Mixing
Are the times the same with a closed and open dam?
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Before I start explaining what I will say regarding the question, read the question before going any further.
Question: Mariam and Leena both live on the bank of the same river, a distance apart. When Mariam visits Leena, she swims downstream (with the current), but then must swim upstream (against the current) to return back home. When the nearby dam is closed, the water between there homes is still and Mariam can visit Leena by swimming in both directions without the help (or hindrance) of any current. Is Mariam’s round trip between her house and Leena’s house quicker when the dam is closed or when it is open?
As you can from the question, it is asking is the round trip faster when the dam is closed or when it is open. I stated that the round trip's time should be the same regardless of the dam is open or closed. If the dam is closed, Mariam's round trip is $2t$ seconds. If the dam is open, the current decrease the amount of time required for Mariam to reach Leena; Mariam saves $x$ seconds; her time will be $t-x$ seconds. But when Mariam goes back home, she uses the $x$ she saved. Her time should be $t+x$ seconds on the way back to home. In theory, the round trip's time should be the same regardless of the dam is open or closed.
Problem: The problem is that I was told this is a good theory, but it is wrong. The person wasn't able to explain properly why it is wrong and as a result, I still remain confused on why it is wrong.
proof-verification word-problem
edited Jul 20 at 15:14
Rodrigo A. Pérez
1,0761712
asked Jul 18 at 23:40
naru sin
19112
add a comment |Â
up vote
4
down vote
favorite
Before I start explaining what I will say regarding the question, read the question before going any further.
Question: Mariam and Leena both live on the bank of the same river, a distance apart. When Mariam visits Leena, she swims downstream (with the current), but then must swim upstream (against the current) to return back home. When the nearby dam is closed, the water between there homes is still and Mariam can visit Leena by swimming in both directions without the help (or hindrance) of any current. Is Mariam’s round trip between her house and Leena’s house quicker when the dam is closed or when it is open?
As you can from the question, it is asking is the round trip faster when the dam is closed or when it is open. I stated that the round trip's time should be the same regardless of the dam is open or closed. If the dam is closed, Mariam's round trip is $2t$ seconds. If the dam is open, the current decrease the amount of time required for Mariam to reach Leena; Mariam saves $x$ seconds; her time will be $t-x$ seconds. But when Mariam goes back home, she uses the $x$ she saved. Her time should be $t+x$ seconds on the way back to home. In theory, the round trip's time should be the same regardless of the dam is open or closed.
Problem: The problem is that I was told this is a good theory, but it is wrong. The person wasn't able to explain properly why it is wrong and as a result, I still remain confused on why it is wrong.
proof-verification word-problem
edited Jul 20 at 15:14
Rodrigo A. Pérez
1,0761712
asked Jul 18 at 23:40
naru sin
19112
Same problem when running around a circular track when the wind is blowing. The resistance is proportional to the square of the velocity difference, so running into and against the wind do not cancel out.
– Weather Vane
Jul 18 at 23:44
Work it out in detail. Say $M$ can swim $v$ mph in still water, and that the current speed (when there is a current) is $w$ (with $w<v$ of course). Say the homes are $1$ mile apart. How long does it take $M$ in still water? How long does it take her with the current? How long does it take her against the current?
– lulu
Jul 18 at 23:46
1
Note: for intuition, it is a real, real bad sign that when $w≥v$ it takes $M$ infinitely long to make it (as she simply can't make progress against the current). Yet your informal argument would have us believe that if $w=v-epsilon$ then she has no problem making the round trip in the usual time.
– lulu
Jul 18 at 23:49
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Before I start explaining what I will say regarding the question, read the question before going any further.
Question: Mariam and Leena both live on the bank of the same river, a distance apart. When Mariam visits Leena, she swims downstream (with the current), but then must swim upstream (against the current) to return back home. When the nearby dam is closed, the water between there homes is still and Mariam can visit Leena by swimming in both directions without the help (or hindrance) of any current. Is Mariam’s round trip between her house and Leena’s house quicker when the dam is closed or when it is open?
