Chapter 11.4, Problem 11.121P

Airplanes A and B are flying at the same altitude and are tracking the eye of hurricane C. The relative velocity of C with respect to A is v_C/A = 350 km/h ⦫ 75°, and the relative velocity of C with respect to B is v_C/B = 400 km/h ⦪ 40°. Determine (a) the relative velocity of B with respect to A, (b) the velocity of A if ground-based radar indicates that the hurricane is moving at a speed of 30 km/h due north, (c) the change in position of C with respect to B during a 15-min interval.

Fig. P11.121

To determine

The relative velocity $\left(v_{B/A}\right)$ of B with respect to A.

Answer to Problem 11.121P

The relative velocity $\left(v_{B/A}\right)$ of B with respect to A is $\underset{\_}{405\text{km/h}}$ at an angle of $\underset{\_}{11.53°}$.

Explanation of Solution

Given Information:

The relative velocity $\left(v_{C/A}\right)$ of C with respect to A is $350\text{km/h}$ at an angle $75°$.

The relative velocity $\left(v_{C/B}\right)$ of C with respect to B is $400\text{km/h}$ at an angle $40°$.

Calculation:

Show the relative velocity vector diagram of airplanes with respect to hurricane as in Figure (1).

Write the relative velocity of hurricane with respect to airplane A:

$\begin{array}{l}{v}_{C/A}=v_C-v_A\\ v_C=v_{C/A}+v_A\end{array}$ (1)

Here, $v_C$ is velocity of hurricane, $v_{C/A}$ is relative velocity of hurricane with respect to airplane A, and $v_A$ is velocity of airplane A.

Write the relative velocity of hurricane with respect to airplane B:

$\begin{array}{l}{v}_{C/B}=v_C-v_B\\ v_C=v_{C/B}+v_B\end{array}$ (2)

Here, $v_C$ is velocity of hurricane, $v_{C/B}$ is relative velocity of hurricane with respect to airplane B and $v_B$ is velocity of airplane B.

Equate equations (1) and (2).

$\begin{array}{l}{v}_{C/A}+v_A=v_{C/B}+v_B\\ v_B-v_A=v_{C/A}-v_{C/B}\end{array}$ (3)

Write the relative velocity of airplane B with respect to airplane A:

$v_{B/A}=v_B-v_A$ (4)

Here, $v_{B/A}$ is relative velocity of airplane B with respect to airplane A.

Substitute equation (4) in equation (3).

$v_{B/A}=v_{C/A}-v_{C/B}$ (5)

Calculate relative velocity B with respective to A by applying law of cosine for Figure (1).

$v_{B/A}^2=v_{C/A}^2+v_{C/B}^2-2v_{C/A}{v}_{C/B}\cos \gamma$

Here, $\gamma$ for angle between the direction of the relative velocity of hurricane with respect to airplane A and the relative velocity of hurricane with respect to airplane B.

Substitute $350\text{km}/\text{h}$ for $v_{C/A}$, $400\text{km}/\text{h}$ for $v_{C/B}$, and $65°$ for $\gamma$.

$\begin{array}{c}{v}_{B/A}^2={\left(350\right)}^2+{\left(400\right)}^2-2\left(350\right)\left(400\right)\left(\cos 65°\right)\\ v_{B/A}=\sqrt{164,166.8864}\\ v_{B/A}=405.175\text{km}/\text{h}\end{array}$

Calculate the angle $\alpha$ by applying the law of sines for Figure (1):

$\frac{\sin \alpha }{v_{C/B}}=\frac{\sin \gamma }{v_{B/A}}$

Here, $\alpha$ is angle between the direction of relative velocity of hurricane with respect to airplane A and relative velocity of airplane B with respect to airplane A.

Substitute $400\text{km}/\text{h}$ for $v_{C/B}$, $405.175\text{km}/\text{h}$ for $v_{B/A}$, and $65°$ for $\gamma$.

