June 10, 2008

Q. 1.7 -- "Problems in General Physics" by I.E. Irodov

$\textbf{Question}$:

Two swimmers leave point $A$on one bank of the river to reach point $B$lying right across on the other bank. One of them crosses the river along the straight line $AB$while the other swims at right angles to the stream and then walks the distance that he has been carried away by the stream to get to point $B$. What was the velocity $u$of his walking if both swimmers reached the destination simultaneously? The stream velocity $v_0$= 2.0 km/hour and the velocity $v'$of each swimmer with respect to the water equals 2.5 km/hour.

$\textbf{Answer}$:

Image diag

Assume that width of river is $w$.

Image diag2

Since the first swimmer crossed the river along the straight line $AB$, his velocity would be such that $v'\sin\phi_1 = v_0$. As a result the first swimmer takes $\frac{w}{v'\cos\phi_1}$time to cross the river.

The second swimmer swims perpendicular to the stream. Time taken by him to cross the river is $\frac{w}{v'}$. As a result, in this time he also goes downstream by a distance of $v_0\frac{w}{v'}$.

Since both the swimmers reach point $B$simultaneously, the time taken by second swimmer to cover the distance on ground is

\begin{displaymath}\frac{w}{v'\cos\phi_1} - \frac{w}{v'}\end{displaymath}
\begin{displaymath}= w\frac{1-\cos\phi_1}{v'\cos\phi_1} \end{displaymath}

In this time, the swimmer must cover the distance by which he was displaced downstream. Hence, his speed on ground:

\begin{displaymath}u = \frac{\frac{wv_0}{v'}}{w\frac{1-\cos\phi_1}{v'\cos\phi_1}} \end{displaymath}

\begin{displaymath}= \frac {v_0\cos\phi_1}{1-\cos\phi_1} \end{displaymath}

Now,

\begin{displaymath}\sin\phi_1 = \frac{v_0}{v'} \end{displaymath}

\begin{displaymath}\Rightarrow \cos\phi_1 = \sqrt{1-\frac{v_0^2}{v'^2}}\end{displaymath}

Using this we get:

\begin{displaymath}u = \frac{v_0\sqrt{1-\frac{v_0^2}{v'^2}}}{1-\sqrt{1-\frac{v_0^2}{v'^2}}} \end{displaymath}

\begin{displaymath}= \frac{v_0}{(1-\frac{v_0^2}{v'^2})^{-\frac{1}{2}} - 1} \end{displaymath}

Using the specified values, we get $u = \frac{2}{(1-(\frac{2}{2.5})^2)^{-\frac{1}{2}}-1} = \frac{2}{\frac{10}{6} - 1} = 3$km/hour.

Posted by Apurva Sharan at 11:23 PM
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