As you can from the question, it is asking is the round trip faster when the dam is closed or when it is open. I stated that the round trip's time should be the same regardless of the dam is open or closed. If the dam is closed, Mariam's round trip is $2t$ seconds. If the dam is open, the current decrease the amount of time required for Mariam to reach Leena; Mariam saves $x$ seconds; her time will be $t-x$ seconds. But when Mariam goes back home, she uses the $x$ she saved. Her time should be $t+x$ seconds on the way back to home. In theory, the round trip's time should be the same regardless of the dam is open or closed.
Problem: The problem is that I was told this is a good theory, but it is wrong. The person wasn't able to explain properly why it is wrong and as a result, I still remain confused on why it is wrong.
proof-verification word-problem
edited Jul 20 at 15:14
Rodrigo A. Pérez
1,0761712
asked Jul 18 at 23:40
naru sin
19112
Before I start explaining what I will say regarding the question, read the question before going any further.
Question: Mariam and Leena both live on the bank of the same river, a distance apart. When Mariam visits Leena, she swims downstream (with the current), but then must swim upstream (against the current) to return back home. When the nearby dam is closed, the water between there homes is still and Mariam can visit Leena by swimming in both directions without the help (or hindrance) of any current. Is Mariam’s round trip between her house and Leena’s house quicker when the dam is closed or when it is open?
As you can from the question, it is asking is the round trip faster when the dam is closed or when it is open. I stated that the round trip's time should be the same regardless of the dam is open or closed. If the dam is closed, Mariam's round trip is $2t$ seconds. If the dam is open, the current decrease the amount of time required for Mariam to reach Leena; Mariam saves $x$ seconds; her time will be $t-x$ seconds. But when Mariam goes back home, she uses the $x$ she saved. Her time should be $t+x$ seconds on the way back to home. In theory, the round trip's time should be the same regardless of the dam is open or closed.
Problem: The problem is that I was told this is a good theory, but it is wrong. The person wasn't able to explain properly why it is wrong and as a result, I still remain confused on why it is wrong.
proof-verification word-problem
edited Jul 20 at 15:14
Rodrigo A. Pérez
1,0761712
asked Jul 18 at 23:40
naru sin
19112
edited Jul 20 at 15:14
Rodrigo A. Pérez
1,0761712
edited Jul 20 at 15:14
Rodrigo A. Pérez
1,0761712
edited Jul 20 at 15:14
Rodrigo A. Pérez
1,0761712
1,0761712
asked Jul 18 at 23:40
naru sin
19112
asked Jul 18 at 23:40
naru sin
19112
asked Jul 18 at 23:40
naru sin
19112
19112
Same problem when running around a circular track when the wind is blowing. The resistance is proportional to the square of the velocity difference, so running into and against the wind do not cancel out.
– Weather Vane
Jul 18 at 23:44
Work it out in detail. Say $M$ can swim $v$ mph in still water, and that the current speed (when there is a current) is $w$ (with $w<v$ of course). Say the homes are $1$ mile apart. How long does it take $M$ in still water? How long does it take her with the current? How long does it take her against the current?
– lulu
Jul 18 at 23:46
1
Note: for intuition, it is a real, real bad sign that when $w≥v$ it takes $M$ infinitely long to make it (as she simply can't make progress against the current). Yet your informal argument would have us believe that if $w=v-epsilon$ then she has no problem making the round trip in the usual time.
– lulu
Jul 18 at 23:49
add a comment |Â
Same problem when running around a circular track when the wind is blowing. The resistance is proportional to the square of the velocity difference, so running into and against the wind do not cancel out.
– Weather Vane
Jul 18 at 23:44
Work it out in detail. Say $M$ can swim $v$ mph in still water, and that the current speed (when there is a current) is $w$ (with $w<v$ of course). Say the homes are $1$ mile apart. How long does it take $M$ in still water? How long does it take her with the current? How long does it take her against the current?
