Question

a ferryboat is traveling in a direction 19.0° north of east with a speed of 5.81 m/s relative to the water. a passenger is walking with a velocity of 2.34 m/s due east relative to the boat. what is (a) the magnitude and (b) the direction of the velocity of the passenger with respect to the water? give the directional angle relative to due east.

Explanation:

Step1: Resolve boat - water velocity components

The velocity of the boat relative to the water $\vec{v}_{bw}=5.81$ m/s at an angle $\theta = 19.0^{\circ}$ north of east.
The x - component of the boat - water velocity is $v_{bw,x}=5.81\cos(19.0^{\circ})$ and the y - component is $v_{bw,y}=5.81\sin(19.0^{\circ})$.
$v_{bw,x}=5.81\cos(19.0^{\circ})\approx5.49$ m/s, $v_{bw,y}=5.81\sin(19.0^{\circ})\approx1.89$ m/s.
The velocity of the passenger relative to the boat $\vec{v}_{pb}=2.34$ m/s due east, so its x - component is $v_{pb,x} = 2.34$ m/s and y - component is $v_{pb,y}=0$ m/s.

Step2: Find the x and y components of passenger - water velocity

The x - component of the passenger relative to the water $v_{pw,x}=v_{bw,x}+v_{pb,x}$.
$v_{pw,x}=5.49 + 2.34=7.83$ m/s.
The y - component of the passenger relative to the water $v_{pw,y}=v_{bw,y}=1.89$ m/s.

Step3: Calculate the magnitude of passenger - water velocity

Use the Pythagorean theorem $v_{pw}=\sqrt{v_{pw,x}^{2}+v_{pw,y}^{2}}$.
$v_{pw}=\sqrt{(7.83)^{2}+(1.89)^{2}}=\sqrt{61.21 + 3.57}=\sqrt{64.78}\approx8.05$ m/s.

Step4: Calculate the direction of passenger - water velocity

Use the formula $\tan\theta=\frac{v_{pw,y}}{v_{pw,x}}$.
$\theta=\arctan(\frac{1.89}{7.83})\approx13.5^{\circ}$ north of east.

Answer:

(a) 8.05 m/s
(b) $13.5^{\circ}$ north of east