# Swimming across a river [closed]

I have come across a physics riddle where 2 swimmers with same speed $c$ are competing against each other in a river, they start at the same spot, first swimmer (s1) is swimming distance $d$ up stream and back, while the other swimmer (s2) is swimming same distance but across and back. the river's stream has velocity of $v$. which one will arrive first at the starting point.

my answer: $$t_1 = 2dc/(c^2-v^2)$$ $$t_2 = 2d/(sqrt(c^2-v^2))$$

comparing both sides we get:

$$t_1 > t_2 : [c > sqrt(c^2-v^2)]$$

the actual answer of the riddle is $t_1<t_2$ they calculated $t_2$ = $2d/c$

how can the velocity of the second swimmer be $2d/c$ ?

it seems they have neglected in their answer the force of the current, and if so the swimmer would not be able to return to the starting point but some point offshore.

shouldn't the velocity be as i calculated $sqrt(c^2-v^2)$ ?

• Hi Mike and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this Meta post on asking homework/exercise questions and this Meta post for "check my work" problems. Commented Jun 18, 2017 at 8:00
• @JohnRennie thanks, i'm way past the age of home work :) its more of a hobby. the question was taken from here link, i was hopping for an explination of the fomula
– Mike
Commented Jun 18, 2017 at 8:01
• @JohnRennie ill clarify my question : i have calculated that the velocity of s2 is sqrt(c^2-v^2) while the answer to the riddle claims to be c this seems false to me as there is no connection to the velocity of the river in their answer
– Mike
Commented Jun 18, 2017 at 8:19

If the second swimmer's aim is simply to cross to the other bank and back to the first bank (not necessarily needing to get back to his starting point) then he can ignore the water flow and direct all his swimming energy in heading across the river. He can just swim a distance of 2d relative to the water and ignore the fact he is drifting downstream.

If his target is to get back to the starting point then he must head into the flow and your calculation is correct.

@IanF1's already given the details, I'll just expand them a bit.

Let's call the two swimmers, $A$ and $B$. Swimmer $A$ goes upstream and then comes back, and you have already calculated, which is simply - $$t_A = \dfrac{l}{c - v} + \dfrac{l}{c + v} = \dfrac{2lc}{c^2 - v^2}$$

Swimmer B goes across the river, and then comes back. Here's the catch - If swimmer $A$ wants to come back to the starting point, he must do so with zero drift. (The condition for zero drift is that the angle b/w the relative velocity of the swimmer with respect to the river - $v_{rel}$) and the absolute velocity of the swimmer(with respect to the earth) should be $\sin^{-1} \Bigg(\dfrac{v_r}{v_{rel}}\Bigg)$, where $v_r$ is the velocity of the river)

Hence, $$t_B = \dfrac{2l}{\sqrt{c^2 - v^2}}$$ Which you have got. (Obtaining the velocity of the swimmer can be done through some simple vector subtraction)

Now, the conclusion that $c > \sqrt{c^2 - v^2}$ is correct; which tells us that swimmer $B$ takes lesser time overall.

But, the solution for $t_2 = \dfrac{2d}{c}$ is correct when the swimmer doesn't care him being drifted by the river, so he won't reach the starting point. That would displace him to a point on the bank, and then walk the distance due to the drift.

This only happens when the swimmer wants to reach in the shortest time, but doesn't care where he lands up.