Question. The motion of a creature in three dimensions can bedescribed by the fo
ID: 1760208 • Letter: Q
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Question. The motion of a creature in three dimensions can bedescribed by the following equations for position in the x -, y -,and z-directions. x(t) = 3t2+5, y(t) = -t2+3t-2, z(t) =2t+1, Find the magnitudes of theacceleration, velocity, and position vectors at times t=0, t=2, andt=-2 .Find the scalar (A dot B) and vectorproducts (A cross B) for the followingvectors A = 4i - 5j +2k B = 3i + 1j– 4k Question. The motion of a creature in three dimensions can bedescribed by the following equations for position in the x -, y -,and z-directions. x(t) = 3t2+5, y(t) = -t2+3t-2, z(t) =2t+1, Find the magnitudes of theacceleration, velocity, and position vectors at times t=0, t=2, andt=-2 .Find the scalar (A dot B) and vectorproducts (A cross B) for the followingvectors A = 4i - 5j +2k B = 3i + 1j– 4kExplanation / Answer
given x(t) =3t2+5, y(t) = -t2+3t-2, z(t) =2t+1 then the components of velocity are vx = 6t;vy = -2t + 3; and vz = 2 and teh components of acceleration are ax = 6;ay = -2 ; az = 0 then the magnitude of acceleration is same at allpoints. therefore the acceleration of teh particle at t = 0 and t = 2is [(6)2 + (-2)2] = 6.32 magnitude of velocity at t = 0 is [(0)2+ (3)2 + (2)2] = 3.6 andat t = 2 is [(12)2 + (1)2 +(2)2] = 12.2 magnitude of displacement at t = 0 is[(5)2 + (-5)2 + (1)2] =7.14 andat t = 2 is [(17)2 + (0)2 +(5)2] = 17.7 A.B = (4)(3) + (-5)(1) +(2)(4) = 15 A x B = [(-5)(-4) -(2)(1)]i - [(4)(-4) - (2)(3)]j +[(4)(1) - (-5)(3)]k = 18i +22 j + 19k = 18i +22 j + 19kRelated Questions
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