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Q62E

Expert-verifiedFound in: Page 124

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**The velocity of the wind relative to the water is crucial to sailboats. Suppose a sailboat is in an ocean current that has a velocity of ** **m/s in a direction ** **east of north relative to the Earth. It encounters a wind that has a velocity of ** **m/s in a direction of ** **south of west relative to the Earth. What is the velocity of the wind relative to the water?**

The velocity of the wind relative to that of water is 68 m/s. The resultant vector has the direction of $53.{3}^{\xb0}$SW .

**Velocity is the rate of change in position of an item in motion as seen from a specific frame of reference and measured by a specific time standard.**

On the ship, you will see that the body is in the y-direction only. The details are given in the sketch. W represents the water, a represents the wind.

Hence the velocity can be calculated by the following equation.

${V}_{aw}={V}_{ae}+{V}_{ew}$

The velocity of the sandal relative to the ship is -17.1 m/s

The component table will be as below:

| X | Y |

V | -2.89 | -3.45 |

V | -1.1 | -1.91 |

Resultant | -3.99 | -5.36 |

Here we need to find the x component and the y component.

$Xcomponent\phantom{\rule{0ex}{0ex}}{V}_{aex}={V}_{i}\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}{V}_{aex}=\left(-4.5\right)\mathrm{cos}{50}^{\xb0}\phantom{\rule{0ex}{0ex}}{V}_{aex}=-2.89$

$Ycomponent\phantom{\rule{0ex}{0ex}}{V}_{aey}={V}_{i}\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}{V}_{aey}=\left(-4.5\right)\mathrm{sin}{50}^{\xb0}\phantom{\rule{0ex}{0ex}}{V}_{aey}=-3.45$

Hence, the velocity of water w.r.t earth:

${V}_{wex}={V}_{we}\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}{V}_{wex}=\left(2.2\right)\mathrm{cos}{30}^{\xb0}\phantom{\rule{0ex}{0ex}}{V}_{wex}=1.10\phantom{\rule{0ex}{0ex}}{V}_{ewx}=-1.10\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

$Ycomponent\phantom{\rule{0ex}{0ex}}{V}_{wey}={V}_{we}\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}{V}_{wey}=\left(2.2\right)\mathrm{sin}{30}^{\xb0}\phantom{\rule{0ex}{0ex}}{V}_{wey}=1.91\phantom{\rule{0ex}{0ex}}{V}_{ewy}=-1.91\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Hence the resultant vector will be

$\stackrel{\rightharpoonup}{R}=\sqrt{{\left(X\right)}^{2}+{\left(Y\right)}^{2}}\phantom{\rule{0ex}{0ex}}\stackrel{\rightharpoonup}{R}=\sqrt{{\left(-3.99\right)}^{2}+{\left(-5.36\right)}^{2}}\phantom{\rule{0ex}{0ex}}\stackrel{\rightharpoonup}{R}=6.68\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Hence the velocity of the wind relative to that of water is 6.68 m/s.

The direction of the velocity will be;

$\begin{array}{rcl}\mathrm{tan}\theta & =& \frac{Y}{X}\\ \mathrm{tan}\theta & =& \frac{5.36}{3.99}\\ \theta & =& 53.3\end{array}$

The resultant vector has the direction of $53.{3}^{\xb0}SW$ .

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