– lulu
Jul 18 at 23:46
1
Note: for intuition, it is a real, real bad sign that when $w≥v$ it takes $M$ infinitely long to make it (as she simply can't make progress against the current). Yet your informal argument would have us believe that if $w=v-epsilon$ then she has no problem making the round trip in the usual time.
– lulu
Jul 18 at 23:49
Same problem when running around a circular track when the wind is blowing. The resistance is proportional to the square of the velocity difference, so running into and against the wind do not cancel out.
– Weather Vane
Jul 18 at 23:44
Same problem when running around a circular track when the wind is blowing. The resistance is proportional to the square of the velocity difference, so running into and against the wind do not cancel out.
– Weather Vane
Jul 18 at 23:44
Work it out in detail. Say $M$ can swim $v$ mph in still water, and that the current speed (when there is a current) is $w$ (with $w<v$ of course). Say the homes are $1$ mile apart. How long does it take $M$ in still water? How long does it take her with the current? How long does it take her against the current?
– lulu
Jul 18 at 23:46
Work it out in detail. Say $M$ can swim $v$ mph in still water, and that the current speed (when there is a current) is $w$ (with $w<v$ of course). Say the homes are $1$ mile apart. How long does it take $M$ in still water? How long does it take her with the current? How long does it take her against the current?
– lulu
Jul 18 at 23:46
1
1
Note: for intuition, it is a real, real bad sign that when $w≥v$ it takes $M$ infinitely long to make it (as she simply can't make progress against the current). Yet your informal argument would have us believe that if $w=v-epsilon$ then she has no problem making the round trip in the usual time.
– lulu
Jul 18 at 23:49
Note: for intuition, it is a real, real bad sign that when $w≥v$ it takes $M$ infinitely long to make it (as she simply can't make progress against the current). Yet your informal argument would have us believe that if $w=v-epsilon$ then she has no problem making the round trip in the usual time.
– lulu
Jul 18 at 23:49
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
3
down vote
accepted
The point is that current changes the velocity of swimming, not the time spent. Let the distance be $d$, the still water swimming speed be $v$ and the current speed be $c$. Swimming with the current is at speed $v+c$ and against the current is at speed $v-c$. If the dam is closed the round trip swimming time is $2frac dv$. If the dam is open the round trip swimming time is $frac dv+c+frac dv-c$ We have
$$frac dv+c+frac dv-c=frac d(v-c)+d(v+c)(v+c)(v-c)\
=frac 2dvv^2-c^2\=2frac dvfrac 11-frac c^2v^2$$
As the final factor is greater than $1$ the time swimming is longer when the dam is open. If $c gt v$ she can't do the round trip at all with the dam open.
answered Jul 18 at 23:48
Ross Millikan
276k21186352
add a comment |Â
up vote
3
down vote
Let the distance traversed in one direction be one length, Mariam's still swimming speed be $v$ lengths per unit time and the river current be $w$ lengths per unit time.
When the dam is open, Mariam takes $frac1v+w+frac1v-w=frac2vv^2-w^2$ time to make a round. When the dam is closed and the river is still, Mariam takes $frac2v=frac2vv^2$ time to make a round. Since $v^2-w^2<v^2$, $frac2vv^2-w^2>frac2v$, so Mariam always takes lesser time when the dam is closed than when the dam is open.
answered Jul 18 at 23:49
Parcly Taxel
33.6k136588
add a comment |Â
up vote
1
down vote
While there are several other excellent answers that formally show why swimming without current is easier, a quick way to gain an intuition of it is to think what happens if the speed of the current exactly matches the speed of the swimmer. Then, swimming with the current in favour will take half the time. But swimming against the current will take ... forever.
