Sheena can row a boat at 2.20 mi/h in still water. She needs to cross a river that is 1.20 mi wide with a current flowing at 1.60 mi/h. Not having her calculator ready, she guesses that to go straight across, she should head upstream at an angle of 25.0° from the direction straight across the river.

Respuesta :

Answer:

Part a)

[tex]v_{bw} = 2.1 mph[/tex]

Part b)

[tex]t = 0.6 h[/tex]

Part c)

[tex]x = 0.4 mile[/tex]

Part d)

[tex]\theta = 46.6 degree[/tex]

Explanation:

Part a)

Velocity of the boat with respect to water stream is given as

[tex]v_{bw} = v_b + v_w[/tex]

[tex]v_{bw} = (1.60 - 2.20sin25) \hat i + 2.20 cos25\hat j[/tex]

so we have

[tex]v_{bw} = 0.67 \hat i + 2 \hat j[/tex]

magnitude of the speed is given as

[tex]v_{bw} = \sqrt{0.67^2 + 2^2}[/tex]

[tex]v_{bw} = 2.1 mph[/tex]

Part b)

Time to cross the river is given as

[tex]t = \frac{y}{v_y}[/tex]

[tex]t = \frac{1.20}{2}[/tex]

[tex]t = 0.6 h[/tex]

Part c)

Distance moved by the boat in downstream is given as

[tex]x = v_x t[/tex]

[tex]x = 0.67 \times 0.6[/tex]

[tex]x = 0.4 mile[/tex]

Part d)

In order to go straight we must net speed along the stream must be zero

so we will have

[tex]vsin\theta = v_w[/tex]

[tex]2.20 sin\theta = 1.60[/tex]

[tex]\theta = 46.6 degree[/tex]

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