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### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# Suppose a pilot flies$\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{km}$ in a direction${\mathbf{60}}^{\mathbf{°}}$ north of east and then flies in a direction role="math" localid="1668679997174" ${\mathbf{15}}^{\mathbf{°}}$north of east as shown in Figure 3.63. Find her total distance $\mathbit{R}$ from the starting point and the direction $\mathbit{\theta }$of the straight-line path to the final position. Discuss qualitatively how this flight would be altered by a wind from the north and how the effect of the wind would depend on both wind speed and the speed of the plane relative to the air mass.

The total distance from the starting point is $64.8\mathrm{km}$, and is directed towards $40.{9}^{°}$, the north of east. If the wind speed is less than the speed of the plane, it is possible to travel to the northeast, but she will be traveling more to the east than when the wind is absent. If the wind speed exceeds the plane's, it is impossible to travel in the northeast direction, and it will be traveling southeast.

See the step by step solution

## Step 1: Triangle law of vector addition

When two vectors are taken along two sides of a triangle, the magnitude and direction of the resultant vector are always in reverse order on the third side of the triangle.

## Step 2: Given data

• The magnitude of the vector is $A$, $A=40\text{km}$.
• The magnitude of the vector is $B$, $B=30\text{km}$.
• The direction of the vector is $A$, ${\theta }_{A}=60°\text{NE}$.
• The direction of the vector is role="math" localid="1668680440938" $B$,${\theta }_{E}=15°\text{NE}$ .

## Step 3: Horizontal components of vectors

The horizontal component of the vector is,

${A}_{x}=A\mathrm{cos}{\theta }_{A}$

Here$A$is the magnitude of a vector $A$and ${\theta }_{A}$is the angle between the horizontal axis and vector $A$.

Substitute values in the above expression, and we get,

$\begin{array}{rcl}{A}_{x}& =& \left(40\text{km}\right)×\mathrm{cos}\left(60°\right)\\ & =& 20\text{km}\\ & & \end{array}$

The horizontal component of the vector $B$ is,

${B}_{x}=B\mathrm{cos}{\theta }_{B}$

Here $B$is the magnitude of the vector $B$and${\theta }_{B}$ is the angle between the horizontal axis and vector $B$.

Substitute values in the above expression, and we get,

$\begin{array}{rcl}{B}_{x}& =& \left(30\text{km}\right)×\mathrm{cos}\left(15°\right)\\ & =& 28.98\text{km}\\ & & \end{array}$

The horizontal component of the resultant vector $R$is,

${R}_{x}={A}_{x}+{B}_{x}$

Substitute values in the above expression, and we get,

$\begin{array}{rcl}{R}_{x}& =& \left(20\text{km}\right)+\left(28.98\text{km}\right)\\ & =& 48.98\text{km}\\ & & \end{array}$

## Step 4: Vertical component of vectors

The vertical component of the vector $A$is,

${A}_{y}=A\mathrm{sin}{\theta }_{A}$

Here$A$ is the magnitude of a vector$A$ and ${\theta }_{A}$is the angle between the horizontal axis and vector A.

Substitute values in the above expression, and we get,

$\begin{array}{rcl}{A}_{y}& =& \left(40\text{km}\right)×\mathrm{sin}\left(60°\right)\\ & =& 34.64\text{km}\\ & & \end{array}$

The vertical component of the vector $B$is,

${B}_{y}=B\mathrm{cos}{\theta }_{B}$

Here $B$is the magnitude of the vector $B$and ${\theta }_{B}$is the angle between the horizontal axis and vector $B$.

Substitute values in the above expression, and we get,

$\begin{array}{rcl}{B}_{y}& =& \left(30\text{km}\right)×\mathrm{sin}\left(15°\right)\\ & =& 7.76\text{km}\\ & & \end{array}$

The vertical component of the resultant vector $R$is,

${R}_{y}={A}_{y}+{B}_{y}$

Substitute values in the above expression, and we get,

$\begin{array}{rcl}{R}_{y}& =& \left(34.64\text{km}\right)+\left(7.76\text{km}\right)\\ & =& 42.4\text{km}\\ & & \end{array}$

## Step 5: Magnitude and direction of the resultant vector

The magnitude of the resultant vector$R$ is,

$R=\sqrt{{R}_{x}^{2}+{R}_{y}^{2}}$

Substitute values in the above expression, and we get,

$\begin{array}{rcl}R& =& \sqrt{{\left(48.98\text{km}\right)}^{2}+{\left(42.4\text{km}\right)}^{2}}\\ & =& 64.8\text{km}\\ & & \end{array}$

The direction of the resultant vector $R$is,

$\theta ={\mathrm{tan}}^{-1}\left(\frac{{R}_{y}}{{R}_{x}}\right)$

Substitute values in the above expression, and we get,

$\begin{array}{rcl}\theta & =& {\mathrm{tan}}^{-1}\left(\frac{42.4\text{km}}{48.98\text{km}}\right)\\ & =& 40.9°\\ & & \end{array}$

Hence, the total distance from the starting point is $64.8\mathrm{km}$and is directed towards $40.9°$the north of the east.

## Step 6: Qualitative description

If the wind speed is less than the speed of the plane, it is possible to travel to the northeast, but she will be traveling more to the east than when the wind is absent. If the wind speed exceeds the plane's, it is impossible to travel in the northeast direction, and it will be traveling southeast.