An interesting addendum: what if the current happened instead to be flowing sideways? In this case, it's obvious that it would hinder the swimmer both ways, because of the need to "waste" some effort compensating for it. And yet, it can be shown that a sideways current, over a round trip, is always better than a current of the same speed aligned with the swimming (favourable on one leg, unfavourable on the other). This is an interesting exercise; and it's also at the basis of one of the famous experiments that helped establish the theory of relativity.
answered Jul 19 at 1:21
Anonymous
4,8033940
add a comment |Â
up vote
0
down vote
The current gains or loses you a fixed distance per second. Unfortunately, you spend more time swimming against the current than with it, so it is a loss in total.
answered Jul 18 at 23:47
Phira
18.3k24197
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The point is that current changes the velocity of swimming, not the time spent. Let the distance be $d$, the still water swimming speed be $v$ and the current speed be $c$. Swimming with the current is at speed $v+c$ and against the current is at speed $v-c$. If the dam is closed the round trip swimming time is $2frac dv$. If the dam is open the round trip swimming time is $frac dv+c+frac dv-c$ We have
$$frac dv+c+frac dv-c=frac d(v-c)+d(v+c)(v+c)(v-c)\
=frac 2dvv^2-c^2\=2frac dvfrac 11-frac c^2v^2$$
As the final factor is greater than $1$ the time swimming is longer when the dam is open. If $c gt v$ she can't do the round trip at all with the dam open.
answered Jul 18 at 23:48
Ross Millikan
276k21186352
add a comment |Â
up vote
3
down vote
accepted
The point is that current changes the velocity of swimming, not the time spent. Let the distance be $d$, the still water swimming speed be $v$ and the current speed be $c$. Swimming with the current is at speed $v+c$ and against the current is at speed $v-c$. If the dam is closed the round trip swimming time is $2frac dv$. If the dam is open the round trip swimming time is $frac dv+c+frac dv-c$ We have
$$frac dv+c+frac dv-c=frac d(v-c)+d(v+c)(v+c)(v-c)\
=frac 2dvv^2-c^2\=2frac dvfrac 11-frac c^2v^2$$
As the final factor is greater than $1$ the time swimming is longer when the dam is open. If $c gt v$ she can't do the round trip at all with the dam open.
answered Jul 18 at 23:48
Ross Millikan
276k21186352
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The point is that current changes the velocity of swimming, not the time spent. Let the distance be $d$, the still water swimming speed be $v$ and the current speed be $c$. Swimming with the current is at speed $v+c$ and against the current is at speed $v-c$. If the dam is closed the round trip swimming time is $2frac dv$. If the dam is open the round trip swimming time is $frac dv+c+frac dv-c$ We have
$$frac dv+c+frac dv-c=frac d(v-c)+d(v+c)(v+c)(v-c)\
=frac 2dvv^2-c^2\=2frac dvfrac 11-frac c^2v^2$$
As the final factor is greater than $1$ the time swimming is longer when the dam is open. If $c gt v$ she can't do the round trip at all with the dam open.
answered Jul 18 at 23:48
Ross Millikan
276k21186352
The point is that current changes the velocity of swimming, not the time spent. Let the distance be $d$, the still water swimming speed be $v$ and the current speed be $c$. Swimming with the current is at speed $v+c$ and against the current is at speed $v-c$. If the dam is closed the round trip swimming time is $2frac dv$. If the dam is open the round trip swimming time is $frac dv+c+frac dv-c$ We have
$$frac dv+c+frac dv-c=frac d(v-c)+d(v+c)(v+c)(v-c)\
=frac 2dvv^2-c^2\=2frac dvfrac 11-frac c^2v^2$$
As the final factor is greater than $1$ the time swimming is longer when the dam is open. If $c gt v$ she can't do the round trip at all with the dam open.
answered Jul 18 at 23:48
Ross Millikan
276k21186352
answered Jul 18 at 23:48
Ross Millikan
276k21186352
answered Jul 18 at 23:48
Ross Millikan
276k21186352
answered Jul 18 at 23:48
Ross Millikan
276k21186352
276k21186352
add a comment |Â
add a comment |Â
up vote
3
down vote
Let the distance traversed in one direction be one length, Mariam's still swimming speed be $v$ lengths per unit time and the river current be $w$ lengths per unit time.
