Are the times the same with a closed and open dam?

Question

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.

Answer 1

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 $2\frac dv$. If the dam is open the round trip swimming time is $\frac d{v+c}+\frac d{v-c}$ We have $$\frac d{v+c}+\frac d{v-c}=\frac {d(v-c)+d(v+c)}{(v+c)(v-c)}\\ =\frac {2dv}{v^2-c^2}\\=2\frac dv\frac 1{1-\frac {c^2}{v^2}}$$ 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.

Answer 2

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.

Answer 3

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 $\frac1{v+w}+\frac1{v-w}=\frac{2v}{v^2-w^2}$ time to make a round. When the dam is closed and the river is still, Mariam takes $\frac2v=\frac{2v}{v^2}$ time to make a round. Since $v^2-w^2<v^2$, $\frac{2v}{v^2-w^2}>\frac2v$, so Mariam always takes lesser time when the dam is closed than when the dam is open.

Answer 4

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.