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 direction of maries velocity? 37° north of east 32° north of east 37° east of north 32° east of east

Explanation:

Step1: Decompose the ship's velocity

The ship's velocity \(v_{ship}\) has \(x\)-component \(v_{ship,x}=v_{ship}\cos45^{\circ}\) and \(y\)-component \(v_{ship,y}=v_{ship}\sin45^{\circ}\). Given \(v_{ship} = 11.18\ m/s\), then \(v_{ship,x}=11.18\times\frac{\sqrt{2}}{2}\approx7.9\ m/s\), \(v_{ship,y}=11.18\times\frac{\sqrt{2}}{2}\approx7.9\ m/s\). Marie's velocity relative to the ship is \(v_{Marie - ship,x}=2.68\ m/s\), \(v_{Marie - ship,y} = 0\ m/s\).

Step2: Calculate Marie's velocity components relative to the observer

Marie's \(x\)-component of velocity relative to the observer \(v_{Marie,x}=v_{ship,x}+v_{Marie - ship,x}=7.9 + 2.68=10.58\ m/s\), \(v_{Marie,y}=v_{ship,y}=7.9\ m/s\).

Step3: Find the direction of Marie's velocity

Using the formula \(\theta=\tan^{- 1}(\frac{v_{Marie,y}}{v_{Marie,x}})\). Substitute \(v_{Marie,x}=10.58\ m/s\) and \(v_{Marie,y}=7.9\ m/s\) into the formula: \(\theta=\tan^{-1}(\frac{7.9}{10.58})\approx37^{\circ}\). The direction is north of east.

Answer:

37° north of east