Chapter 3, Problem 29P

(a)

To determine

The unit vector expression for the velocity of hurricane during first $3$ hours.

Answer to Problem 29P

The unit vector expression for the velocity of hurricane during first $3$ hours is $\left(-20.5\text{km}/\text{h}\right)\widehat{i}+\left(35.50\text{km}/\text{h}\right)\widehat{j}$.

Explanation of Solution

Given information:

The velocity of hurricane for the first $3$ hours is $41.0\text{km}/\text{h}$ with a direction of $60.0°$ north of west and the velocity of hurricane for next $1.50$ hour is $25.0\text{km}/\text{h}$ with a direction along north.

Consider the east direction as positive $x$ direction.

The unit vector expression of velocity of hurricane for first $3$ hours is given as,

${\overrightarrow{\text{v}}}_1=\left(-\left|{\overrightarrow{\text{v}}}_{\text{1}}\right|\cos 60.0°\right)\widehat{i}+\left(\left|{\overrightarrow{\text{v}}}_{\text{1}}\right|\sin 60.0°\right)\widehat{j}$ (I)

Substitute $41.0\text{km}/\text{h}$ for $\left|{\overrightarrow{\text{v}}}_{\text{1}}\right|$ in above expression in equation (I).

$\begin{array}{c}{\overrightarrow{\text{v}}}_1=\left(-\left(41.0\text{km}/\text{h}\right)\cos 60.0°\right)\widehat{i}+\left(\left(41.0\text{km}/\text{h}\right)\sin 60.0°\right)\widehat{j}\\ =\left(-20.5\text{km}/\text{h}\right)\widehat{i}+\left(35.50\text{km}/\text{h}\right)\widehat{j}\end{array}$

Conclusion:

Therefore, the unit vector expression for the velocity of hurricane during first $3$ hours is $\left(-20.5\text{km}/\text{h}\right)\widehat{i}+\left(35.50\text{km}/\text{h}\right)\widehat{j}$.

(b)

To determine

The unit vector expression for the new velocity of the hurricane.

Answer to Problem 29P

The unit vector expression for the new velocity of hurricane is $\left(25.0\text{km}/\text{h}\right)\widehat{j}$.

Explanation of Solution

Given information:

The velocity of hurricane for the first $3$ hours is $41.0\text{km}/\text{h}$ with a direction of $60.0°$ north of west and the velocity of hurricane for next $1.50$ hour is $25.0\text{km}/\text{h}$ with a direction along north.

The unit vector expression for the new velocity of the hurricane is,

${\overrightarrow{\text{v}}}_{\text{2}}=\left(\left|{\overrightarrow{\text{v}}}_{\text{2}}\right|\right)\widehat{j}$

Substitute $25.0\text{km}/\text{h}$ for $\left|{\overrightarrow{\text{v}}}_{\text{2}}\right|$ in above expression.

${\overrightarrow{\text{v}}}_{\text{2}}=\left(25.0\text{km}/\text{h}\right)\widehat{j}$

Conclusion:

Therefore, the unit vector expression for the new velocity of hurricane is $\left(25.0\text{km}/\text{h}\right)\widehat{j}$.

(c)

To determine

The unit vector expression for the displacement of the hurricane during first $3$ hours.

Answer to Problem 29P

The unit vector expression for the displacement of the hurricane during first $3$ hours is $\left(-61.5\text{km}\right)\widehat{i}+\left(106.5\text{km}\right)\widehat{j}$.

Explanation of Solution

Given information:

The velocity of hurricane for the first $3$ hours is $41.0\text{km}/\text{h}$ with a direction of $60.0°$ north of west and the velocity of hurricane for next $1.50$ hour is $25.0\text{km}/\text{h}$ with a direction along north.

Formula to determine the unit vector expression for the displacement during first $3$ hours is,

${\overrightarrow{\text{d}}}_1=\left({\overrightarrow{\text{v}}}_1\right)\left(t\right)$ (II)

• ${\overrightarrow{\text{v}}}_1$ is the velocity of hurricane during first $3$ hours.
• $t$ is the time.
• ${\overrightarrow{\text{d}}}_1$ is the displacement of hurricane for $3$ hours.

Substitute $\left(-20.5\text{km}/\text{h}\right)\widehat{i}+\left(35.50\text{km}/\text{h}\right)\widehat{j}$ for ${\overrightarrow{\text{v}}}_1$ and $3.00\text{hours}$ for $t$ in equation (II).

$\begin{array}{c}{\overrightarrow{\text{d}}}_1=\left(\left(-20.5\text{km}/\text{h}\right)\widehat{i}+\left(35.50\text{km}/\text{h}\right)\widehat{j}\right)\left(3.00\text{hours}\right)\\ =\left(-61.5\text{km}\right)\widehat{i}+\left(106.5\text{km}\right)\widehat{j}\end{array}$

Conclusion:

Therefore, the unit vector expression for the displacement of the hurricane during first $3$ hours is $\left(-61.5\text{km}\right)\widehat{i}+\left(106.5\text{km}\right)\widehat{j}$.

