Chapter 3, Problem 79P

(a)

To determine

The displacement of the plane from its starting point.

Answer to Problem 79P

The person S’s speed with respect to the starting point on the bank is $1.80\text{mi/h}$ .

Explanation of Solution

The person S’s motion is shown in Figure 1.

Write an expression to calculate the x component of the speed of person S with respect to the starting point on the bank.

$v_{Sbx}=v\cos \theta +{v^{\prime }}_x$ (I)

Here, $v_{Sbx}$ is the x component of the speed of person S with respect to the starting point on the bank, $v$ is the speed of the person s, $\theta$ is the angle and ${v^{\prime }}_x$ is the x component of the speed of stream.

Write an expression to calculate the y component of the speed of the person S.

$v_{Sby}=v\sin \theta +{v^{\prime }}_y$ (II)

Here, $v_{Sby}$ is the y component of the speed of person S with respect to the starting point on the bank and ${v^{\prime }}_y$ is the y component of the speed of stream.

Write an expression to calculate the speed of the person with respect to the starting point on the bank.

$v_{Sb}=\sqrt{v_{Sbx}^2+v_{Sby}^2}$ (III)

Conclusion:

Substitute $3.00\text{mi/h}$ for $v$, $0\text{mi/h}$ for ${v^{\prime }}_x$ , and $60.0°$ for $\theta$  in equation (I) to find $v_{Sbx}$.

$\begin{array}{l}{v}_{Sbx}=\left(3.00\text{mi/h}\right)\cos \left(60.0°\right)+0\text{mi/h}\\ =\left(3.00\text{mi/h}\right)\left(\frac{1}{2}\right)\\ =1.50\text{mi/h}\end{array}$

Substitute $3.00\text{mi/h}$ for $v$, $-1.60\text{mi/h}$ for ${v^{\prime }}_y$ , and $60.0°$ for $\theta$  in equation (II) to find $v_{Sby}$.

$\begin{array}{c}{v}_{Sby}=\left(3.00\text{mi/h}\right)\sin \left(60.0°\right)+-1.60\text{mi/h}\\ =\left(3.00\text{mi/h}\right)\left(\frac{\sqrt{3}}{2}\right)-1.60\text{mi/h}\\ =1.00\text{mi/h}\end{array}$

Substitute $1.50\text{mi/h}$ for $v_{Sbx}$, and $1.00\text{mi/h}$ for $v_{Sby}$ in equation (III) to find $v_{Sb}$.

$\begin{array}{c}{v}_{Sb}=\sqrt{{\left(1.50\text{mi/h}\right)}^2+{\left(1.00\text{mi/h}\right)}^2}\\ =\sqrt{3.25}\text{mi/h}\\ =\text{1}\text{.80}\text{mi/h}\end{array}$

Thus, the person S’s speed with respect to the starting point on the bank is $1.80\text{mi/h}$ .

(b)

To determine

The direction at which the jetliner have flown directly to the same destination.

Answer to Problem 79P

The time taken to cross the river is $48.0\text{min}$.

Explanation of Solution

Write an expression to calculate the time taken to cross the river.

$\Delta t=\frac{\Delta x}{v\cos \theta }$ (IV)

Here, $\Delta t$ is the time taken to cross the river and $\Delta x$ is the width of the river.

Conclusion:

Substitute $1.20\text{mi}$ for $\Delta x$, $3.00\text{mi/h}$ for $v$, and $60.0°$ for $\theta$ in equation (IV) to find $\Delta t$.

$\begin{array}{c}\Delta t=\frac{1.20\text{mi}}{\left(3.00\text{mi/h}\right)\cos \left(60.0°\right)}\\ =\frac{1.20\text{mi}}{\left(3.00\text{mi/h}\right)\cos \left(60.0°\right)}\\ =\frac{1.20\text{mi}}{\left(3.00\text{mi/h}\right)\left(\frac{1}{2}\right)}\\ =\left(0.800\text{h}\right)\left(\frac{60\text{min}}{1\text{hr}}\right)\\ =48.0\text{min}\end{array}$

Thus, the time taken to cross the river is $48.0\text{min}$.

(c)

To determine

The time taken for the trip.

Answer to Problem 79P

The upstream or downstream of the person S from the starting point the person S reaches at the opposite bank is $0.800\text{mi}\text{upstream}$.

Explanation of Solution

Write an expression to calculate the upstream or downstream of the person S from the starting point the person S reaches at the opposite bank.

$\Delta y=v\sin \theta \Delta t$ (V)

Here, $\Delta y$ is the upstream or downstream of the person S from the starting point the person S reaches at the opposite bank.

Conclusion:

Substitute $0.800\text{h}$ for $\Delta t$, $3.00\text{mi/h}$ for $v$, and $60.0°$ for $\theta$ in equation (V) to find $\Delta y$.

$\begin{array}{c}\Delta y=\left(3.00\text{mi/h}\right)\sin \left(60.0°\right)\left(0.800\text{h}\right)\\ =\left(3.00\text{mi/h}\right)\left(\frac{\sqrt{3}}{2}\right)\left(0.800\text{h}\right)\\ =0.800\text{mi}\text{upstream}\end{array}$

Thus, the upstream or downstream of the person S from the starting point the person S reaches at the opposite bank is $0.800\text{mi}\text{upstream}$.

(d)

To determine

The time taken for the direct flight.

Answer to Problem 79P

The upstream angle the person S should go in order to go straight across is $32.2°\text{upstream}$.

Explanation of Solution

The upstream component of her velocity relative to the water should be equal in magnitude to the velocity of the current relative to the bank.

Write an expression to calculate the launch angle $\theta$ at which the maximum range occurs.$\begin{array}{c}{v}_{Sby}=0\\ v\sin \theta \text{'}+{v^{\prime }}_y=0\end{array}$ (VI)

Here, $\theta \text{'}$ is upstream angle the person S should go in order to go straight across.

Rearrange the equation (VI) to find $\theta \text{'}$.

$\theta \text{'}={\sin }^{-1}\left(\frac{-{v^{\prime }}_y}{v}\right)$ (VII)

Conclusion:

Substitute $3.00\text{mi/h}$ for $v$, and $-1.60\text{mi/h}$ for ${v^{\prime }}_y$   in equation (VII) to find $v_{Sby}$.

$\begin{array}{c}\theta \text{'}={\sin }^{-1}\left(\frac{-\left(-1.60\text{mi/h}\right)}{3.00\text{mi/h}}\right)\\ ={\sin }^{-1}\left(\frac{1.60\text{mi/h}}{3.00\text{mi/h}}\right)\\ =32.2°\text{upstream}\end{array}$

Thus, the upstream angle the person S should go in order to go straight across is $32.2°\text{upstream}$.