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A particle moves along a straight line such that its acceleration is a = (6 t^2-4)m/s^2 where t is in seconds.When t = 0 the particle is located 5 m to the left of the origin, and when t = 2, it is 19 m to the left of the origin.I think the correct equation to use is s = s0 + v0t + 1/2at^2s0 = 5mt = 4a = the function already givenv0 = ?
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