Question

A person on a cruise ship is doing laps on the promenade deck. on one portion of the track the person is moving north with a speed of 3.8m/s relative to the ship. the ship moves east with a speed of 12m/s relative to the water. what is the direction of motion of the person relative to the water?

Answer 1

The two components of the motion (3.8 m/s north and 12 m/s east) correspond to the two sides of a right triangle, where 3.8 is the length of the vertical side while 12 is the length of the horizontal side. Therefore, the angle which gives the direction of motion is given by

$\theta = arctan (\frac{v_y}{v_x})$

where vx is the horizontal velocity and vy is the vertical velocity. Substituting numbers into the equation, we find

$\theta= arctan (\frac{3.8}{12})=arctan(0.317)=17.6^{\circ}$

so, 17.6 degrees north of east.

Answer 2

The resultant motion is given by pithagoras, since the two components (north and east) are perpendicular to each other.
They are asking you about the direction so you have to use trigonometry, finding that the direction is Ф=arctan(3.8/12)=17.57° north of east.