# Measuring the length of a moving Beam in two different directions

• kimiko333
In summary, the relationship between t1 and t2 is that t1 is the time it takes for the walker to pass by the beam while moving with the beam, and t2 is the minimum time it takes to pass by the beam. However, because the distance the walker travels on the ground changes with each trip, the two times are not always the same.
kimiko333 said:
but why is this so? I'm supposed to get 12. And I get 12 with the second equation
It's just totally wrong! You multiply something bigger that ##v## by ##10## and get ##3v##? Come on!

PeroK said:
It's just totally wrong! You multiply something bigger that ##v## by ##10## and get ##3v##? Come on!
Oh Jesus. I forgot to multiply v by 10. Aaaahhhh. Told you... I'm tired :D

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?

kimiko333 said:
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?
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).

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.

PeroK said:
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).

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.
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)

kimiko333 said:
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)
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.

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.

berkeman
PeroK said:
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.

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.
Great answers! Thank you for the explanations! :) I'm grateful!

berkeman and PeroK
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.

kimiko333
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##

kuruman and PeroK
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.

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