Question

1. a boat sets out from the north shore of a 200 m wide east - flowing river. the boat always faces due south but the current carries it 300 m downstream while crossing. the trip takes 2.0 min. assume three significant digits. (a) what is the boats displacement during the trip? (ans. 360 m 56° e of s) (b) what is the boats average velocity during the trip? (ans. 3.00 m/s 56° e of s) (c) if the boats velocity relative to the water is 1.7 m/s s, what is the velocity of the current? (ans. 2.5 m/s e)

Explanation:

Step1: Calculate displacement (for part a)

Displacement \(d\) is the straight - line distance from start to finish. Using the Pythagorean theorem \(d=\sqrt{d_{y}^{2}+d_{x}^{2}}\), where \(d_{y} = 200m\) (south - ward) and \(d_{x}=300m\) (east - ward).
\(d=\sqrt{200^{2}+300^{2}}=\sqrt{40000 + 90000}=\sqrt{130000}\approx360m\)
The direction \(\theta=\tan^{-1}(\frac{d_{x}}{d_{y}})=\tan^{-1}(\frac{300}{200})\approx56^{\circ}\) east of south.

Step2: Calculate average velocity (for part b)

Average velocity \(v=\frac{d}{t}\). Time \(t = 2.0min=2\times60s = 120s\).
\(v=\frac{360}{120}=3.0m/s\). The direction is the same as the displacement direction (\(56^{\circ}\) east of south)

Step3: Calculate current velocity (for part c)

The boat's velocity relative to water \(v_{bw}=1.7m/s\) (south - ward). Let the current velocity be \(v_{w}\) (east - ward).
The time to cross the river \(t=\frac{d_{y}}{v_{bw}}=\frac{200}{1.7}\approx117.65s\)
The current velocity \(v_{w}=\frac{d_{x}}{t}\). Substituting \(d_{x} = 300m\) and \(t=\frac{200}{1.7}s\)
\(v_{w}=\frac{300}{\frac{200}{1.7}}=\frac{300\times1.7}{200}=2.55\approx2.5m/s\) (east - ward)

Answer:

a. The boat's displacement is \(360m\) at \(56^{\circ}\) east of south.
b. The boat's average velocity is \(3.0m/s\) at \(56^{\circ}\) east of south.
c. The velocity of the current is \(2.5m/s\) east.