Question

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

Answer 1

Answer:0.3568 m/s

Explanation:

Given

Velocity of a canoe is 0.4 m/s southeast relative to earth

Representing in vector form

$V_{c}=\left ( 0.4cos45\right )\hat{i}-\left ( 0.4sin45\right )\hat{j}$

Also Velocity of river $\left ( V_r\right )=0.5\hat{i}$

thus velocity of canoe relative to the river

$V_{cr}=V_c-V_r$

$V_{cr}=\frac{0.4}{\sqrt{2}}\hat{i}-\frac{0.4}{\sqrt{2}}\hat{j}-0.5\hat{i}$

$V_{cr}=\left ( \frac{0.4\sqrt{2}-1}{2}\right )\hat{i}-\left ( \frac{0.4\sqrt{2}}{2}\right )\hat{j}$

magnitude of Velocity$=\sqrt{\left ( \frac{0.4\sqrt{2}-1}{2}\right )^2+\left ( \frac{0.4\sqrt{2}}{2}\right )}$

$V_{net}=0.3568 m/s$

$tan\theta =\frac{\frac{0.4\sqrt{2}}{2}}{\frac{0.4\sqrt{2}-1}{2}}$

$\theta =52.47 ^{\circ}$ with east direction in clockwise direction