Answer:
The velocity of the canoe relative to the river is 0.385 m/s, S37.26⁰W
Explanation:
Given;
velocity of the canoe relative to the earth, [tex]V_{r/e} = 0.33 \ m/s[/tex]
velocity of the river relative to the earth, [tex]V_{r/e} = 0.54 \ m/s[/tex]
The velocity of the canoe relative to the river is calculated as;
[tex]V_{(c/r)x} = V_{(c/e)x}- V_{(r/e)x} \ \ ----(1)\\\\V_{(c/r)y} = V_{(c/e)y}- V_{(r/e)y} \ \ ----(2)[/tex]
The x - component of the velocity of the canoe relative to the earth;
[tex]V_{(c/e)x} = 0.33 \times cos \ 45^0\\\\V_{(c/e)x} = 0.2333 \ m/s[/tex]
The y-component of the velocity of the canoe relative to the earth;
[tex]V_{(c/e)y} = 0.33 \times sin \ 45^0\\\\V_{(c/e)y} = 0.2333 \ m/s[/tex]
Note: velocity of the river relative to the earth has only x-component = 0.54 m/s
Apply equation (1) and (2) to calculate the velocity of the canoe relative to the river;
[tex]V_{(c/r)}x = 0.2333 - 0.54 = -0.3067 \ m/s\\\\V_{(c/r)}y = 0.2333 - 0 = 0.2333 \ m/s\\\\The \ resultant \ velocity;\\\\V_{c/r} = \sqrt{(-0.3067)^2 + (0.2333)^2} \\\\V_{c/r} = 0.385 \ ms/\\\\The \ direction:\\\\\theta = tan^{-1} (\frac{0.2333}{0.3067} ) = 37.26^0 \ south \ west \ of \ the \ river[/tex]
