Question

The water in a river flows uniformly at a constant speed of $\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{\text{m/s}}$ between parallel banks role="math" localid="1663689914645" $\mathbf{80}\mathbf{}\mathbf{\text{km}}$apart. You are to deliver a package across the river, but you can swim only at $\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{\text{m/s}}$. (a) If you choose to minimize the time you spend in the water, in what direction should you head? (b) How far downstream will you be carried? (c) If you choose to minimize the distance downstream that the river carries you, in what direction should you head?

(d) How far downstream will you be carried?

Short answer

(a) Perpendicular to stream.

(b) The distance he will be carried is $1.33\text{m}$.

(c)The swimmer must swim at ${37}^{\circ }$upstream from perpendicular direction to minimize the distance carried away downstream.

(d) The distance carried by downstream is $107\text{m}$.

Step-by-step solution

Definition of downstream.

Downstream refers to the direction of or proximity to a stream's mouth.

Find the direction chosen to minimize the time spend in the water.

(a)

To save time, the swimmer must swim as fast as he can towards the other bank, or as fast as he can perpendicular to the bank. That is feasible if he begins swimming perpendicular to the bank or river flow, that is, if he begins swimming directly towards the other bank.

Find the distance of the downstream you will carry.

(b)

If he swims to the other bank, his speed will be $\mathrm{v}=1.50\mathrm{m}/\mathrm{s}$.

As a result, the time it takes him to reach the other bank is

$\begin{array}{c}t=\frac{80\mathrm{m}}{1.50\mathrm{m}/\mathrm{s}}\\ =53.3\mathrm{s}\end{array}$

The water moves at a pace of ${\mathrm{v}}_{\mathrm{w}}=2.50\mathrm{m}/\mathrm{s}$.

As a result, the distance he will be carried is

$\begin{array}{c}d=v_{w}t\\ =(2.50\mathrm{m}/\mathrm{s})(53.3\mathrm{s})\\ =133\mathrm{m}\end{array}$

Find the direction chosen to minimize the distance downstream that the river carries.

(c)

Consider the necessity for the swimmer to swim at an angle $\theta$above the direct line.

As a result, the velocity's perpendicular component is

${\mathrm{v}}_{\perp }=(1.50\mathrm{m}/\mathrm{s})\mathrm{cos\theta }$

and the parallel component of the velocity is

${\mathrm{v}}_{\parallel }=(2.50\mathrm{m}/\mathrm{s})-(1.50\mathrm{m}/\mathrm{s})\mathrm{sin\theta }$

Hence, the total time taken to reach the other bank is

$t=\frac{(80\mathrm{m})}{(1.50\mathrm{m}/\mathrm{s})\cos \theta }$

Hence, the total distance travelled in

$\begin{array}{c}y=\left[\right(2.50\mathrm{m}/\mathrm{s})-(1.50\mathrm{m}/\mathrm{s}\left)\sin \theta \right]\frac{(80\mathrm{m})}{(1.50\mathrm{m}/\mathrm{s})\cos \theta }\\ =\frac{133\mathrm{m}}{\cos \theta }-(80\mathrm{m})\tan \theta \end{array}$

Now, differentiating with$\theta$we have

$\frac{dy}{d\theta }=\frac{\left(133\mathrm{m}\right)\sin \theta }{{\cos }^2\theta }-(80\mathrm{m}){\sec }^2\theta$

Now, for downstream to be minimum,

$\begin{array}{c}\frac{dy}{d\theta }=0\\ \frac{\left(133\mathrm{m}\right)\sin \theta }{{\cos }^2\theta }-(80\mathrm{m}){\sec }^2\theta =0\\ \sin \theta =\frac{80\mathrm{m}}{133\mathrm{m}}\\ \theta =\arcsin \left(\frac{80\mathrm{m}}{133\mathrm{m}}\right)\\ \theta =37.0°\end{array}$

Hence, the swimmer must swim at ${37}^{\circ }$upstream from perpendicular direction to minimize the distance carried away downstream.

Find the distance of the downstream you will carry.

(d)

The time taken by the swimmer, in this case, to cross the river is

$\begin{array}{c}t=\frac{(80\mathrm{m})}{(1.50\mathrm{m}/\mathrm{s})\cos \left(37.0°\right)}\\ =66.8\mathrm{s}\end{array}$

Hence, the distance carried by downstream is

$\begin{array}{c}d=\left[(2.50\mathrm{m}/\mathrm{s})-(1.50\mathrm{m}/\mathrm{s})\sin \left(37.0°\right)\right](66.8\mathrm{s})\\ =107\mathrm{m}\end{array}$