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(A) The velocity $v$ of the particle is the rate of change of displacement with respect to time $t$,given by $v = \frac{dS}{dt}$.Given $S = 2t^3 + 2t^

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$\frac{1}{3}$

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(A) The velocity $v$ of the particle is the rate of change of displacement with respect to time $t$,given by $v = \frac{dS}{dt}$.Given $S = 2t^3 + 2t^2 - 2t - 3$.Differentiating with respect to $t$,we get $v = \frac{d}{dt}(2t^3 + 2t^2 - 2t - 3) = 6t^2 + 4t - 2$.$A$ particle changes its direction when its velocity becomes zero.Setting $v = 0$,we have $6t^2 + 4t - 2 = 0$.Dividing by $2$,we get $3t^2 + 2t - 1 = 0$.Factoring the quadratic equation: $3t^2 + 3t - t - 1 = 0 \implies 3t(t + 1) - 1(t + 1) = 0$.$(3t - 1)(t + 1) = 0$.This gives $t = \frac{1}{3}$ or $t = -1$.Since time cannot be negative,we take $t = \frac{1}{3}$ seconds.

If the displacement $S$ of a particle travelling along a straight line in $t$ seconds is given by $S = 2t^3 + 2t^2 - 2t - 3$,then the time taken (in seconds) by the particle to change its direction is

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