Chapter 1, Problem 59P

To determine

The final location of the car..

Answer to Problem 59P

The final location of the car relative to the origin is $2.3\text{km}$$34°\text{north}\text{of}\text{east}$

Explanation of Solution

Write the first displacement of the car in vector notation by considering East as positive x coordinate and North as positive y coordinate.

$A=A_x\cos \theta \widehat{i}+A_j\cos \theta \widehat{j}$

Here, $A$ is the first displacement of the car, $\theta$ is the angle measured with respect to x axis, $A_x$ and $A_y$ are the displacement along x  and y axis respectively.

Write the second displacement of the car in vector notation.

 $B=B_x\cos {\theta }^{\prime }\widehat{i}+B_j\sin {\theta }^{\prime }\widehat{j}$

Here, $B$ is the first displacement of the car, ${\theta }^{\prime }$ is the angle measured with respect to x axis, $B_x$ and $B_y$ are the displacement along x  and y axis respectively.

Write the expression to find the distance of the car relative to origin.

    $C=\sqrt{C_x^2+C_y^2}$        (I)

Here, $C$ is the final distance of the car relative to origin, $C_x$ is the final distance car moved from the origin along x , and $C_y$ is the final distance car moved from the origin along y.

Write the net displacement along x and y  axis.

$C_x=A_x\cos \theta +B_x\cos {\theta }^{\prime }$        (II)

$C_y=A_y\sin \theta +B_y\sin {\theta }^{\prime }$        (III)

Write the expression to find the angle.

    $\theta ={\tan }^{-1}\left(\frac{C_y}{C_x}\right)$        (IV)

Conclusion:

Substitute $0$ for $A_x$, $3.1\text{km}$ for $A_y$, and $53°$ for ${\theta }^{\prime }$ in (II) to find $C_x$

    $\begin{array}{c}{C}_x=\left(0\right)+\left(3.1\text{km}\right)\cos \left(53°\right)\\ =1.9\text{km}\end{array}$

Substitute $1.2\text{km}$ for $A_x$, $3.1\text{km}$ for $A_y$, $-90°$ for $\theta$, and $53°$ for ${\theta }^{\prime }$ in (III) to find $C_y$

    $\begin{array}{c}{C}_y=\left(1.2\text{km}\right)\sin \left(-90°\right)+\left(3.1\text{km}\right)\sin \left(53°\right)\\ =1.3\text{km}\end{array}$

Substitute $1.9\text{km}$ for $C_x$ and $1.3\text{km}$ for $C_y$ in (I).

    $\begin{array}{c}C=\sqrt{{\left(1.9\text{km}\right)}^2+{\left(1.3\text{km}\right)}^2}\\ =2.3\text{km}\end{array}$

Substitute $1.9\text{km}$ for $C_x$ and $1.3\text{km}$ for $C_y$ in (IV) to find $\theta$

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{1.3\text{km}}{1.9\text{km}}\right)\\ =34°\end{array}$

Therefore, the final location of the car relative to the origin is $2.3\text{km}$$34°\text{north}\text{of}\text{east}$