The motion of a particle along a straight lin | vedclass

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(D) Given the position equation: $x = 8 + 12t - t^3$. Velocity $v$ is the derivative of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{d

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$12$

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(D) Given the position equation: $x = 8 + 12t - t^3$. Velocity $v$ is the derivative of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(8 + 12t - t^3) = 12 - 3t^2$. When the velocity becomes zero: $12 - 3t^2 = 0 \implies 3t^2 = 12 \implies t^2 = 4 \implies t = 2 \, s$. Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(12 - 3t^2) = -6t$. At $t = 2 \, s$,the acceleration is $a = -6(2) = -12 \, m/s^2$. Retardation is the magnitude of negative acceleration,so retardation = $12 \, m/s^2$.

The motion of a particle along a straight line is described by the equation $x = 8 + 12t - t^3$,where $x$ is in meters and $t$ is in seconds. The retardation of the particle when its velocity becomes zero is .......... $m/s^2$.

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