qz2sol_6839sp10 - velocity onto the x and y axes we get v...

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TA: Tomoyuki Nakayama Tuesday, January 26 th , 2010 PHY 2048: Physic 1, Discussion Section 6839 Quiz 2 (Homework Set #3) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ A 150-m-wide river flows due east at a uniform speed of 6.00 m/s. A boat with a speed of 10.0 m/s relative to the water leaves the south bank pointed in a direction 20.0º west of north. a) Take +x direction to the east and +y direction to the north, and express the velocity of the boat relative to the water and velocity of the flow of the river in unit vector notation. The velocity of the boat relative to the water makes 110º (= 20º + 90º) to the +x axis. Projecting the
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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 unit-vector 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 = tan-1 (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.

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