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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 1 i + 20sin 1 j = 11.5 i + 16.4 j The velocity of B with respect to A is v BA = 30cos 2 i + 30sin 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 = tan-1 (3.7/(-15.7)) = -13.3 + 180 = 167, 13 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 25 north of east....
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- Summer '08