$\begin{array}{l}\frac{\sin \alpha }{400}=\frac{\sin 65°}{405.175}\\ \sin \alpha =\frac{\sin 65°}{405.175}\times 400\\ \alpha =63.474°\end{array}$

Write the angle subtended by $v_{B/A}$ with the horizontal as:

$\theta =75°-\alpha$

Here, angle subtended by $v_{B/A}$ with the horizontal is $\theta$.

Substitute $63.474°$ for $\alpha$.

$\begin{array}{c}\theta =75°-63.474°\\ =11.526°\\ \approx 11.53°\end{array}$

Therefore, the relative velocity $\left(v_{B/A}\right)$ of B with respect to A is $\underset{\_}{405\text{km/h}}$ at an angle of $\underset{\_}{11.53°}$.

To determine

The velocity $\left(v_A\right)$ of A if ground based radar indicates that the hurricane is moving at a speed of $30\text{km/h}$ due north.

Answer to Problem 11.121P

The velocity $\left(v_A\right)$ of A if ground based radar indicates that the hurricane is moving at a speed of $30\text{km/h}$ due north is $\underset{\_}{379\text{km/h}}$ at an angle $\underset{\_}{76.17°}$.

Explanation of Solution

Given Information:

The relative velocity $\left(v_{C/A}\right)$ of C with respect to A is $350\text{km/h}$ at an angle $75°$.

The relative velocity $\left(v_{C/B}\right)$ of C with respect to B is $400\text{km/h}$ at an angle $40°$.

Calculation:

Rewrite Equation (1):

$v_A=v_C-v_{C/A}$ (6)

Write vector $v_{C/A}$ in terms of components by resolving the vector into horizontal and vertical component.

$v_{C/A}=-v_{C/A}\cos 75i-v_{C/A}\sin 75j$ (7)

Choose direction of $v_C$ in the direction of unit vector of $j$ since it moves in the direction of due north. Rewrite Equation (6) using Equation (7) as:

$v_A=v_{C}j-\left[-v_{C/A}\cos 75i-v_{C/A}\sin 75j\right]$ (8)

Substitute $30\text{km}/\text{h}$ for $v_C$ and $350\text{km}/\text{h}$ for $v_{C/A}$ in Equation (11) to find $v_C$.

$\begin{array}{c}{v}_A=30j-\left[-350\cos 75i-350\sin 75j\right]\\ =30j+90.587i+338.07j\\ =\left(90.587\text{km}/\text{h}\right)i+\left(368.07\text{km}/\text{h}\right)j\end{array}$

Calculate the resultant velocity $\left(v_A\right)$ of airplane A with respect to ground as:

$v_A=\sqrt{{\left(v_A\right)}_x^2+{\left(v_A\right)}_y^2}$

Here, ${\left(v_A\right)}_x$ is  x-component of $v_A$ and ${\left(v_A\right)}_y$ is y-component of $v_A$.

Substitute $90.587\text{km}/\text{h}$ for ${\left(v_A\right)}_x$ and $368.07\text{km}/\text{h}$ for ${\left(v_A\right)}_y$.

$\begin{array}{c}{v}_A=\sqrt{{\left(90.587\right)}^2+{\left(368.07\right)}^2}\\ =\sqrt{143,681.5295}\\ =379.0535\\ \approx 379\text{km}/\text{h}\end{array}$

Calculate the angle $\left({\theta }_A\right)$ of resultant velocity of airplane A with horizontal with respect to ground using the relation:

${\theta }_A={\tan }^{-1}\left(\frac{{\left(v_A\right)}_y}{{\left(v_A\right)}_x}\right)$

Substitute $90.587\text{km}/\text{h}$ for ${\left(v_A\right)}_x$ and $368.07\text{km}/\text{h}$ for ${\left(v_A\right)}_y$.

$\begin{array}{c}{\theta }_A={\tan }^{-1}\left(\frac{368.07}{90.587}\right)\\ ={\tan }^{-1}\left(4.0632\right)\\ =76.17°\end{array}$

Therefore, the velocity $\left(v_A\right)$ of A if ground based radar indicates that the hurricane is moving at a speed of $30\text{km/h}$ due north is $\underset{\_}{379\text{km/h}}$ at an angle $\underset{\_}{76.17°}$.

