Answer to Question #241974 in Physics for Kristine

Question

A canoe on a river flowing 0.50 𝑚/𝑠 east has a velocity of 0.40 𝑚/𝑠 southeast relative to the earth. Find the magnitude and the direction of the canoe’s velocity relative to the river. 

Answer 1

Given:

${\bf u}=0.50\hat i\:\rm m/s$

${\bf V}=\frac{0.40}{\sqrt{2}}(\hat i-\hat j)\:\rm m/s$

The velocity of a canoe relative to the earth

$${\bf V}={\bf u}+{\bf v}$$

Hence, the velocity of a canoe relative to the river

$${\bf v}={\bf V}-{\bf u}$$

$$=\frac{0.40}{\sqrt{2}}(\hat i-\hat j)-0.50\hat i=-0.217\hat i-0.283\hat j$$

Magnitude:

$$v=\sqrt{0.217^2+0.283^2}=0.36\:\rm m/s$$

Direction:

$$\theta=\tan^{-1}\frac{-0.283}{-0.217}=53^{\circ}\quad \rm S\;of\; W$$