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please any on help me for this question ý need urgently.

The position of a particle moving along the x-axis is given by x = 3.49t2 – 2.21t3, where x is in meters and t is in seconds. What is the position of the particle when it achieves its maximum speed in the positive x-direction?

  • PHYSICS -

    x = 3.49t^2 – 2.21t^3
    V = dx/dt = 6.98 t - 6.63t^2
    At maximum V, dV/dt = 0
    6.98 = 13.26 t
    t = 0.5264 s @ maximum V
    X @t=0.5264 = 0.6447 m

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