Chapter 3, Problem 88P

(a)

To determine

The direction dolphin must head.

Answer to Problem 88P

The dolphin must head $\underset{\_}{30.0°}$ north of west.

Explanation of Solution

Consider the positive x direction to be west, and positive y direction to be north.

Figure 1 represents the relative motion of the dolphin with respect to the bay and uniform water current.

Write the expression for velocity of dolphin with respect to bay.

    ${\stackrel{\rightharpoonup }{v}}_{\text{db}}={\stackrel{\rightharpoonup }{v}}_{\text{dw}}+{\stackrel{\rightharpoonup }{v}}_{\text{wb}}$        (I)

Here, ${\stackrel{\rightharpoonup }{v}}_{\text{db}}$ is the velocity of dolphin with respect to bay, ${\stackrel{\rightharpoonup }{v}}_{\text{dw}}$ is the velocity of dolphin with respect to the water current, and ${\stackrel{\rightharpoonup }{v}}_{\text{wb}}$ is the velocity of the water current with respect to the bay.

Write the expression for component of velocity of the dolphin with respect to bay in the x direction.

    $v_{\text{dbx}}=v_{\text{dwx}}+v_{\text{wbx}}$

Here, $v_{\text{dbx}}$ is the velocity of the dolphin with respect to bay in the x direction, $v_{\text{dwx}}$ is the velocity of the dolphin with respect to the water current in the x direction, and $v_{\text{wbx}}$ is the velocity of the water current with respect to bay in the y direction.

From Figure 1 the above equation is written as

    $\begin{array}{c}{v}_{\text{dbx}}=v_{\text{dw}}\cos \theta -v_{\text{wb}}\cos 45°\\ =v_{\text{dw}}\cos \theta -\frac{v_{\text{wb}}}{\sqrt{2}}\end{array}$        (II)

Write the expression for component of velocity of the dolphin with respect to bay in the y direction.

    $v_{\text{dby}}=v_{\text{dwy}}+v_{\text{wby}}$

Here, $v_{\text{dby}}$ is the velocity of the dolphin with respect to bay in the y direction, $v_{\text{dwy}}$ is the velocity of the dolphin with respect to the water current in the y direction, and $v_{\text{wby}}$ is the velocity of the water current with respect to bay in the y direction

From Figure 1 the above equation can be written as

    $\begin{array}{c}{v}_{\text{dby}}=v_{\text{dw}}\sin \theta -v_{\text{wb}}\sin 45°\\ =v_{\text{dw}}\sin \theta -\frac{v_{\text{wb}}}{\sqrt{2}}\end{array}$        (III)

Since $v_{\text{dby}}=0$, the above is reduced to

    $\begin{array}{c}0=v_{\text{dw}}\sin \theta -\frac{v_{\text{wb}}}{\sqrt{2}}\\ \sin \theta =\frac{v_{\text{wb}}}{v_{\text{dw}}\sqrt{2}}\\ \theta ={\sin }^{-1}\left(\frac{v_{\text{wb}}}{v_{\text{dw}}\sqrt{2}}\right)\end{array}$        (IV)

Conclusion:

Substitute $2.83\text{m}/\text{s}$ for $v_{\text{wb}}$, and $4.00\text{m}/\text{s}$ for $v_{\text{dw}}$ in equation (IV), to find $\theta$.

    $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{2.83\text{m}/\text{s}}{\left(4.00\text{m}/\text{s}\right)\sqrt{2}}\right)\\ =30.0°\end{array}$

The dolphin should head $30.0°$ north of west.

Therefore, the dolphin must head $\underset{\_}{30.0°}$ north of west.

(b)

To determine

The time taken by the dolphin to swim $0.80\text{km}$ distant home.

Answer to Problem 88P

The time taken by the dolphin to swim $0.80\text{km}$ distant home is $\underset{\_}{5.47\times {10}^{-4}\text{s}}$.

Explanation of Solution

Write the expression for time taken by the dolphin to swim $0.80\text{km}$ distant home.

    $\Delta t=\frac{\Delta x}{v_{\text{dbx}}}$        (V)

Here, $\Delta t$ is the time taken by the dolphin to swim $0.80\text{km}$ distant home, $\Delta x$ is the distance travelled by the dolphin, and $v_{\text{dbx}}$ is the component of velocity of the dolphin with respect to bay in the x direction.

From subpart (a), $v_{\text{dbx}}$ is derived as $v_{\text{dw}}\cos \theta -\frac{v_{\text{wb}}}{\sqrt{2}}$.

Use the above condition in equation (V).

    $\Delta t=\frac{\Delta x}{v_{\text{dw}}\cos \theta -\frac{v_{\text{wb}}}{\sqrt{2}}}$        (VI)

Conclusion:

Substitute $4.00\text{m}/\text{s}$ for $v_{\text{dw}}$, $2.83\text{m}/\text{s}$ for $v_{\text{wb}}$ $0.80\text{km}$ for $\Delta x$, and $30.0°$ for $\theta$ in equation (VI), to find $\Delta t$.

    $\begin{array}{c}\Delta t=\frac{0.80\text{km}\times \frac{1\text{m}}{{10}^{-3}\text{km}}}{\left(4.00\text{m}/\text{s}\right)\cos 30.0°-\frac{2.83\text{m}/\text{s}}{\sqrt{2}}}\\ =\frac{0.80\times {10}^{-3}\text{m}}{3.46\text{m}/\text{s}-2.00\text{m}/\text{s}}\\ =5.47\times {10}^{-4}\text{s}\end{array}$

Therefore, the time taken by the dolphin to swim $0.80\text{km}$ distant home is $\underset{\_}{5.47\times {10}^{-4}\text{s}}$.