The motion of a particle along a straight lin | vedclass

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(B) The position of the particle is given by $x = 8 + 12t - t^3$.Velocity $v$ is the rate of change of position with respect to time: $v = \frac{dx}{d

Vedclass · Accepted Answer

$12$

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(B) The position of the particle is given by $x = 8 + 12t - t^3$.Velocity $v$ is the rate of change of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(8 + 12t - t^3) = 12 - 3t^2$.Set the velocity to zero to find the time: $12 - 3t^2 = 0 \implies 3t^2 = 12 \implies t^2 = 4 \implies t = 2 \ s$ (since $t > 0$).Acceleration $a$ is the rate of change of velocity: $a = \frac{dv}{dt} = \frac{d}{dt}(12 - 3t^2) = -6t$.At $t = 2 \ s$,the acceleration is $a = -6(2) = -12 \ m s^{-2}$.Retardation is the magnitude of negative acceleration,which is $12 \ m s^{-2}$.

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

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