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Unformatted text preview: river with respect to the ground and the velocity of the boat with respect to water are, respectively, v WG = 3 i , v BW = 8cos(-105º) i + 8sin(-105º) j = -2.07 i – 7.72 j The velocity of the boat with respect to the ground is v BG = v BW + v WG = 0.93 i – 7.72 j The magnitude and direction of v BG are v BG = √ (.93 2 +(-7.72) 2 ) = 7.78 m/s, θ = tan-1 ((-.7.72)/.93) = -83.1º, 6.9º east of south. Note that v BG has a positive x component and negative y component. It is in the 4 th quadrant. b) How long does the boat take to cross the river? To cross the river, the boat must move 180 m in the negative y direction. Therefore, t = Δ y/v y = 23.3 s...
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This note was uploaded on 01/25/2011 for the course PHY 2048 taught by Professor Field during the Summer '08 term at University of Florida.
- Summer '08