Chapter 3, Problem 49P

(a)

To determine

To draw the initial and final velocity vectors.

Answer to Problem 49P

The diagram of the initial and final velocity vectors is

Explanation of Solution

Velocity is rate of change of displacement of an object. It is a vector and has both magnitude and direction. The initial and final velocity vectors of the person is drawn in figure 1.

In figure, ${\overrightarrow{v}}_i$ is the initial velocity vector. It has magnitude of $192\text{ km/h}$ and is directed along the north direction. ${\overrightarrow{v}}_f$ is the final velocity vector. It has magnitude $240\text{ km/h}$ and is at angle $45°$ west of north.

Conclusion:

Therefore, the initial and final velocity vectors is drawn in figure 1.

(b)

To determine

The change in velocity.

Answer to Problem 49P

The change in velocity is $170\text{ km/h at }7°\text{ south of west}$ .

Explanation of Solution

Take the positive x-axis to be directed along west direction and the positive y-axis to be directed along north direction.

Write the expression for the initial velocity vector.

  ${\overrightarrow{v}}_i=192\text{ km/h, north}$

Write the expression for the x component of ${\overrightarrow{v}}_i$ .

  $v_{ix}=0$        (I)

Here, ${\overrightarrow{v}}_{ix}$ is the x component of ${\overrightarrow{v}}_i$ .

Write the expression for the y component of ${\overrightarrow{v}}_i$ .

  $v_{iy}=192\text{ km/h}$        (II)

Here, $v_{iy}$ is the y component of ${\overrightarrow{v}}_i$ .

Write the expression for the final velocity vector.

  ${\overrightarrow{v}}_f=240\text{ km/h, 45}°\text{ west of north}$

Here, ${\overrightarrow{v}}_f$ is the final velocity vector.

Write the expression for the x component of ${\overrightarrow{v}}_f$ .

  $v_{fx}=\left(240\text{ km/h}\right)\sin 45°$        (III)

Here, $v_{fx}$ is the x component of ${\overrightarrow{v}}_i$ .

Write the expression for the y component of ${\overrightarrow{v}}_f$ .

  $v_{fy}=\left(240\text{ km/h}\right)\cos 45°$        (IV)

Here, $v_{fy}$ is the y component of ${\overrightarrow{v}}_i$

Write the equation for the x-component of the change in velocity.

  $\Delta v_x=v_{fx}-v_{ix}$

Here, $\Delta v_x$ is the x-component of the change in velocity.

Put equations (I) and (III) in the above equation.

  $\begin{array}{c}\Delta v_x=\left(240\text{ km/h}\right)\sin 45°-0\\ =\left(240\text{ km/h}\right)\sin 45°\end{array}$        (V)

Write the equation for the y-component of the change in velocity.

  $\Delta v_y=v_{fy}-v_{iy}$

Here, $\Delta v_y$ is the y-component of the change in velocity.

Put equations (II) and (IV) in the above equation.

  $\Delta v_y=\left(240\text{ km/h}\right)\cos 45°-192\text{ km/h}$        (VI)

Write the equation for the magnitude of the change in velocity.

  $\Delta v=\sqrt{{\left(\Delta v_x\right)}^2+{\left(\Delta v_y\right)}^2}$        (VII)

Here, $\Delta v$ is the magnitude of the change in velocity.

Write the equation for the direction of the change in velocity.

  $\theta ={\tan }^{-1}\frac{\Delta v_y}{\Delta v_x}$        (VIII)

Here, $\theta$ is the angle the change in velocity makes with the x-axis.

Conclusion:

Put equations (V) and (VI) in equation (VII) to find $\Delta v$.

  $\begin{array}{c}\Delta v=\sqrt{{\left(\left(240\text{ km/h}\right)sin45°\right)}^2+{\left(\left(240\text{ km/h}\right)\cos 45°-192\text{ km/h}\right)}^2}\\ =171.1\text{ km/h}\\ \approx \text{170 km/h}\end{array}$

Put equations (V) and (VI) in equation (VIII) to find $\theta$.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{\left(240\text{ km/h}\right)\cos 45°-192\text{ km/h}}{\left(240\text{ km/h}\right)\sin 45°}\right)\\ =-7.4°\\ \approx -7°\end{array}$

The negative sign shows that the angle is clockwise or the change in velocity is at $7°$ south of west.

Therefore, the change in velocity is $170\text{ km/h at }7°\text{ south of west}$.

(c)

To determine

The average acceleration during the trip.

Answer to Problem 49P

The average acceleration during the trip is $570{\text{km/h}}^2\text{ at }7°\text{ south of west}$ .

Explanation of Solution

Write the equation for the magnitude of the average acceleration.

  $a_{\text{av}}=\frac{\Delta v}{\Delta t}$        (IX)

Here, $a_{\text{av}}$ is the magnitude of the average acceleration and $\Delta t$ is the time interval.

Conclusion:

Substitute $170\text{ km/h}$ for $\Delta v$ and $3.0\text{ h}$ for $\Delta t$ in equation (IX) to find $a_{\text{av}}$.

  $\begin{array}{c}{a}_{\text{av}}=\frac{170\text{ km/h}}{3.0\text{ h}}\\ =57{\text{ km/h}}^2\end{array}$

The direction of acceleration will be the same as that of change in velocity.

Therefore, the average acceleration during the trip is $570{\text{km/h}}^2\text{ at }7°\text{ south of west}$.