Suppose that the acceleration vector for a point mass M as a function of the time...

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Suppose that the acceleration vector for a point mass M as a function of the time t is given by: ā = ( 4 cos (2t) ) i + (12 t + 2) +(3 e') k, the initial position is: , = r(0) = -i+3* and the initial velocity is: v,= v(0) = -i + 3k. You are to compute the velocity and position of M (Be careful. Remember once you make a mistake, the rest is wrong.) 19. (5 pts.) The velocity of the point mass is v(t) =, А. В. С. D. E. АВ. АС. AD. AE. ВС. BD. BE. v(t) = 20. ( 5 pts.) The position of the point mass is 7(t) =. A. B. C. D. E. AB. AC. AD. AE. BC. BD. BE. F(t) = Possible answers for this page. A.( 2 sin (2t) - 1) i + (4t° + 2t ) j + (3 e' ) k B. (2 cos (2t) ) i + (t + 2t ) j + (3 e' ) k C. (-sin (2t) - 1) i + (2t +t)j + (3 e' ) k D. (2 cos (2t) ) i + (t' + 2t ) j + (3 e' ))k E. (-2 sin (2t) - 1) i + (4t + 2t ) j + (3 e' ) k AB. (- cos (2t) -t) i +(t+t) +(3 e') k AC ( 2 sin (2t) - t) i +(4t2 + 2t ) j + (3 e') k AD (- cos (2t) - t) i +(*+t)j + (3 e') k AE. (- sin (2t) - t) i + (4 t +2) j - (3 e') & BC (- cos (2t) - t) i + (t? +t)j +(3 e) & BD. ( 4 sin (2t) - t) i + (4t + 2t ) j +(3 e')k BE. None of the above. Suppose that the acceleration vector for a point mass M as a function of the time t is given by: ā = ( 4 cos (2t) ) i + (12 t + 2) +(3 e') k, the initial position is: , = r(0) = -i+3* and the initial velocity is: v,= v(0) = -i + 3k. You are to compute the velocity and position of M (Be careful. Remember once you make a mistake, the rest is wrong.) 19. (5 pts.) The velocity of the point mass is v(t) =, А. В. С. D. E. АВ. АС. AD. AE. ВС. BD. BE. v(t) = 20. ( 5 pts.) The position of the point mass is 7(t) =. A. B. C. D. E. AB. AC. AD. AE. BC. BD. BE. F(t) = Possible answers for this page. A.( 2 sin (2t) - 1) i + (4t° + 2t ) j + (3 e' ) k B. (2 cos (2t) ) i + (t + 2t ) j + (3 e' ) k C. (-sin (2t) - 1) i + (2t +t)j + (3 e' ) k D. (2 cos (2t) ) i + (t' + 2t ) j + (3 e' ))k E. (-2 sin (2t) - 1) i + (4t + 2t ) j + (3 e' ) k AB. (- cos (2t) -t) i +(t+t) +(3 e') k AC ( 2 sin (2t) - t) i +(4t2 + 2t ) j + (3 e') k AD (- cos (2t) - t) i +(*+t)j + (3 e') k AE. (- sin (2t) - t) i + (4 t +2) j - (3 e') & BC (- cos (2t) - t) i + (t? +t)j +(3 e) & BD. ( 4 sin (2t) - t) i + (4t + 2t ) j +(3 e')k BE. None of the above.

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