qz3sol_0132Hf10

# qz3sol_0132Hf10 - the north In unit-vector notation the...

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TA: Tomoyuki Nakayama Monday, September 20, 2010 PHY 2048: Physic 1, Discussion Section 0132H 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 20.0 m/s, at the angle θ 1 = 55.0º. Relative to the guard, B has accelerated from rest at a constant rate of 0.500 m/s 2 at the angle θ 2 = 25.0º. At a certain time during the acceleration, B has a speed of 30.0 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 = 20cos θ 1 i + 20sin θ 1 j = 11.5 i + 16.4 j The velocity of B with respect to A is v BA = 30cos θ 2 i + 30sin θ 2 j = 27.2 i + 12.7 j The velocity of P relative to B is v PB = v PA – v BA = -15.7 i + 3.7 j The magnitude of v PB is v PB = √ ((-15.7) 2 +3.7 2 ) = 16.1 m/s. The direction is θ = tan-1 (3.7/(-15.7)) = -13.3 + 180 = 167º, 13º north of west. 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 velocity 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 25º north of east....
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