(d)

To determine

The unit vector expression for the displacement of the hurricane during the later $1.50$ hours.

Answer to Problem 29P

The unit vector expression for the displacement of the hurricane during the later $1.50$ hours is $\left(37.5\text{km}\right)\widehat{j}$.

Explanation of Solution

Given information:

The velocity of hurricane for the first $3$ hours is $41.0\text{km}/\text{h}$ with a direction of $60.0°$ north of west and the velocity of hurricane for next $1.50$ hour is $25.0\text{km}/\text{h}$ with a direction along north.

Formula to determine the unit vector expression for the displacement during the later $1.50$ hours is,

${\overrightarrow{\text{d}}}_2=\left({\overrightarrow{\text{v}}}_2\right)\left(t\right)$ (III)

• ${\overrightarrow{\text{v}}}_1$ is the velocity of hurricane during the later $1.50$ hours.
• $t$ is the time.
• ${\overrightarrow{\text{d}}}_1$ is the displacement of hurricane for the later $1.50$ hours.

Substitute $\left(25.0\text{km}/\text{h}\right)\widehat{j}$ for ${\overrightarrow{\text{v}}}_2$ and $1.50\text{hours}$ for $t$ in equation (III).

$\begin{array}{c}{\overrightarrow{\text{d}}}_1=\left(\left(25.0\text{km}/\text{h}\right)\widehat{j}\right)\left(1.50\text{hours}\right)\\ =\left(37.5\text{km}\right)\widehat{j}\end{array}$

Conclusion:

Therefore, the unit vector expression for the displacement of the hurricane during the later $1.50$ hours is $\left(37.5\text{km}\right)\widehat{j}$.

(e)

To determine

The net displacement of the hurricane after $4.50\text{hours}$.

Answer to Problem 29P

The net displacement of the hurricane after $4.50\text{hours}$ is $157\text{km}$.

Explanation of Solution

Section 1:

To determine: The unit vector expression for the net displacement after $4.50\text{hours}$.

Answer: The unit vector expression for the net displacement after $4.50\text{hours}$ is $\left(-61.5\text{km}\right)\widehat{i}+\left(144\text{km}\right)\widehat{j}$.

Given information:

The velocity of hurricane for the first $3$ hours is $41.0\text{km}/\text{h}$ with a direction of $60.0°$ north of west and the velocity of hurricane for next $1.50$ hour is $25.0\text{km}/\text{h}$ with a direction along north.

Formula to determine the unit vector expression for the net displacement after $4.50\text{hours}$ is,

$\overrightarrow{\text{d}}={\overrightarrow{\text{d}}}_1+{\overrightarrow{\text{d}}}_2$ (IV)

Substitute $\left(-61.5\text{km}\right)\widehat{i}+\left(106.5\text{km}\right)\widehat{j}$ for ${\overrightarrow{\text{d}}}_1$ and $\left(37.5\text{km}\right)\widehat{j}$ for ${\overrightarrow{\text{d}}}_2$ in equation (IV).

$\begin{array}{c}\overrightarrow{\text{d}}=\left(-61.5\text{km}\right)\widehat{i}+\left(106.5\text{km}\right)\widehat{j}+\left(37.5\text{km}\right)\widehat{j}\\ =\left(-61.5\text{km}\right)\widehat{i}+\left(144\text{km}\right)\widehat{j}\end{array}$

Section 2:

To determine: The magnitude of the net displacement after $4.50\text{hours}$.

Answer: The magnitude of the net displacement after $4.50\text{hours}$ is $157\text{km}$.

Given information:

The velocity of hurricane for the first $3$ hours is $41.0\text{km}/\text{h}$ with a direction of $60.0°$ north of west and the velocity of hurricane for next $1.50$ hour is $25.0\text{km}/\text{h}$ with a direction along north.

From part (d), the unit vector expression for the net displacement after $4.50\text{hours}$ is given as,

$\overrightarrow{\text{d}}=\left(-61.5\text{km}\right)\widehat{i}+\left(144\text{km}\right)\widehat{j}$

Formula to calculate the magnitude of the net displacement after $4.50\text{hours}$ is,

$\left|\overrightarrow{\text{d}}\right|=\sqrt{{\left(d_{\text{x}}\right)}^2+{\left(d_{\text{y}}\right)}^2}$ (V)

Substitute $-61.5\text{km}$ for $d_{\text{x}}$ and $144\text{km}$ for $d_{\text{y}}$ in equation (V).

$\begin{array}{c}\left|\overrightarrow{\text{d}}\right|=\sqrt{{\left(-61.5\text{km}\right)}^2+{\left(144\text{km}\right)}^2}\\ =156.58\text{km}\\ \approx 157\text{km}\end{array}$

Conclusion:

Therefore, the magnitude of the net displacement after $4.50\text{hours}$ is $157\text{km}$