Question

1. a person can swim 0.65 m/s in still water. she heads directly south across a river 130 m wide and lands at a point 88 m w downstream. (a) determine the velocity of the water relative to the ground. (b) determine the swimmers velocity relative to earth. (c) determine the direction she should swim to land at a point directly south of the starting point.

Explanation:

Step1: Calculate time taken to cross the river

The swimmer's velocity in still - water in the south - direction $v_{s}=0.65\ m/s$ and the width of the river $d = 130\ m$. Using the formula $t=\frac{d}{v_{s}}$, we have $t=\frac{130}{0.65}=200\ s$.

Step2: Calculate velocity of water relative to the ground

The swimmer is displaced $x = 88\ m$ downstream in time $t = 200\ s$. The velocity of the water relative to the ground $v_{w}=\frac{x}{t}$. So, $v_{w}=\frac{88}{200}=0.44\ m/s$.

Step3: Calculate swimmer's velocity relative to Earth

The swimmer's velocity relative to Earth is the resultant of the velocity of the swimmer in still - water and the velocity of the water. Using the Pythagorean theorem $v=\sqrt{v_{s}^{2}+v_{w}^{2}}$. Substituting $v_{s}=0.65\ m/s$ and $v_{w}=0.44\ m/s$, we get $v=\sqrt{(0.65)^{2}+(0.44)^{2}}=\sqrt{0.4225 + 0.1936}=\sqrt{0.6161}\approx0.785\ m/s$. The direction $\theta=\tan^{- 1}(\frac{v_{w}}{v_{s}})=\tan^{-1}(\frac{0.44}{0.65})\approx34.1^{\circ}$ west of south.

Step4: Calculate the direction to swim to land directly south

Let the angle $\alpha$ be the angle the swimmer should make with the south direction. We know that $\sin\alpha=\frac{v_{w}}{v_{s}}$. Substituting $v_{w}=0.44\ m/s$ and $v_{s}=0.65\ m/s$, we get $\sin\alpha=\frac{0.44}{0.65}$, so $\alpha=\sin^{-1}(\frac{0.44}{0.65})\approx42.1^{\circ}$ east of south.

Answer:

(a) $0.44\ m/s$
(b) $0.785\ m/s$, $34.1^{\circ}$ west of south
(c) $42.1^{\circ}$ east of south