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Sagot :
Relative velocity is defined as the velocity an object X has in the rest frame of another object Y
- The velocity of the canoe relative to the river is approximately 0.482 m/s
- The direction of the canoe relative to the river is approximately 41.33°
Reason:
The given parameter are;
The velocity of the boat, v = 0.450 m/s southeast relative to Earth
The velocity of the river, v = 0.680 m/s East relative to the Earth
The magnitude of the velocity of the canoe relative to the river = Required
Solution:
The x-component of the velocity of the canoe, vₓ, is given as follows;
vₓ = 0.45 × cos (45°) ≈ 0.31819805153
The x-component of the velocity of the canoe relative the river, [tex]v_{xr}[/tex], is given as follows;
[tex]v_{xr}[/tex] = 0.31819805153 - 0.680 = -0.36180194847
The y-component of the velocity of the canoe, [tex]v_y[/tex] = 0.45 × sin (45°) ≈ 0.31819805153
The magnitude of the velocity of the canoe relative to the river, [tex]v_{cr}[/tex], is given as follows;
√(0.31819805153² + (-0.36180194847)²) ≈ 0.4818201427
[tex]v_{cr}[/tex] = [tex]\sqrt{0.31819805153^2 + (-0.36180194847)^2} \approx 0.482[/tex]
Therefore;
- The velocity of the canoe relative to the river, [tex]v_{cr}[/tex] ≈ 0.482 m/s
[tex]The \ direction,\ \theta = \ arcsin\left (\dfrac{0.45 \times sin(45^{\circ})}{\sqrt{0.31819805153^2 + (-0.36180194847)^2} } \right) \approx 41.33^{\circ}[/tex]
- The direction of the canoe relative to the river, θ ≈ 41.33°
Learn more here:
https://brainly.com/question/24655105
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