When the dam is open, Mariam takes $frac1v+w+frac1v-w=frac2vv^2-w^2$ time to make a round. When the dam is closed and the river is still, Mariam takes $frac2v=frac2vv^2$ time to make a round. Since $v^2-w^2<v^2$, $frac2vv^2-w^2>frac2v$, so Mariam always takes lesser time when the dam is closed than when the dam is open.
answered Jul 18 at 23:49
Parcly Taxel
33.6k136588
add a comment |Â
up vote
3
down vote
Let the distance traversed in one direction be one length, Mariam's still swimming speed be $v$ lengths per unit time and the river current be $w$ lengths per unit time.
When the dam is open, Mariam takes $frac1v+w+frac1v-w=frac2vv^2-w^2$ time to make a round. When the dam is closed and the river is still, Mariam takes $frac2v=frac2vv^2$ time to make a round. Since $v^2-w^2<v^2$, $frac2vv^2-w^2>frac2v$, so Mariam always takes lesser time when the dam is closed than when the dam is open.
answered Jul 18 at 23:49
Parcly Taxel
33.6k136588
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Let the distance traversed in one direction be one length, Mariam's still swimming speed be $v$ lengths per unit time and the river current be $w$ lengths per unit time.
When the dam is open, Mariam takes $frac1v+w+frac1v-w=frac2vv^2-w^2$ time to make a round. When the dam is closed and the river is still, Mariam takes $frac2v=frac2vv^2$ time to make a round. Since $v^2-w^2<v^2$, $frac2vv^2-w^2>frac2v$, so Mariam always takes lesser time when the dam is closed than when the dam is open.
answered Jul 18 at 23:49
Parcly Taxel
33.6k136588
Let the distance traversed in one direction be one length, Mariam's still swimming speed be $v$ lengths per unit time and the river current be $w$ lengths per unit time.
When the dam is open, Mariam takes $frac1v+w+frac1v-w=frac2vv^2-w^2$ time to make a round. When the dam is closed and the river is still, Mariam takes $frac2v=frac2vv^2$ time to make a round. Since $v^2-w^2<v^2$, $frac2vv^2-w^2>frac2v$, so Mariam always takes lesser time when the dam is closed than when the dam is open.
answered Jul 18 at 23:49
Parcly Taxel
33.6k136588
answered Jul 18 at 23:49
Parcly Taxel
33.6k136588
answered Jul 18 at 23:49
Parcly Taxel
33.6k136588
answered Jul 18 at 23:49
Parcly Taxel
33.6k136588
33.6k136588
add a comment |Â
add a comment |Â
up vote
1
down vote
While there are several other excellent answers that formally show why swimming without current is easier, a quick way to gain an intuition of it is to think what happens if the speed of the current exactly matches the speed of the swimmer. Then, swimming with the current in favour will take half the time. But swimming against the current will take ... forever.
An interesting addendum: what if the current happened instead to be flowing sideways? In this case, it's obvious that it would hinder the swimmer both ways, because of the need to "waste" some effort compensating for it. And yet, it can be shown that a sideways current, over a round trip, is always better than a current of the same speed aligned with the swimming (favourable on one leg, unfavourable on the other). This is an interesting exercise; and it's also at the basis of one of the famous experiments that helped establish the theory of relativity.
answered Jul 19 at 1:21
Anonymous
4,8033940
add a comment |Â
up vote
1
down vote
While there are several other excellent answers that formally show why swimming without current is easier, a quick way to gain an intuition of it is to think what happens if the speed of the current exactly matches the speed of the swimmer. Then, swimming with the current in favour will take half the time. But swimming against the current will take ... forever.
