Let us say , a swimmer has to swim upstream and downstream a river , between two points which are at distance $L$ from each other . The swimmer can swim in still-water at velocity $V_s$ , and the velocity of the river is $V_r$.
Now in the frame or reference of the river bank (or,ground) , his velocity upstream is $V_s-V_r$, and downstream is $V_s+V_r$, so the total time taken is
$$T_1=L \Bigg[\frac{1}{V_s+V_r} + \frac{1}{V_s - V_r}\Bigg]$$
Now if we observe the motion in the frame of reference of the river , the swimmer's velocity is Vs only , and as the distance is same , the time taken in this case is
$$T_2=\frac{2L}{V_s}$$
But shouldn't the time come the same ,regardless of the frame of reference ? I've tried putting in some values of $V_s$ and $V_r$ also and they are certainly not equal .
Please help me out , and point out where am I making a mistake and what should be the correct way of solving in different frames if references . Thank you .
Let's label the starting point $O$, the goal $A$, the swimmer $S$, and the reference frame of the river $R.$ Let's also define the positive direction as being downstream, so that $V_{R/O} = V_r.$
If we accept that the river has some velocity with respect to $O$ and $A$, then points $O$ and $A$ have an equal and opposite velocity in the $R$ reference frame. So, let $V_{A/R} = -V_r$. Also, the swimmer is swimming upstream, which is the negative direction, so let $V_{S/R} = -V_s.$
Now, the distance between the swimmer and $A$ can be written as $X_{A/S} = X_{A/R} - X_{S/R},$ where $X_{A/R}$ and $X_{S/R}$ are the positions of $A$ and $S$ with respect to $R.$ If you're confused where this comes from, I suggest drawing a diagram.
Taking derivatives with respect to time on both sides gives us $V_{A/S} = V_{A/R} - V_{S/R} = -V_r - (-V_s) = V_s - V_r,$ matching our result when measuring from the fixed frame.
Now, after the swimmer turns around they will have a new velocity $V_{S/R} = V_r,$ as they are now moving in the positive direction.
Now we can do $X_{O/S} = X_{O/R} - X_{S/R},$ so $V_{O/S} = V_{O/R} - V_{S/R} = -V_r - V_s,$ which matches our result when measuring from the fixed frame. (the sign is negative here because from the swimmer's perspective $O$ is travelling upstream, which is negative)