- #1
RiccardoVen
- 118
- 2
Hi,
I'm an amateur italian physics addict, so please sorry for my english, eventually.
I'm reading carefully the Taylor & Wheeler 2nd edition, and I'm trying to do
all the problems before jumping to the next chapter.
Practice problem 3-1, page 78: let's talk about point b. Here we have two floats
which are towed at constant speed at 1/3 m/s through a lake.
I'm assuming the swimmer is swimming with constant 1 m/s velocity respect to
water, as pointed out few lines above for the "stationary" pool ( all in all the lake
is a sort of big pool, so I guess the 1 m/s still holds ).
That said, we have to compute the total swimming time from A -> B and B -> A,
keeping into account the floats are actually moving.
I'm able, as you will see, with my reasoning to reach the correct answer, but the intermdiate
results have not exactly the same meaning as the answers at the end of this holy book.
First of all: as you've noticed, I'm italian, so may be I've misunderstood the word "towed".
I'm actually assuming this means within the lake there are no flows or water current ( otherwise
the problem would have been probably stated a river and not a lake ). So to me, "towed" means
someone/thing is actually towing with constant speed the 2 floats, keeping them at a fixed
30m distance apart.
this is important, since actually there's not any constant water flow which could be added to
swimming motion.
that said, we have 2 situations: 1) the swimmer swims in the same direction as the floats are moving.
2) after having reached the B float, the swimmer come back to float A.
I've used the same approach Feynman adopted for the hat falling into a river ( hope you know
about this problem, but I'm quite sure yes ):
1) the floats and the swimmer start together, heading the same direction. So, actually, we can
"forget" about the floats moving, imagining them at rest. This is achievable thinking about the
swimmer as moving with 1 - 1/3 = 2/3 m/s relative speed related to the flows.
So this answers also to the first question of b: the relative speed of the swimmer respect to
the floats is 2/3 m/s.
This also means we could sit in the "floats frame" seeing the swimmer heading point B with 2/3
m/s speed. Since the floats are at rest for us, we can answer to the second question about
distance: since the floats are kept at 30m apart during their motions, the swimmer has still
to cover this distance.
that said, the time needed by the swimmer to go from A -> B is actually:
t1 = x / ( 2/3 ) = 90 / 2 = 45s
If we compare this time with the pool time ( point a) ) for going from A -> B, i.e. 30s we
can see it's slower. This makes sense, since the relative speed between the swimmer and the
floats is now just 2/3, instead of 1 as in the pool case, so the swimmer takes more time
to cover the distance.
2) Now the swimmer is coming back to B from A. Again, sitting at rest within the floats frame,
we now modify the relative speed like this: 1 + 1/3 = 4/3 m/s.
We have to think about this as for relative collisions: if you are colliding with something which
heading towards you with its own speed, you have to SUM the speeds. This is just the case.
So actually, since the distance is always 30m in this case as well, we have:
t2 = x / ( 4/3) = 90 / 4 = 22.5s
So in this case the swimmer takes less time than the 30s pool case, since it's moving
faster than 1m/s.
The total trip time is actually: t1 + t2 = 67.5s
If you look at Taylor & Wheeler result you can see it's correct. But actually the 2 times are swapped, leading to me in troubles.
As you can see, the answer is talking about a "current", so the 22.5 is actually the time
when the swimmer swims with the current, while 45 when it's swimming against it.
This is unfortunately the opposite of my results, and I cannot understand them properly.
Of course the Taylor & Wheeler result makes sense if there would a river with a water current:
of course, looking from the shore, the swimmer would take less time when she swims with the
current than against, but this is true if points A & B are kept fixed with the shore, and not
moving as in our case.
There's is not current at all in our case, just 2 frames moving with 1/3 and 1 m/s against the water.
So, probably, I've misunderstood the word "towed". So the first question is: have I missed
the problem completely?
2) forgetting eventually about my misunderstanding about the "towed" word, may you confirm
my reasoning still holds? ( i.e. imagining the water is at rest and the floats are simply "moved"
at constant speed respect to the water?
Thanks for you help in advance
Regards
Ricky
EDIT: I'm pasting here the part of the exercice we are talking about:
"Check how very different the story is for the swimmer plowing along at 1 m/s with respect to the water.
a) How long does it take her to swim down the length of a 30mt pool and back again?
b) How long does it take her to swim from float A to float B and back again when the
two float are still 30mt apart, but now are being towed through a lake at 1/3 m/s?
Discussion: When the swimmer is swimming in the same direction in which floats are
being towed, what is her speed relative to the floats?
And how great is the distance she has to travel expressed in the frame of reference
of the floats? So, how long does it take to travel that leg of her trip?
Then consider the same three questions for the return trip".
And their correct answer is: "a) 60 sec b) 45 seconds against the current, 22.5 seconds
with the current, 67.5 seconds round trip."