An interesting addendum: what if the current happened instead to be flowing sideways? In this case, it's obvious that it would hinder the swimmer both ways, because of the need to "waste" some effort compensating for it. And yet, it can be shown that a sideways current, over a round trip, is always better than a current of the same speed aligned with the swimming (favourable on one leg, unfavourable on the other). This is an interesting exercise; and it's also at the basis of one of the famous experiments that helped establish the theory of relativity.
answered Jul 19 at 1:21
Anonymous
4,8033940
add a comment |Â
up vote
1
down vote
up vote
1
down vote
While there are several other excellent answers that formally show why swimming without current is easier, a quick way to gain an intuition of it is to think what happens if the speed of the current exactly matches the speed of the swimmer. Then, swimming with the current in favour will take half the time. But swimming against the current will take ... forever.
An interesting addendum: what if the current happened instead to be flowing sideways? In this case, it's obvious that it would hinder the swimmer both ways, because of the need to "waste" some effort compensating for it. And yet, it can be shown that a sideways current, over a round trip, is always better than a current of the same speed aligned with the swimming (favourable on one leg, unfavourable on the other). This is an interesting exercise; and it's also at the basis of one of the famous experiments that helped establish the theory of relativity.
answered Jul 19 at 1:21
Anonymous
4,8033940
While there are several other excellent answers that formally show why swimming without current is easier, a quick way to gain an intuition of it is to think what happens if the speed of the current exactly matches the speed of the swimmer. Then, swimming with the current in favour will take half the time. But swimming against the current will take ... forever.
An interesting addendum: what if the current happened instead to be flowing sideways? In this case, it's obvious that it would hinder the swimmer both ways, because of the need to "waste" some effort compensating for it. And yet, it can be shown that a sideways current, over a round trip, is always better than a current of the same speed aligned with the swimming (favourable on one leg, unfavourable on the other). This is an interesting exercise; and it's also at the basis of one of the famous experiments that helped establish the theory of relativity.
answered Jul 19 at 1:21
Anonymous
4,8033940
answered Jul 19 at 1:21
Anonymous
4,8033940
answered Jul 19 at 1:21
Anonymous
4,8033940
answered Jul 19 at 1:21
Anonymous
4,8033940
4,8033940
add a comment |Â
add a comment |Â
up vote
0
down vote
The current gains or loses you a fixed distance per second. Unfortunately, you spend more time swimming against the current than with it, so it is a loss in total.
answered Jul 18 at 23:47
Phira
18.3k24197
add a comment |Â
up vote
0
down vote
The current gains or loses you a fixed distance per second. Unfortunately, you spend more time swimming against the current than with it, so it is a loss in total.
answered Jul 18 at 23:47
Phira
18.3k24197
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The current gains or loses you a fixed distance per second. Unfortunately, you spend more time swimming against the current than with it, so it is a loss in total.
answered Jul 18 at 23:47
Phira
18.3k24197
The current gains or loses you a fixed distance per second. Unfortunately, you spend more time swimming against the current than with it, so it is a loss in total.
answered Jul 18 at 23:47
Phira
18.3k24197
answered Jul 18 at 23:47
Phira
18.3k24197
answered Jul 18 at 23:47
Phira
18.3k24197
answered Jul 18 at 23:47
Phira
18.3k24197
18.3k24197
add a comment |Â
add a comment |Â
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Same problem when running around a circular track when the wind is blowing. The resistance is proportional to the square of the velocity difference, so running into and against the wind do not cancel out.
– Weather Vane
Jul 18 at 23:44
Work it out in detail. Say $M$ can swim $v$ mph in still water, and that the current speed (when there is a current) is $w$ (with $w<v$ of course). Say the homes are $1$ mile apart. How long does it take $M$ in still water? How long does it take her with the current? How long does it take her against the current?
– lulu
Jul 18 at 23:46
1
Note: for intuition, it is a real, real bad sign that when $w≥v$ it takes $M$ infinitely long to make it (as she simply can't make progress against the current). Yet your informal argument would have us believe that if $w=v-epsilon$ then she has no problem making the round trip in the usual time.
– lulu
Jul 18 at 23:49