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(B) Given position function: $x = -3t^3 + 18t^2 + 16t$.Velocity $v$ is the first derivative of position with respect to time: $v = \frac{dx}{dt} = \fr

Vedclass · Accepted Answer

$52$

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(B) Given position function: $x = -3t^3 + 18t^2 + 16t$.Velocity $v$ is the first derivative of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(-3t^3 + 18t^2 + 16t) = -9t^2 + 36t + 16$.Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(-9t^2 + 36t + 16) = -18t + 36$.Set acceleration to zero to find the time: $-18t + 36 = 0 \implies 18t = 36 \implies t = 2 \,s$.Substitute $t = 2 \,s$ into the velocity equation: $v = -9(2)^2 + 36(2) + 16$.$v = -9(4) + 72 + 16 = -36 + 72 + 16 = 52 \,m/s$.

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