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A canoe has a velocity of 0.450 m/s southeast relative to the earth. The canoe is on a river that is flowing at 0.680 m/s east relative to the earth.

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°

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