (v+3)

Simple algebra gives v = 15.

Thus neither the length of the car nor the time interval it takes for 40 and 60 cars to pass the man matters. Only how fast the man was walking and how many cars passed him in each direction matter.

This explanation works assuming point masses, which a lot of basic physics relies on. If you start bringing length into it, you're dealing with rigid body motion, and relativity starts coming in to play.

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Actually, the solution is 15 mph. The question is how fast are the cars going.

For extra credit: The problem does not state how long the observation took. If we assume it took 2.5 minutes to watch the trolley cars go past, how long is a trolley car?

1 trolley car = 44 feet

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heh notice how this guy got his answer for his homework question and then never returned..

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ooh!!!

i thought it was trolley cars going one way, then (veichle vroom vroom) cars goin the other way

then there wouldnt be enough information.

thats very smart also the real answer

I can explain this :

Let the speed of trolley car be xmph

now 40 trolleys are moving in the direction where the man is moving

so relative speed of one trolley withrespect to man is (x+3)mph

40 trolleys it wil be 40(x+3)

similarly 60 trolleys are moving in oppsite direction

so relative speed of 60 trolleys wrt man wil be 60(x-3)

since it is told that the by the time 40 trolleys go in mans direction 60 trolleys pass him by that time

so 40(x+3) = 60 (x-3)

solving x= 15mph

gud que......

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It must be assumed that the cars headed in opposite directions are traveling at the same speed. Therefore, if the walker is making up for 20 cars per hour (or equivalent ratio), then he/she is offsetting the remainding 20 cars (60-40). The trains' speed must then be equivalent to the walker's speed by a product of three. Nine miles per hour.

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This explanation works assuming point masses, which a lot of basic physics relies on. If you start bringing length into it, you're dealing with rigid body motion, and relativity starts coming in to play.

u only have to assume the man as point mass... the cars can be of any length and relativity won't be a matter of concern as long as speed of cars (say x) and that of man (3mph) <<<<< speed of light ... in fact, the cars length L shud be >0.

The equation is 40L / (x-3) = 60L / (x+3) (i.e the common time taken by cars on both sides to cover the relative distances)

which gives x = 15 mph if L is not equal to 0.

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u only have to assume the man as point mass... the cars can be of any length and relativity won't be a matter of concern as long as speed of cars (say x) and that of man (3mph) <<<<< speed of light ... in fact, the cars length L shud be >0.

The equation is 40L / (x-3) = 60L / (x+3) (i.e the common time taken by cars on both sides to cover the relative distances)

which gives x = 15 mph if L is not equal to 0.

I wish I could have posted here alot sooner, like.. last year. Basically most people here are idiots.

"Cars" does not equal automobiles.. they are trolley cars, don't read it verbatim, it was already suggested they are trolley cars. I don't know how many times I read posts where everyone analyses the stated question to the extreme to prove pointless arguments. (however, there have been several creative solutions based on those derivations)

and for those "Math" Majors (who are probably in their 1st years)..

The answer cannot be 15mph, You are already presented a situation which obviously shows 2 trolleys moving at different speeds, and the question asks for the speed in which they are both moving. That means trolleyA speed cannot NOT Equal TolleyB speed meaning 15 mph is the wrong answer. 15 mph is the basis, but plug that back into your equations to get 18mph (opposite direction) & 12mph (in his direction) answers . I think only 1 person posting here was smart enough to do that.

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I wish I could have posted here alot sooner, like.. last year. Basically most people here are idiots.

"Cars" does not equal automobiles.. they are trolley cars, don't read it verbatim, it was already suggested they are trolley cars. I don't know how many times I read posts where everyone analyses the stated question to the extreme to prove pointless arguments. (however, there have been several creative solutions based on those derivations)

and for those "Math" Majors (who are probably in their 1st years)..

The answer cannot be 15mph, You are already presented a situation which obviously shows 2 trolleys moving at different speeds, and the question asks for the speed in which they are both moving. That means trolleyA speed cannot NOT Equal TolleyB speed meaning 15 mph is the wrong answer. 15 mph is the basis, but plug that back into your equations to get 18mph (opposite direction) & 12mph (in his direction) answers . I think only 1 person posting here was smart enough to do that.

The Answer is 15mph. We are to tell the actual speed of trolley cars not the relative speed w.r.t man.

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Basically most people here are idiots.

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OK, this is the best answer i could come up with given the limited information.

First we need to assume that the trolley cars are going the same speed the whole day and the distance between each car is uniform.

{we'll need to define some variables to continue, d1 = the distance between cars, d2 = the distance the boy has walked, t1 is the time it takes a car to travel d1 miles and t2 is the time it takes the boy to travel d2 miles, and r = the speed of the cars}

For the 40th car and 60th car to pass him at the same time, the 50th car in both dirrections would have have to be passing each other where the boy started. This means that the d2 = 10 * d1.

Using d1 = r * t1 and d2 = 3 * t2, we get 10 * r * t1 = 3 * t2

which leads to r = (3 * t2) / (10 * t1).

Now all we need is the ratio between t1 and t2.

The boy took t2 to move d2 miles. But remember, at this momment, the 50th car going his direction is just passing his starting point. So t2 = 50 * t1.

We now have r = (3 * 50 * t1) / (10 * t1) or r = 15.

So the cars are moving 15 mph.

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to fast?..

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OK, this is the best answer i could come up with given the limited information.

First we need to assume that the trolley cars are going the same speed the whole day and the distance between each car is uniform.

{we'll need to define some variables to continue, d1 = the distance between cars, d2 = the distance the boy has walked, t1 is the time it takes a car to travel d1 miles and t2 is the time it takes the boy to travel d2 miles, and r = the speed of the cars}

For the 40th car and 60th car to pass him at the same time, the 50th car in both dirrections would have have to be passing each other where the boy started. This means that the d2 = 10 * d1.

