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A particle moves along a straight line with the equation of motion  s= t^6-2t^5 Find the value of t (other than 0 ) at which the acceleration is equal to zero.

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  • ?
    Lv 7
    1 month ago

    s(t) = t⁶ - 2t⁵

    v(t) = ds(t)/dt = 6t⁵ - 10t⁴

    a(t) = dv(t)/dt = 30t⁴ - 40t³

    a(t) = t³(30t - 40) = 0

    t³ = 0 → t = 0 (multiplicity = 3)

    30t - 40 = 0 → t = 4/3

    Ans: Acceleration is zero at t = 0 and t = 4/3

  • 1 month ago

    acceleration = a= d^2s/dt^2 = 30t^4 - 40t^3 = 0

    3t - 4 = 0 -->  t = 4/3

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