Chapter 6, Problem 66A

a.

To determine

Velocity of boat relative to shore.

Answer to Problem 66A

  $v_{bs}=5\text{m/s}$ and $\theta =53.13°$

Explanation of Solution

Given:

The given figure is:

  

Velocity of boat relative to water is $v_{bw}=4\text{m/s}$ perpendicular to shore

Velocity of river relative to shore is $v_{ws}=3\text{m/s}$

Formula used:

The resultant of two vector is,

  $C^2=A^2+B^2$

Where, C is resultant vector, A and B are two perpendicular vectors

Calculation:

Here, $C=v_{bs}$ , $A=v_{bw}$ and $B=v_{ws}$

So, velocity of motorboat relative to shore is given as,

  $\begin{array}{c}{v}_{bs}=\sqrt{v_{bw}{}^2+v_{ws}{}^2}\\ =\sqrt{4^2+3^2}\\ =5\text{m/s}\end{array}$

And direction of boat is

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{v_{bw}}{v_{ws}}\right)\\ ={\tan }^{-1}\left(\frac{4}{3}\right)\\ =53.13°\end{array}$

Conclusion:

Velocity of boat relative to shore is $v_{bs}=5\text{m/s}$ and direction is $\theta =53.13°$ from shore.

b.

To determine

Component of velocity parallel to shore and perpendicular to shore.

Answer to Problem 66A

Boat velocity parallel to shore is 3 m/s and boat velocity perpendicular to shore is 4 m/s.

Explanation of Solution

Given:

The given figure is:

  

Velocity of boat relative to water is $v_{bw}=4\text{m/s}$

Velocity of river is $v_{ws}=3\text{m/s}$

Formula used:

The x -component and y -component of a vector R are:

  $\begin{array}{l}{R}_x=R\cos \theta \\ R_y=R\sin \theta \end{array}$

Calculation:

Component of boat velocity parallel to shore is

  $\begin{array}{c}{V}_x=V\cos \theta \\ =5\times \cos \left(53.13°\right)\\ =3\text{m}/\text{sec}\end{array}$

Component of boat velocity perpendicular to shoreis

  $\begin{array}{c}{V}_y=V\sin \theta \\ =5\times \sin \left(53.13°\right)\\ =3.9999\text{m}/\text{sec}\\ \approx 4\text{m}/\text{sec}\end{array}$

Conclusion:

Therefore, boat velocity parallel to shore is 3 m/s and boat velocity perpendicular to shore is 4 m/s.