Using d1 = r * t1 and d2 = 3 * t2, we get 10 * r * t1 = 3 * t2

which leads to r = (3 * t2) / (10 * t1).

Now all we need is the ratio between t1 and t2.

The boy took t2 to move d2 miles. But remember, at this momment, the 50th car going his direction is just passing his starting point. So t2 = 50 * t1.

We now have r = (3 * 50 * t1) / (10 * t1) or r = 15.

So the cars are moving 15 mph.

You must also assume that the lengths of both sets of trolleys are identical. If one set was 100ft long and the other set was 200ft long it would invalidate your math.

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Ok, assuming your answer is correct. Lets say 1 set of cars was 1 mile long and it took 1 minute to pass the man. That means the cars are traveling more than a mile a minute which is well over 15 mph.

It only doesn't matter if the both trolley's are identical in length and speed. The OP doesn't give that information and you cannot assume one or the other. Lets say train two was stopped and you were walking at 3 mph, how fast is train one going (in your direction) given the same car ratio?

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Who said there were an equal amount of trolleys coming from either direction? Perhaps one trolley has 4 cars and one has 6.

Edited by Llam4
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Who said there were an equal amount of trolleys coming from either direction? Perhaps one trolley has 4 cars and one has 6.

In many if not most cities, inherent in the word "trolley" is "an equal number coming from either direction" (and usually powerered by overhead electricity). This is because there are a set number of trollies and they go back and forth on the same route (on different tracks or sides of the road). Even if there is only one multi-car trolley, it will also pass in both directions. I suppose it is possible that there are cities with trolleys going against each other on circular routes, where a different number could be on each track, but I've never seen that.

For the nit-pickers, during rush hour, the number of trolleys are increased. While the trolleys are being added, more trolleys will be coming from the "reservoir". Most of the day, however, the system is in steady state and the number of trolleys passing a point over an extended period of time will be the same for both directions. (This reasoning could start to break down if the trolley system extends significantly outside of a given city and cars jump to different routes; that is outside the norm.)

Edited by Kev_a_Swing_Dancer
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Assuming they're all moving at the same speed, 15mph works. But that is not stated so there isn't enough parameters as one could assume anything.

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3 friends spent the nite drinking to celebrate their reunion. They then decided to share a hotel room. The clerk charged them \$30 Total and gave them a room on the third floor. After they went up to their room, the desk clerk realized today was to be a Special Price of \$27 total . So instead of costing them \$30 ea., it was supposed to be \$25 total.
He asked the bellboy to return \$5 to the customers. On the way up to their room, the bellboy decided to only return \$3 of the \$5 for easier calculation by the customer, so instead of costing them \$30 (\$10 ea.) , it now cost them \$27 (\$9 ea.).

So if the price dropped from \$10 ea.(\$30 total) to \$9 ea. (\$27 total) and the bellboy kept \$2 for himself, what happened to the other \$1?

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60-40= 20 car lengths;    therefor man walked only 10 car lengths.  ;    mans rate is 3 mph;     t = d / speed ;   Now assume 10 cl to be   1 mile . /  3 mph =.333 hrs;

traveling in same direction train traveled = 40 + 10 = 50 lts  =5 miles ;   s = d / t   5 m/ .33 hrs = 15 mph

train traveling opposite direction =60 - 10 = 50 lts  = 5m;  5m / .33 hrs = 15 mph

So let us assume 10cl = 0.5 miles ;   0.5 / 3mph = 016666667 hrs ;  50 cl = 2.5 m;  2.5 m / 0.16666666667 = 15 mph

Regardless of what value in miles  is assumed for 10 car lengths the answer is 15 mph

60-40= 20 car lengths;    therefor man walked only 10 car lengths.  ;    mans rate is 3 mph;     t = d / speed ;   Now assume 10 cl to be   1 mile . /  3 mph =.333 hrs;

traveling in same direction train traveled = 40 + 10 = 50 lts  =5 miles ;   s = d / t   5 m/ .33 hrs = 15 mph

train traveling opposite direction =60 - 10 = 50 lts  = 5m;  5m / .33 hrs = 15 mph

So let us assume 10cl = 0.5 miles ;   0.5 / 3mph = 016666667 hrs ;  50 cl = 2.5 m;  2.5 m / 0.16666666667 = 15 mph

Regardless of what value in miles  is assumed for 10 car lengths the answer is 15 m

time = distance / speed;  "t " will be in hours.

speed = distance / time

Regardless of what value in miles  is assumed for 10 car lengths the answer is 15 mph

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60-40= 20 car lengths;    therefor man walked only 10 car lengths.  ;    mans rate is 3 mph;     t = d / speed ;   We have to look at the 1st car in each  direction .as a unit of length Now assume 10 cl to be   1 mile   1m /  3 mph =.333 hrs;

traveling in same direction train traveled = 40 + 10 = 50 lts  =5 miles ;   s = d / t   5 m/ .33 hrs = 15 mph

train traveling opposite direction =60 - 10 = 50 lts  = 5m;     5m / .33 hrs = 15 mph

So let us assume a different value for  10cl = 0.5 miles ;   0.5 / 3mph = 016666667 hrs ;  50 cl = 2.5 m;     2.5 m / 0.16666666667 = 15 mph

Regardless of what value in miles  is assumed for 10 car lengths the answer is 15 mph

If you cannot assume both are traveling at the same speed,then it is unsolvable.  or actually a solutions for every increment of time he walks i.e. he walks 15 seconds and in that time 40 passes one way and 60 passes the other way...Obviously they would really be flying and one going approximately 50% faster

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