Question

A ferry is crossing a river. The ferry is
headed due north with a speed of 2.5 m/s
relative to the water and the river’s velocity
is 3.5 m/s to the east. Find the direction in which the ferry is moving_measured from due east with counterclockwise postive)

Answer 1

A ferry is crossing a river. The ferry is headed due north with a speed of $2.5\mathrm{m / s}$ relative to the water and the river's velocity is $3.5\mathrm{m / s}$ to the east. Find the direction in which the ferry is moving_measured from due east with counterclockwise positive)

Solution:

$\mathrm{V}_{\mathrm{f,r}} = 2.5 \frac{\mathrm{m}}{\mathrm{s}} -$ velocity of the ferry relative to the water;

$\mathrm{V}_{\mathrm{river}} = 3.5 \frac{\mathrm{m}}{\mathrm{s}} - \mathrm{velocity}$ of the river;

$\alpha$

- the angle between east direction and the direction of the ferry's motion Formula for the relative velocity of the track:

$\overline{\mathrm{V}}_{\mathrm{f,r}} = \overline{\mathrm{V}}_{\mathrm{ferry}} - \overline{\mathrm{V}}_{\mathrm{river}}$

$\overline{\mathrm{V}}_{\text {ferry }} = \overline{\mathrm{V}}_{\mathrm{f,r}} + \overline{\mathrm{V}}_{\text {river }}$

From the right triangle ABC:

$\tan \alpha = \frac{\mathrm{V}_{f,r}}{\mathrm{V}_{river}} \Rightarrow \alpha = \arctan \left(\frac{\mathrm{V}_{f,r}}{\mathrm{V}_{river}}\right) = \arctan \left(\frac{2.5 \frac{\mathrm{m}}{\mathrm{s}}}{3.5 \frac{\mathrm{m}}{\mathrm{s}}}\right) = 36{}^{\circ}$

Answer: the angle between east direction and the direction of the ferry's motion is $36{}^{\circ}$