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(C) The distance $S$ is given by the function $S(t) = t^3 - t^2 - t + 3$. The velocity $v$ of the particle is the rate of change of distance with resp

Vedclass · Accepted Answer

$2$

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(C) The distance $S$ is given by the function $S(t) = t^3 - t^2 - t + 3$. The velocity $v$ of the particle is the rate of change of distance with respect to time, given by $v = \frac{dS}{dt}$. $v = \frac{d}{dt}(t^3 - t^2 - t + 3) = 3t^2 - 2t - 1$. The particle comes to rest when its velocity $v = 0$. Setting $v = 0$, we get $3t^2 - 2t - 1 = 0$. Factoring the quadratic equation: $3t^2 - 3t + t - 1 = 0 \Rightarrow 3t(t - 1) + 1(t - 1) = 0 \Rightarrow (3t + 1)(t - 1) = 0$. Since time $t$ cannot be negative, we take $t = 1$ second. Now, substitute $t = 1$ into the distance equation $S(t)$: $S(1) = (1)^3 - (1)^2 - (1) + 3 = 1 - 1 - 1 + 3 = 2$ metres. Thus, the distance travelled by the particle when it comes to rest is $2$ metres.

$A$ particle moving from a fixed point on a straight line travels a distance $S$ metres in $t$ seconds. If $S = t^3 - t^2 - t + 3$, then the distance (in metres) travelled by the particle when it comes to rest is:

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