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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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How do you figure that. There are no time values giving outside of the man walking at 3mph. At what intervals did each car pass him in minutes. hours, days, one at a time five at a time Obviously the cars that passed him moving in the same direction are going faster than than he is. The cars that are moving towards him could be moving at a much slower speed than he is but will at some point pass each other as they are moving towards each other again when in minutes. hours, or days. Better refigure this one

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It's a homework problem people. You are looking way too deep into it. Would you like the wind direction also? How about the time of day? Or which hemisphere it's in? Or maybe if he's walking on grass or cement, because that can also change the distance that each stride was. I know, was it high tide or low tide? Sometimes the most simple answer is the correct one.

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if he was standing still, for every 50 carts going by in one direction, 50 will go into the other...

the 3mph must explain the 10 cards difference out of 50 = 1/5

==> Carts Speed = 5 x 3mph = 15

I'm not sure this solution can be valid. With the information given, all we know is that the trolly's frequency is 4/6ths the cars frequency. Nevermind the fact that the man is traveling at "3" mph, we don't know the time interval at which trolly's/cars pass him...

Consider this scenario for clarification: The man is walking at 3 mph heading west. A car passes him going east at "X" mph every 10 seconds, a trolly passes him going west every 15 seconds at "Y" mph. This qualifies the only given restraint: that in any given time "t" a ratio 40 trollys, per 60 cars passes the man. It doesn't matter what speeds the man, cars, and trollys are traveling.

Currently I don't believe this problem is solvable, hope this helps...

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Walking toward the 60 cars adds 3 mph to their observed speed and takes 3 mph from the observed speed of the 40 cars.

60 cars pass at x+3 mph

40 cars pass at x-3 mph

So the ratio is 3/2 or 1.5

or 1.5 times x-3 is x+3

1.5(x-3) = x+3

1.5x-4.5 = x+3

.5x=7.5

X= 15 mph

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Walking toward the 60 cars adds 3 mph to their observed speed and takes 3 mph from the observed speed of the 40 cars.

60 cars pass at x+3 mph

40 cars pass at x-3 mph

So the ratio is 3/2 or 1.5

or 1.5 times x-3 is x+3

1.5(x-3) = x+3

1.5x-4.5 = x+3

.5x=7.5

X= 15 mph

This would be correct only if it is assumed the trolley cars and the automobiles were traveling at the same speed.

If they cars and autos are not traveling at the same speed then there would not be enough information to solve the problem.

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Walking toward the 60 cars adds 3 mph to their observed speed and takes 3 mph from the observed speed of the 40 cars.

60 cars pass at x+3 mph

40 cars pass at x-3 mph

So the ratio is 3/2 or 1.5

or 1.5 times x-3 is x+3

1.5(x-3) = x+3

1.5x-4.5 = x+3

.5x=7.5

X= 15 mph

Hence, the answers are 18mph and 12mph, right? Hehe, I'm sure it's just a little slip, other than that, good explanation.

Remember sometimes your "answer" is not the solution to the problem.

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Hence, the answers are 18mph and 12mph, right? Hehe, I'm sure it's just a little slip, other than that, good explanation.

Remember sometimes your "answer" is not the solution to the problem.

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?

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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.

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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.

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.

Edited by dwmead03
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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.

Okay, a real reply this time.

Let's say that each trolley is 1 mile long and that it passes the man in 1 minute. So the trolley is going at a speed (observed by the man) of 1 mile-per-minute or 60mph. Let's say that these are the trolleys going in the opposite direction. So then the trolleys going in the same direction as the man must be going at 40mph (since 40 of these pass for every 60 in the other direction). This means that there is an observed difference of 20mph between the trolleys, which are actually going at the same speed from a stationary point of reference. To account for this, the man would have to be walking at 10mph, which contradicts one of the specified elements of the question: the man is walking at 3mph.

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Okay, a real reply this time.

Let's say that each trolley is 1 mile long and that it passes the man in 1 minute. So the trolley is going at a speed (observed by the man) of 1 mile-per-minute or 60mph. Let's say that these are the trolleys going in the opposite direction. So then the trolleys going in the same direction as the man must be going at 40mph (since 40 of these pass for every 60 in the other direction). This means that there is an observed difference of 20mph between the trolleys, which are actually going at the same speed from a stationary point of reference. To account for this, the man would have to be walking at 10mph, which contradicts one of the specified elements of the question: the man is walking at 3mph.

Yes, that example is false. But lets say the second trolley is traveling at 54 mph, with a car length of 40/54 of a mile. So after 1 hour and 54 miles the man will have seen 40 cars one way and 60 cars the other way. With a point of reference of 3mph. Correct?

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Yes, that example is false. But lets say the second trolley is traveling at 54 mph, with a car length of 40/54 of a mile. So after 1 hour and 54 miles the man will have seen 40 cars one way and 60 cars the other way. With a point of reference of 3mph. Correct?

I think the point was that the trolleys are all the same dimensions, though it's not specified in the OP.

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Okay, a real reply this time.

Let's say that each trolley is 1 mile long and that it passes the man in 1 minute. So the trolley is going at a speed (observed by the man) of 1 mile-per-minute or 60mph. Let's say that these are the trolleys going in the opposite direction. So then the trolleys going in the same direction as the man must be going at 40mph (since 40 of these pass for every 60 in the other direction). This means that there is an observed difference of 20mph between the trolleys, which are actually going at the same speed from a stationary point of reference. To account for this, the man would have to be walking at 10mph, which contradicts one of the specified elements of the question: the man is walking at 3mph.

Now that I think about it, this statement is wrong. Why are the trains going the same speed from a stationary point of reference? From stationary they would be going 63 mph and 37 mph. Or when the man is walking it would be 60 mph and 40mph to him. This example IS valid and proves the 15 mph answer wrong.

Edited by dwmead03
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Now that I think about it, this statement is wrong. Why are the trains going the same speed from a stationary point of reference? From stationary they would be going 63 mph and 37 mph. Or when the man is walking it would be 60 mph and 40mph to him. This example IS valid and proves the 15 mph answer wrong.

I believe the intent of the OP was that the trolleys are identical and traveling at the same speed, just in opposite directions. A man is walking at 3mph and notes the number of trolleys passing him in either direction. If the trolleys are not all the same size and not going at the same speed, then there is not one solution to the problem.

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I believe the intent of the OP was that the trolleys are identical and traveling at the same speed, just in opposite directions. A man is walking at 3mph and notes the number of trolleys passing him in either direction. If the trolleys are not all the same size and not going at the same speed, then there is not one solution to the problem.

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.

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