I'm an amateur italian physics addict, so please sorry for my english, eventually.
I'm reading carefully the Taylor & Wheeler 2nd edition, and I'm trying to do
all the problems before jumping to the next chapter.
Practice problem 3-1, page 78: let's talk about point b. Here we have two floats
which are towed at constant speed at 1/3 m/s through a lake.
I'm assuming the swimmer is swimming with constant 1 m/s velocity respect to
water, as pointed out few lines above for the "stationary" pool ( all in all the lake
is a sort of big pool, so I guess the 1 m/s still holds ).
That said, we have to compute the total swimming time from A -> B and B -> A,
keeping into account the floats are actually moving.
I'm able, as you will see, with my reasoning to reach the correct answer, but the intermdiate
results have not exactly the same meaning as the answers at the end of this holy book.
First of all: as you've noticed, I'm italian, so may be I've misunderstood the word "towed".
I'm actually assuming this means within the lake there are no flows or water current ( otherwise
the problem would have been probably stated a river and not a lake ). So to me, "towed" means
someone/thing is actually towing with constant speed the 2 floats, keeping them at a fixed
30m distance apart.
this is important, since actually there's not any constant water flow which could be added to
swimming motion.
that said, we have 2 situations: 1) the swimmer swims in the same direction as the floats are moving.
2) after having reached the B float, the swimmer come back to float A.
I've used the same approach Feynman adopted for the hat falling into a river ( hope you know
about this problem, but I'm quite sure yes ):
1) the floats and the swimmer start together, heading the same direction. So, actually, we can
"forget" about the floats moving, imagining them at rest. This is achievable thinking about the
swimmer as moving with 1 - 1/3 = 2/3 m/s relative speed related to the flows.
So this answers also to the first question of b: the relative speed of the swimmer respect to
the floats is 2/3 m/s.
This also means we could sit in the "floats frame" seeing the swimmer heading point B with 2/3
m/s speed. Since the floats are at rest for us, we can answer to the second question about
distance: since the floats are kept at 30m apart during their motions, the swimmer has still
to cover this distance.
that said, the time needed by the swimmer to go from A -> B is actually:
t1 = x / ( 2/3 ) = 90 / 2 = 45s
If we compare this time with the pool time ( point a) ) for going from A -> B, i.e. 30s we
can see it's slower. This makes sense, since the relative speed between the swimmer and the
floats is now just 2/3, instead of 1 as in the pool case, so the swimmer takes more time
to cover the distance.
2) Now the swimmer is coming back to B from A. Again, sitting at rest within the floats frame,
we now modify the relative speed like this: 1 + 1/3 = 4/3 m/s.
We have to think about this as for relative collisions: if you are colliding with something which
heading towards you with its own speed, you have to SUM the speeds. This is just the case.
So actually, since the distance is always 30m in this case as well, we have:
t2 = x / ( 4/3) = 90 / 4 = 22.5s
So in this case the swimmer takes less time than the 30s pool case, since it's moving
faster than 1m/s.
The total trip time is actually: t1 + t2 = 67.5s
If you look at Taylor & Wheeler result you can see it's correct. But actually the 2 times are swapped, leading to me in troubles.
As you can see, the answer is talking about a "current", so the 22.5 is actually the time
when the swimmer swims with the current, while 45 when it's swimming against it.
This is unfortunately the opposite of my results, and I cannot understand them properly.
Of course the Taylor & Wheeler result makes sense if there would a river with a water current:
of course, looking from the shore, the swimmer would take less time when she swims with the
current than against, but this is true if points A & B are kept fixed with the shore, and not
moving as in our case.
There's is not current at all in our case, just 2 frames moving with 1/3 and 1 m/s against the water.
So, probably, I've misunderstood the word "towed". So the first question is: have I missed
the problem completely?
2) forgetting eventually about my misunderstanding about the "towed" word, may you confirm
my reasoning still holds? ( i.e. imagining the water is at rest and the floats are simply "moved"
at constant speed respect to the water?
Thanks for you help in advance
Regards
Ricky
EDIT: I'm pasting here the part of the exercice we are talking about:
"Check how very different the story is for the swimmer plowing along at 1 m/s with respect to the water.
a) How long does it take her to swim down the length of a 30mt pool and back again?
b) How long does it take her to swim from float A to float B and back again when the
two float are still 30mt apart, but now are being towed through a lake at 1/3 m/s?
Discussion: When the swimmer is swimming in the same direction in which floats are
being towed, what is her speed relative to the floats?
And how great is the distance she has to travel expressed in the frame of reference
of the floats? So, how long does it take to travel that leg of her trip?
Then consider the same three questions for the return trip".
And their correct answer is: "a) 60 sec b) 45 seconds against the current, 22.5 seconds
with the current, 67.5 seconds round trip."
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