Chapter 6.3, Problem 35SSC

a.

To determine

Velocity of current.

Answer to Problem 35SSC

  $v_{ws}=7.96\text{m/s}\text{towads east}$

Explanation of Solution

Given:

River is flowing towards east.

Direction of boat, $\theta =53°$ north-west

Velocity of boat relative to shore, $v_{bs}=6\text{m/s}$ towards north

Formula used:

The resultant of two perpendicular vectors is

  $R=\sqrt{x^2+y^2}$

Angle of resultant is

  $\theta ={\tan }^{-1}\frac{x}{y}$

Calculation:

From the given situation,

  

  $\begin{array}{c}\tan \theta =\frac{v_{ws}}{v_{bs}}\\ v_{ws}=v_{bs}\tan \theta \\ =6\times \tan 53°\\ =7.96\text{m/s}\text{towads east}\end{array}$

Conclusion:

Velocity of water relative to shore is $v_{ws}=7.96\text{m/s}\text{towads east}$

b.

To determine

Speed of boat relative to water.

Answer to Problem 35SSC

  $v_{bw}=9.97\text{m/s}$

Explanation of Solution

Given:

River is flowing towards east.

Direction of boat, $\theta =53°$ north-west

Velocity of boat relative to shore, $v_{bs}=6\text{m/s}$ towards north

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:

From the given condition

  

  $\begin{array}{c}\cos \theta =\frac{v_{bs}}{v_{bw}}\\ v_{bw}=\frac{v_{bs}}{\cos \theta }\\ =\frac{6}{\cos 53°}\\ =9.97\text{m/s}\end{array}$

Conclusion:

Speed of boat relative to water is $v_{bw}=9.97\text{m/s}$ .