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 $sqrt(c^2-v^2)$ ?