Unformatted text preview: velocity onto the x and y axes, we get v BW = v BW cosθ i + v BW sinθy j = 10cos110º i + 10sin110º j = 3.42 m/s i + 9.40 m/s j The river flows in the +x direction. Therefore, v WG = 6.00 m/s i b) What are the magnitude and direction of the velocity of the boat relative to the ground? The velocity of the boat relative to the ground is the sum of v BW and v WG . v BG = v BW + v WG = 3.42 i + 9.40 j + 6.00 i = 2.58 m/s i + 9.40 m/s j Now we convert the unitvector notation to the magnitude and angle. v BG = √(v x 2 +v y 2 ) = √(2.58 2 + 9.40 2 ) = 9.75 m/s θ = tan1 (v y /v x ) = 74.7º (74.7º north of east) c) How long does the boat take to cross the river? The boat reaches the north bank when the y displacement of the boat is equal to the width of the river. Therefore, we get t = W/v y = 150/9.40 = 16.0 s...
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This note was uploaded on 01/15/2012 for the course PHY 2048 taught by Professor Field during the Fall '08 term at University of Florida.
 Fall '08
 Field
 Physics, Work

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