Chapter 6.3, Problem 29PP

a.

To determine

Person’s speed relative to water.

Answer to Problem 29PP

  $v_{pw}=3.93\text{m/s}$

Explanation of Solution

Given:

Velocity of boat relative to water is $v_{bw}=3.8\text{m/s}$ in east.

Velocity of person relative to boat is $v_{pb}=1.3\text{m/s}$ towards boat.

Formula Used:

The resultant (relative velocity) when the velocities are perpendicular can be obtained by:

  $v_{pw}=\sqrt{v_{pb}{}^2+v_{bw}{}^2}$

Where, $v_{pw}$ is the velocity of person relative to water, $v_{pb}$ is the velocity of person relative to boat and $v_{bw}$ is the velocity of boat with respect to water.

Calculation:

From the given condition,

  

The velocity of person relative to water is given as

  $\begin{array}{l}{v}_{pw}{}^2=v_{pb}{}^2+v_{bw}{}^2\\ v_{pw}=\sqrt{v_{pb}{}^2+v_{bw}{}^2}\\ =\sqrt{{3.8}^2+{1.3}^2}\\ =4.02\text{m/s}\end{array}$

b.

To determine

Direction of person relative to ground.

Answer to Problem 29PP

Direction of person relative to ground is ${18.88}^o$ south east.

Explanation of Solution

Given:

Speed of boat relative to water is $v_{bw}=3.8\text{m/s}$ in east.

Speed of person relative to boat is $v_{pb}=1.3\text{m/s}$ towards boat.

Formula Used:

The direction can be obtained by:

  $\theta ={\tan }^{-1}\left(\frac{v_{pb}}{v_{bw}}\right)$

Calculations:

From the given condition,

  

Now, direction of person is given as

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{v_{pb}}{v_{bw}}\right)\\ ={\tan }^{-1}\left(\frac{1.3}{3.8}\right)={18.88}^o\end{array}$

So, direction of person relative to ground is ${18.88}^o$ south east.