Asked by Alice
Find the position vector of a particle whose acceleration vector is a = (6t, 2) with an initial velocity vector (0, 0) and initial position vector (3, 0)
Please I need your help with explanation. Thank you
Please I need your help with explanation. Thank you
Answers
Answered by
oobleck
this is just straightforward integration and evaluation at given points to determine the constant.
acceleration: a = (6t,2)
velocity: v = (3t^2,2t) + c
since v(0) = 0, c = (0,0), and thus v = (3t^2,2t)
position: s = (t^3,t^2) + c
since s(0) = (3,0), c = (3,0) and thus
s(t) = (t^3+3, t^2)
acceleration: a = (6t,2)
velocity: v = (3t^2,2t) + c
since v(0) = 0, c = (0,0), and thus v = (3t^2,2t)
position: s = (t^3,t^2) + c
since s(0) = (3,0), c = (3,0) and thus
s(t) = (t^3+3, t^2)
Answered by
Alice
What else I need to do to get the answer, please?
Answered by
oobleck
huh? I gave you the position vector. Was there something else you wanted????
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