## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 6 - Section 6.6 - Vectors - Exercise Set - Page 784: 113

#### Answer

The plane’s true speed relative to the ground is $269.08\ \text{miles/hr}$ and its compass heading relative to the ground is $278.34{}^\circ$.

#### Work Step by Step

Let the velocity of the plane with respect to earth be ${{\mathbf{v}}_{1}}$. The velocity of the plane with respect to wind is ${{\mathbf{v}}_{2}}$. The velocity of wind with respect to earth is $\mathbf{u}$. The formula to calculate the speed of the plane with respect to earth is ${{\mathbf{v}}_{1}}=\mathbf{u}+{{\mathbf{v}}_{2}}$ The speed of the plane with respect to the wind is \begin{align} & {{\mathbf{v}}_{2}}=\left\| {{\mathbf{v}}_{2}} \right\|\cos \theta \mathbf{i}+\left\| {{\mathbf{v}}_{2}} \right\|\sin \theta \mathbf{j} \\ & =240\cos 170{}^\circ \mathbf{i}+240\sin 170{}^\circ \mathbf{j} \\ & =-236.35\mathbf{i}+41.67\mathbf{j} \end{align} Here, $\theta$ is the angle of the plane’s speed with respect to the wind from the horizontal axis in the anticlockwise direction. The speed of the wind with respect to the earth is \begin{align} & \mathbf{u}=\left\| \mathbf{u} \right\|\cos \varphi \mathbf{i}+\left\| \mathbf{u} \right\|\sin \varphi \mathbf{j} \\ & =30\cos 185{}^\circ \mathbf{i}+30\sin 185{}^\circ \mathbf{j} \\ & =-29.88\mathbf{i}-2.61\mathbf{j} \end{align} Here, $\varphi$ is the angle of the wind’s speed with respect to earth from the horizontal axis in the anticlockwise direction. So, the speed of the plane with respect to the earth is \begin{align} & {{\mathbf{v}}_{1}}=\mathbf{u}+{{\mathbf{v}}_{2}} \\ & =\left( -29.88\mathbf{i}-2.61\mathbf{j} \right)+\left( -236.35\mathbf{i}+41.67\mathbf{j} \right) \\ & =-266.23\mathbf{i}+39.06\mathbf{j} \end{align} The magnitude of the plane’s speed with respect to the earth is \begin{align} & {{\mathbf{v}}_{1}}=\sqrt{{{\left( -266.23 \right)}^{2}}+{{\left( 39.06 \right)}^{2}}} \\ & =269.08\ \text{miles/hr} \end{align} The direction of a plane’s speed with respect to the earth is \begin{align} & \tan \phi =\frac{39.06}{-266.23} \\ & =188.34{}^\circ \end{align} Here, $\phi$ is the angle of the plane’s speed with respect to the earth from the horizontal axis in the anticlockwise direction. Therefore, the compass heading relative to the ground is $270{}^\circ +8.34{}^\circ =278.34{}^\circ$.

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