The velocity of a boat with respect to the water is $4.0 \mathrm{~m} / \mathrm{s}$ at an angle of 30.0 degrees NORTH of EAST. The velocity of the water with respect to the ground is $3.0 \mathrm{~m} / \mathrm{s}$ at an angle of 90.0 degrees NORTH of EAST. What is the velocity of the boat with respect to the ground?
A) $5.8 \mathrm{~m} / \mathrm{s}$ at 36 degrees NORTH of EAST
B) $5.8 \mathrm{~m} / \mathrm{s}$ at 55 degrees NORTH of EAST
C) $6.1 \mathrm{~m} / \mathrm{s}$ at 61 degrees NORTH of EAST
D) $6.1 \mathrm{~m} / \mathrm{s}$ at 36 degrees NORTH of EAST
E) $6.1 \mathrm{~m} / \mathrm{s}$ at 55 degrees NORTH of EAST

Question

Quizplus · Accepted Answer

To find the velocity of the boat relative to the ground, we add the velocity of the boat relative to the water to the velocity of the water relative to the ground vectorially. The boat's velocity can be decomposed into $4.0 \cos(30^\circ) \, \mathrm{m/s}$ EAST and $4.0 \sin(30^\circ) \, \mathrm{m/s}$ NORTH. Adding the water's velocity of $3.0 \, \mathrm{m/s}$ NORTH gives a total EAST component of $4.0 \cos(30^\circ) \, \mathrm{m/s}$ and a total NORTH component of $4.0 \sin(30^\circ) + 3.0 \, \mathrm{m/s}$. Calculating these gives an EAST component of $3.46 \, \mathrm{m/s}$ and a NORTH component of $5.0 \, \mathrm{m/s}$. The magnitude of the resultant velocity is $\sqrt{(3.46)^2 + (5.0)^2} \approx 6.1 \, \mathrm{m/s}$, and the angle $\theta$ from EAST is given by $\tan^{-1}(5.0/3.46) \approx 55^\circ$.

The Velocity of a Boat with Respect to the Water 4.0 m/s4.0 \mathrm{~m} / \mathrm{s}4.0 m/s

The Velocity of a Boat with Respect to the Water $4.0 \mathrm{~m} / \mathrm{s}$