You can put this solution on YOUR website! If 𝑠(𝑡)=(1/2)t^4-5t^3+12t^2, find the velocity of the object when its acceleration is zero.
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The 1st derivative is speed.
𝑠(𝑡)=(1/2)t^4-5t^3+12t^2
𝑠'(𝑡)= 2t^3 - 15t^2 + 24t
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The 2nd derivative is accel:
𝑠"(𝑡)= 6t^2 - 30t + 24
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6t^2 - 30t + 24 = 0
t^2 - 5t + 4 = 0
t = -1, -4
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𝑠'(𝑡)= 2t^3 - 15t^2 + 24t
Sub the values t = -1 and t = -4