Question

1 marie is jogging on the deck of a large cruise ship. from the ships reference frame, marie is jogging east at a speed of 2.68 meters per second. to a stationary outside observer, the boat is moving directly northeast (45° north of east) at a speed of 11.18 meters per second. to the stationary outside observer, what is the magnitude of maries velocity? 13.2 m/s 13.9 m/s 11.5 m/s 11.9 m/s

Explanation:

Step1: Find the east - component of the boat's velocity

The boat is moving at a speed \(v_{boat}=11.18\ m/s\) at \(45^{\circ}\) north of east.
The east - component of the boat's velocity \(v_{boat,x}=v_{boat}\cos45^{\circ}\)

$$v_{boat,x}=11.18\times\frac{\sqrt{2}}{2}\approx11.18\times0.707 = 7.91\ m/s$$

Step2: Find the north - component of the boat's velocity

The north - component of the boat's velocity \(v_{boat,y}=v_{boat}\sin45^{\circ}\)

$$v_{boat,y}=11.18\times\frac{\sqrt{2}}{2}\approx7.91\ m/s$$

Step3: Find the east - component of Marie's velocity relative to the outside observer

Marie's velocity relative to the ship is \(v_{Marie,ship,x} = 2.68\ m/s\) east.
The east - component of Marie's velocity relative to the outside observer \(v_{x}=v_{boat,x}+v_{Marie,ship,x}\)

$$v_{x}=7.91 + 2.68=10.59\ m/s$$

Step4: Find the magnitude of Marie's velocity relative to the outside observer

Since \(v_{y}=v_{boat,y} = 7.91\ m/s\) (Marie has no north - south motion relative to the ship), using the Pythagorean theorem \(v=\sqrt{v_{x}^{2}+v_{y}^{2}}\)

$$v=\sqrt{(10.59)^{2}+(7.91)^{2}}=\sqrt{112.1481 + 62.5681}=\sqrt{174.7162}\approx13.2\ m/s$$

Answer:

13.2 m/s