Chapter 3, Problem 79P

To determine

The velocity of truck with respect to car.

Answer to Problem 79P

The relative velocity of track with respect to car is $63\text{km}/\text{hr}$ at $40°$ south -west direction.

Explanation of Solution

Take north as positive y axis, south as negative y-axis, west as negative x-axis and east as positive x-axis.

Write the velocity vector for car moving in north direction.

    ${\overrightarrow{v}}_{\text{car}}=\left(110\text{km}/\text{hr}\right)\widehat{j}$

Here, ${\overrightarrow{v}}_{\text{car}}$ is the velocity vector of car.

It is given that truck is moving at $35°$ degree west of north. It is equivalent to $125°$ from east direction (+ve -x-axis) in counterclockwise direction.

Write the velocity vector for truck moving in north-west direction.

    $\begin{array}{c}{\overrightarrow{v}}_{\text{truck}}=\left[\left(85\text{km}/\text{hr}\right)\left(\cos 125°\right)\right]\widehat{i}+\left[\left(85\text{km}/\text{hr}\right)\left(\sin 125°\right)\right]\widehat{j}\\ =\left(-48.7\text{km}/\text{hr}\right)\widehat{i}+\left(69.6\text{km}/\text{hr}\right)\widehat{j}\end{array}$

Here, ${\overrightarrow{v}}_{\text{truck}}$ is the velocity vector of truck.

Write the expression for relative velocity of truck with respect to car.

    ${\overrightarrow{v}}_{t-c}={\overrightarrow{v}}_{\text{truck}}-{\overrightarrow{v}}_{\text{car}}$

Here, ${\overrightarrow{v}}_{t-c}$ is the relative velocity of truck with respect to car.

Write the expression to find the magnitude of ${\overrightarrow{v}}_{t-c}$.

    $\left|{\overrightarrow{v}}_{t-c}\right|=\sqrt{x^2+y^2}$

Here, $\left|{\overrightarrow{v}}_{t-c}\right|$ is the magnitude of ${\overrightarrow{v}}_{t-c}$, $x$ is the x-component of ${\overrightarrow{v}}_{t-c}$ and y is the y-component of ${\overrightarrow{v}}_{t-c}$.

Write the expression to find the direction of ${\overrightarrow{v}}_{t-c}$.

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

Here ${\theta }_{t-c}$ is the direction of ${\overrightarrow{v}}_{t-c}$.

Conclusion:

Substitute $\left(-48.7\text{km}/\text{hr}\right)\widehat{i}+\left(69.6\text{km}/\text{hr}\right)\widehat{j}$ for ${\overrightarrow{v}}_{\text{truck}}$ and $\left(110\text{km}/\text{hr}\right)\widehat{j}$ for ${\overrightarrow{v}}_{\text{car}}$ in the equation for ${\overrightarrow{v}}_{t-c}$.

    $\begin{array}{c}{\overrightarrow{v}}_{t-c}=\left(-48.7\text{km}/\text{hr}\right)\widehat{i}+\left(69.6\text{km}/\text{hr}\right)\widehat{j}-\left(110\text{km}/\text{hr}\right)\widehat{j}\\ =\left(-48.7\text{km}/\text{hr}\right)\widehat{i}-\left(40.4\text{km}/\text{hr}\right)\widehat{j}\end{array}$

Substitute $-48.7\text{km}/\text{hr}$ for $x$ and $\left(-40.4\text{km}/\text{hr}\right)$ for y in the equation for $\left|{\overrightarrow{v}}_{t-c}\right|$.

    $\begin{array}{c}\left|{\overrightarrow{v}}_{t-c}\right|=\sqrt{{\left(-48.7\text{km}/\text{hr}\right)}^2+{\left(-40.4\text{km}/\text{hr}\right)}^2}\\ =\sqrt{8688}\text{km}/\text{hr}\\ =63\text{km}/\text{hr}\end{array}$

Substitute $-48.7\text{km}/\text{hr}$ for $x$ and $-40.4\text{km}/\text{hr}$ for y in the equation for ${\theta }_{t-c}$. Note that the x and y components are negative. It means that the angle comes in third quadrant.

    $\begin{array}{c}{\theta }_{t-c}={\tan }^{-1}\left(\frac{-40.4\text{km}/\text{hr}}{-48.7\text{km}/\text{hr}}\right)\\ ={\tan }^{-1}\left(0.83\right)\end{array}$

Consider the angle corresponding to $a{n}^{-1}\left(0.83\right)$ in third quadrant.

    ${\theta }_{t-c}=220°$

It is measured counterclockwise direction from the east direction (+ve x-axis). It is equivalent to say that relative velocity is directed at $40°$ south west.

Thus, the relative velocity of track with respect to car is $63\text{km}/\text{hr}$ at $40°$ south -west direction.