Question

a ferryboat is traveling in a direction 26.0° north of east with a speed of 6.18 m/s relative to the water. a passenger is walking with a velocity of 2.64 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. current attempt in progress attempts: 0 of 3 used save for later submit answer using multiple attempts will impact your score. 20% score reduction after attempt 2

Explanation:

Step1: Resolve boat velocity components

The velocity of the boat relative to the water has components. The x - component $v_{bx}=6.18\cos(26.0^{\circ})$ and the y - component $v_{by}=6.18\sin(26.0^{\circ})$. The velocity of the passenger relative to the boat is $v_{px}=2.64$ (in the x - direction, east) and $v_{py} = 0$.
$v_{px}=2.64$ m/s, $v_{bx}=6.18\cos(26.0^{\circ})\approx6.18\times0.899 = 5.556$ m/s, $v_{by}=6.18\sin(26.0^{\circ})\approx6.18\times0.438 = 2.707$ m/s.
The x - component of the passenger's velocity relative to the water $v_x=v_{px}+v_{bx}=2.64 + 5.556=8.196$ m/s, and the y - component $v_y = v_{by}=2.707$ m/s.

Step2: Calculate magnitude of velocity

Use the Pythagorean theorem to find the magnitude of the passenger's velocity relative to the water $v=\sqrt{v_x^{2}+v_y^{2}}$.
$v=\sqrt{(8.196)^{2}+(2.707)^{2}}=\sqrt{67.18+7.33}=\sqrt{74.51}\approx8.63$ m/s.

Step3: Calculate direction of velocity

Use the formula $\theta=\tan^{- 1}(\frac{v_y}{v_x})$ to find the direction of the velocity relative to the east.
$\theta=\tan^{-1}(\frac{2.707}{8.196})\approx\tan^{-1}(0.3303)\approx18.0^{\circ}$ north of east.

Answer:

(a) 8.63
(b) 18.0° north of east