A boat is travelling at 30km/h from island A to island B. The bearing from island A to island B is 045 degrees. The distance between the two islands is 50km. There is a steady current of 5km/h at a bearing of 315 degrees.

B. On what course must the boat travel to offset the flow of current?
Please give the answer in only one equations form so that let me understand easily.

1 answer

To find the course the boat must travel to offset the flow of the current, you can use vector addition.

Let:

  • \( \vec{V_b \) = velocity of the boat relative to the water
  • \( \vec{V_c} \) = velocity of the current
  • \( \vec{V_r} \) = resultant velocity of the boat (the intended path from A to B)

Given:

  • The bearing from A to B is 045 degrees, which translates to \( \theta_r = 45^\circ \)
  • The speed of the boat \( |V_b| = 30 , \text{km/h} \)
  • The current speed \( |V_c| = 5 , \text{km/h} \)
  • The current bearing is 315 degrees, translating to \( \theta_c = 315^\circ \)

You can express the velocities in terms of their components using trigonometric functions:

\[ \vec{V_b} = |V_b| \cdot (\cos(\theta_b), \sin(\theta_b)) = (30 \cdot \cos \theta_b, 30 \cdot \sin \theta_b) \] \[ \vec{V_c} = |V_c| \cdot (\cos(315^\circ), \sin(315^\circ)) = (5 \cdot \cos(315^\circ), 5 \cdot \sin(315^\circ)) \]

Since \( \cos(315^\circ) = \frac{\sqrt{2}}{2} \) and \( \sin(315^\circ) = -\frac{\sqrt{2}}{2} \):

\[ \vec{V_c} = (5 \cdot \frac{\sqrt{2}}{2}, 5 \cdot -\frac{\sqrt{2}}{2}) = \left(\frac{5\sqrt{2}}{2}, -\frac{5\sqrt{2}}{2}\right) \]

For the resultant velocity of the boat to go from A to B, we want:

\[ \vec{V_r} = \vec{V_b} + \vec{V_c} \]

The resultant velocity should have a direction of 045 degrees:

\[ \vec{V_r} = |V_r| \cdot (\cos(45^\circ), \sin(45^\circ)) = \left(\frac{50\sqrt{2}}{2}, \frac{50\sqrt{2}}{2}\right) \]

Thus, separating into components we get the following equations for the x and y components:

  1. \( 30 \cdot \cos \theta_b + \frac{5\sqrt{2}}{2} = \frac{50\sqrt{2}}{2} \)
  2. \( 30 \cdot \sin \theta_b - \frac{5\sqrt{2}}{2} = \frac{50\sqrt{2}}{2} \)

These two equations will allow you to solve for \( \theta_b \), the required course angle.

To encapsulate all that in one equation that represents \( \theta_b \) directly, we can express it in the following form:

\[ \tan \theta_b = \frac{\left( \frac{50\sqrt{2}}{2} + \frac{5\sqrt{2}}{2} \right)}{30 - \frac{5\sqrt{2}}{2}} \]

This equation allows you to solve directly for the course angle of the boat considering the offset from the current.