Question

Two swimmers, Chris and Sarah, start together at the same point on the bank of a wide stream that flows with a speed $\mathit{v}$. Both move at the same speed$\mathit{c}$(where $\mathit{c}\mathbf{>}\mathit{v}$) relative to the water. Chris swims downstream a distance $\mathit{L}$ and then upstream the same distance. Sarah swims so that her motion relative to the Earth is perpendicular to the banks of the stream. She swims the distance $\mathit{L}$ and then back the same distance, with both swimmers returning to the starting point. In terms of $\mathit{L}\mathbf{,}\mathit{c}$and $\mathit{v}$, find the time intervals required (a) for Chris's round trip and (b) for Sarah's round trip. (c) Explain which swimmer returns first.

Short answer

a) The time interval for Chris’s round trip is $\frac{\left(\frac{2L}{c}\right)}{\left(1-\frac{v^2}{c^2}\right)}$.

b) The time interval for Sarah’s trip is $\frac{\left(\frac{2L}{c}\right)}{\sqrt{1-\frac{v^2}{c^2}}}$.

c) Sarah returns first to the starting point.

Step-by-step solution

Definition of acceleration.

The rate at which an object's velocity changes with respect to time is called acceleration. Accelerations are measured in terms of vectors. The orientation of the net force applied on an item determines the orientation of its acceleration.

The time interval required for Chris’s round trip.

(a)

Speed of the stream is$V$

Speed of Chris (or) Sarah is $C$

Chris now walks downstream for a distance of $\text{'}L\text{'}$then upstream for the same distance, returning to the beginning position.

Then, total time taken by Chris is

$\begin{array}{c}{t}_{\mathrm{Chris}}=t_{\mathrm{diown}}+t_{\mathrm{up}}\\ =\frac{L}{c+v}+\frac{L}{c-v}\\ =\frac{2cL}{c^2-v^2}\\ =\frac{2cL}{c^2\left[1-\frac{v^2}{c^2}\right]}\\ =\frac{\left(\frac{2L}{c}\right)}{\left(1-\frac{v^2}{c^2}\right)}\end{array}$

The time interval required for Sarah’s round trip.

(b)

Sarah now travels the same '$L$' distance and returns to the beginning position by travelling the same distance, making her motion perpendicular to the earth.

The time spent in this situation is then

$\begin{array}{c}{t}_{\mathrm{Sumal}}=\frac{2L}{\sqrt{c^2-v^2}}\\ =\frac{\left(\frac{2L}{c}\right)}{\sqrt{1-\frac{v^2}{c^2}}}\end{array}$

Explain which swimmer returns first.

(c)

Using the results of (a) and (b):

Here, $\sqrt{1-\frac{v^2}{c^2}}>1-\frac{v^2}{c^2}$

Then, $t_{\mathrm{Saruh}}<t_{\mathrm{Chris}}$

Thus, time taken by Sarah is small. So, Sarah returns first to the starting point.