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How do you get ##L = 3##?kimiko333 said:
Measuring the length of a moving Beam in two different directions
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The discussion revolves around measuring the length of a moving beam while walking in two different directions, leading to equations involving the speeds of the walker and the beam. Participants clarify the relationship between the time taken to pass the beam in both directions and the speeds involved, ultimately deriving a ratio of the speeds. The conversation highlights the challenge of solving the problem due to multiple unknowns, emphasizing the importance of understanding the relationships rather than just plugging in values. A breakthrough occurs when the ratio of speeds is established, allowing for a clearer path to the solution. The final conclusion is that while the length of the beam can be determined, the specific speeds remain undetermined, illustrating the complexity of the problem.
- #32
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The length of the beam cannot be less than both 10 and 15 steps.kimiko333 said:
- #33
kimiko333
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If I substitute 10 into the first equation, I get:PeroK said:
##L=t_1*(\frac {v} {5} +v)##
##L=2v+v=3v##
- #34
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I'm not impressed by that!kimiko333 said:
- #35
kimiko333
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PeroK said:
but why is this so? I'm supposed to get 12. And I get 12 with the second equation- #36
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It's just totally wrong! You multiply something bigger that ##v## by ##10## and get ##3v##? Come on!kimiko333 said:
- #37
kimiko333
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Oh Jesus. I forgot to multiply v by 10. Aaaahhhh. Told you... I'm tired :DPeroK said:
- #38
kimiko333
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Now it's 12, okay. But if I substitute in, I get 12v, not just simply 12. How do I know, that the answer is in steps?
- #39
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That's the units you used when you set ##t_1 = 10##. That's time in units of "time for a step". In those units ##v = 1## (meaning ##1## step per time for a step).kimiko333 said:
When you come to study Special Relativity, you'll find this is an amusing analogy to setting the speed of light ##c = 1##!
Alternatively, you could have left time in unspecified units, whereby:
##L = \frac{6}{5}vt_1##
And ##vt_1 = 10## steps.
- #40
kimiko333
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Ooooh, okay. Totally makes sense! Great! Thank you! :D Now I just have to figure out, why is it that I can easily solve problems in electrodynamics, thermodynamics, etc, but don't know how to deal with these... :D (I'm a teacher-student, studying to be an English and Physics teacher)PeroK said:
- #41
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You started off with the assumption that the quantities could be determined. But, actually, the only things that could be determined were the relationships between quantities. You end up with 12 steps, but that's true whatever the speed of the walker and beam. All we've found is that one is 5 times the other - but, they could be anything.kimiko333 said:
This is the sort of problem that undermines the "plug-and-chug" approach - too many unknowns, you said. Well, everything is still an unknown, because we don't know how long a step is or how fast anything is moving.
Finally, there is a difference between a problem being elementary and being difficult. This is an elementary problem that really made you think. You might have an advanced EM problem that involves advanced concepts but only requires you to plug the given numbers into the correct formula - so it's advanced but easy.
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kimiko333
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Great answers! Thank you for the explanations! :) I'm grateful!PeroK said:
- #43
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PS if the walker takes ##s_1## steps walking in the same direction as the beam is moving; and, ##s_2## steps walking in the opposite direction; then, the length of the beam in steps is $$\frac{2s_1s_2}{s_1 + s_2}$$
We can check that for ##s_1 = 15## and ##s_2 = 10##, we do indeed get the answer of ##12## steps.
In general, I'm rarely content to solve a single numerical problem. I usually want to find a general formula like the one above.
We can check that for ##s_1 = 15## and ##s_2 = 10##, we do indeed get the answer of ##12## steps.
In general, I'm rarely content to solve a single numerical problem. I usually want to find a general formula like the one above.
- #44
Steve4Physics
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I know it’s a bit late, but I think there are 2 possible solutions depending on which is faster, the walker or the wagon. The original question seems to allow both possibilities.
Take the unit of distance as 1-step and the unit of time as the time-for-1-step.
Speed of walker = 1.
Speed of wagon = v.
There are 2 cases:
Case 1; wagon is faster than walker: v>1
Case 2: wagon is slower than walker: v<1
___________________________
Case 1; wagon is faster than walker: v>1
When walker and beam move in the same direction:
Front of beam passes observer at ##t=0##
Back of beam passes observer at ##t= t_1##
##t_1 = \frac {L}{v-1}## (since speed of walker relative to beam is v-1.)
When walker and beam move in opposite directions:
##t_2 = \frac {L}{v+1}## (since speed of walker relative to beam is v+1.)
Eliminating v gives:
##L = \frac {2t_1t_2}{t_1 – t_2}##
Since the number of steps is the same as the number of time units:
##L = \frac{2s_1s_2}{s_1 – s_2}##
For ##s_1 = 15## and ##s_2 = 10##
##L = \frac {2·15·10}{15 – 10} = 60##
____________________________________
Case 2: wagon is slower than walker: v<1
Observer passes back of beam at ##t = 0##
Observer passes front of beam at ##t = t_1##
##t_1 = \frac {L}{1-v}## (since speed of walker relative to beam is 1-v.)
When walker and beam move in the opposite directions:
##t_2 = \frac {L}{v+1}## (since speed of walker relative to beam is v+1.)
Eliminating v gives:
##L = \frac {2t_1t_2}{t_1 + t_2}##
Since the number of steps is the same as the number of time units:
##L = \frac{2s_1s_2}{s_1 + s_2}##
For ##s_1 = 15## and ##s_2 = 10##
##L = \frac {2·15·10}{15 + 10} = 12##
Take the unit of distance as 1-step and the unit of time as the time-for-1-step.
Speed of walker = 1.
Speed of wagon = v.
There are 2 cases:
Case 1; wagon is faster than walker: v>1
Case 2: wagon is slower than walker: v<1
___________________________
Case 1; wagon is faster than walker: v>1
When walker and beam move in the same direction:
Front of beam passes observer at ##t=0##
Back of beam passes observer at ##t= t_1##
##t_1 = \frac {L}{v-1}## (since speed of walker relative to beam is v-1.)
When walker and beam move in opposite directions:
##t_2 = \frac {L}{v+1}## (since speed of walker relative to beam is v+1.)
Eliminating v gives:
##L = \frac {2t_1t_2}{t_1 – t_2}##
Since the number of steps is the same as the number of time units:
##L = \frac{2s_1s_2}{s_1 – s_2}##
For ##s_1 = 15## and ##s_2 = 10##
##L = \frac {2·15·10}{15 – 10} = 60##
____________________________________
Case 2: wagon is slower than walker: v<1
Observer passes back of beam at ##t = 0##
Observer passes front of beam at ##t = t_1##
##t_1 = \frac {L}{1-v}## (since speed of walker relative to beam is 1-v.)
When walker and beam move in the opposite directions:
##t_2 = \frac {L}{v+1}## (since speed of walker relative to beam is v+1.)
Eliminating v gives:
##L = \frac {2t_1t_2}{t_1 + t_2}##
Since the number of steps is the same as the number of time units:
##L = \frac{2s_1s_2}{s_1 + s_2}##
For ##s_1 = 15## and ##s_2 = 10##
##L = \frac {2·15·10}{15 + 10} = 12##
- #45
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Good thinking outside the proverbial box. The statement of the problem invokes the stereotypical picture of a fast walker catching up with a slow-moving (ox)cart. It would have been different if the beam were on a flatbed truck on the highway. Either formulation would have two solutions.