To determine

The change in position $\left(\Delta r_{C/B}\right)$ C with respect to B during a 15 min interval.

Answer to Problem 11.121P

The change in position $\left(\Delta r_{C/B}\right)$ C with respect to B during a 15 min interval is $\underset{\_}{100\text{km}}$ at an angle $\underset{\_}{40°}$.

Explanation of Solution

Given Information:

The relative velocity $\left(v_{C/A}\right)$ of C with respect to A is $350\text{km/h}$ at an angle $75°$.

The relative velocity $\left(v_{C/B}\right)$ of C with respect to B is $400\text{km/h}$ at an angle $40°$.

Calculation:

Write the position of airplane B as:

$\begin{array}{c}{v}_B=\frac{r_B-{\left(r_0\right)}_B}{t}\\ r_B=v_{B}t+{\left(r_0\right)}_B\end{array}$ (8)

Here, $r_B$ is final position of airplane B, ${\left(r_0\right)}_B$ is initial position of airplane B , $v_B$ is velocity of airplane B and t is time.

Write the position of hurricane as:

$\begin{array}{c}{v}_C=\frac{r_C-{\left(r_0\right)}_C}{t}\\ r_C=v_{C}t+{\left(r_0\right)}_C\end{array}$ (9)

Here, ${\left(r_0\right)}_C$ is final position of hurricane is $r_C$, initial position of hurricane and $v_C$ is velocity of hurricane.

Write the position of hurricane with respect to airplane B as:

$r_{C/B}=r_C-r_B$ (10)

Here, $r_{C/B}$ is position of hurricane with respect to airplane B.

Use Equations (8) and (9) to rewrite Equation (10)

$\begin{array}{c}{r}_{C/B}=v_{C}t+{\left(r_0\right)}_C-v_{B}t-{\left(r_0\right)}_B\\ =\left(v_C-v_B\right)t+\left[{\left(r_0\right)}_C-{\left(r_0\right)}_B\right]\\ =v_{C/B}t+\left[{\left(r_0\right)}_C-{\left(r_0\right)}_B\right]\end{array}$ (11)

Write the change in position of hurricane with respect to B as:

$\Delta r_{C/B}={\left(r_{C/B}\right)}_{t_2}-{\left(r_{C/B}\right)}_{t_1}$ (12)

Here, $\Delta r_{C/B}$ is change in position of hurricane with respect to B, ${\left(r_{C/B}\right)}_{t_2}$ position of hurricane with respect to B at an instant of time $t_2$ and ${\left(r_{C/B}\right)}_{t_1}$ is position of hurricane with respect to B at an instant of time $t_1$.

Rewrite Equation (11) using Equation (12) as:

$\begin{array}{c}\Delta r_{C/B}={\left(r_{C/B}\right)}_{t_2}-{\left(r_{C/B}\right)}_{t_1}\\ =v_{C/B}{t}_2+\left[{\left(r_0\right)}_C-{\left(r_0\right)}_B\right]-v_{C/B}{t}_1-\left[{\left(r_0\right)}_C-{\left(r_0\right)}_B\right]\\ =v_{C/B}\left(t_2-t_1\right)\\ =v_{C/B}\left(\Delta t\right)\end{array}$ (13)

Calculate the change in position $\left(\Delta r_{C/B}\right)$ C with respect to B during a 15 min interval:

Substitute $400\text{km}/\text{h}$ for $v_{C/B}$ and 15 min for $\Delta t$ in Equation (13)

$\begin{array}{c}\Delta r_{C/B}=400\text{km}/\text{h}\left(15\min \right)\left(\frac{1\text{h}}{60\min }\right)\\ =400\left(\frac{1}{4}\right)\text{km}\\ \text{=100}\text{km}\end{array}$

Therefore, the change in position $\left(\Delta r_{C/B}\right)$ C with respect to B during a 15 min interval is $\underset{\_}{100\text{km}}$ at an angle $\underset{\_}{40°}$.