qz3sol_7193f10 - the north. In unit-vector notation, the...

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TA: Tomoyuki Nakayama Monday, September 20, 2010 PHY 2048: Physic 1, Discussion Section 7193 Quiz 3 (Homework Set #4) 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. ________________________________________________________________________________ In the overhead view of the figure below right, Jeeps P and B race along straight lines, across flat terrain, and past stationary border guard A . Relative to the guard, P travels at a constant speed of 30.0 m/s, at the angle θ 1 = 65.0º. Relative to the guard, B has accelerated from rest at a constant rate of 0.500 m/s 2 at the angle θ 2 = 35.0º. At a certain time during the acceleration, B has a speed of 20.0 m/s. a) At that time, what are the magnitude and direction of the velocity of P relative to B ? We take our +x axis to the east and +y axis to
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Unformatted text preview: the north. In unit-vector notation, the velocity of P with respect to stationary guard A is v PA = 30cos 1 i + 30sin 1 j = 12.7 i + 27.2 j The velocity of B with respect to A is v BA = 20cos 2 i + 20sin 2 j = 16.4 i + 11.5 j The velocity of P relative to B is v PB = v PA v BA = -3.7 i + 15.7 j The magnitude of v PB is v PB = ((-3.7) 2 +15.7 2 ) = 16.1 m/s. The direction is = tan-1 (15.7/(-3.7)) = -76.7 + 180 = 103, 13 west of north. Note that v PB has a negative x component and positive y component, so it is in the 2 nd quadrant. b) At that time, what are the magnitude and direction of the acceleration of B relative to P ? Since P moves at a constant velocity, its acceleration is zero. Thus we have a BP = a BA a PA = a BA It has a magnitude of 0.500 m/s 2 and is directed 35 north of east....
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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.

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qz3sol_7193f10 - the north. In unit-vector notation, the...

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