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(A) The position of the particle is given by $x = 40 + 12t - t^3$.Velocity $v$ is the rate of change of position with respect to time: $v = \frac{dx}{

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$16$

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(A) The position of the particle is given by $x = 40 + 12t - t^3$.Velocity $v$ is the rate of change of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(40 + 12t - t^3) = 12 - 3t^2$.For the particle to come to rest,its velocity must be zero: $v = 0$.$12 - 3t^2 = 0 \implies 3t^2 = 12 \implies t^2 = 4 \implies t = 2 	ext{ s}$ (since time cannot be negative).The distance traveled by the particle from $t = 0$ to $t = 2$ is given by the change in position:$x(0) = 40 + 12(0) - (0)^3 = 40 	ext{ m}$.$x(2) = 40 + 12(2) - (2)^3 = 40 + 24 - 8 = 56 	ext{ m}$.Distance traveled = $x(2) - x(0) = 56 - 40 = 16 	ext{ m}$.

$A$ particle moves along a straight line $OX$. At a time $t$ (in seconds),the distance $x$ (in metres) of the particle from $O$ is given by $x = 40 + 12t - t^3$. How long would the particle travel before coming to rest?

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