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(C) The position of the particle is given by $s(t) = t^2 - 6t + 5$.To find the average speed, we need the total distance covered.First, find the veloc

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$3$

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(C) The position of the particle is given by $s(t) = t^2 - 6t + 5$.To find the average speed, we need the total distance covered.First, find the velocity $v(t) = \frac{ds}{dt} = 2t - 6$.The particle stops when $v(t) = 0$, which occurs at $t = 3 	ext{ s}$.At $t = 0 	ext{ s}$, $s(0) = 5 	ext{ m}$.At $t = 3 	ext{ s}$, $s(3) = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4 	ext{ m}$.At $t = 6 	ext{ s}$, $s(6) = (6)^2 - 6(6) + 5 = 5 	ext{ m}$.Distance covered from $t = 0$ to $t = 3 	ext{ s}$ is $|s(3) - s(0)| = |-4 - 5| = 9 	ext{ m}$.Distance covered from $t = 3$ to $t = 6 	ext{ s}$ is $|s(6) - s(3)| = |5 - (-4)| = 9 	ext{ m}$.Total distance = $9 + 9 = 18 	ext{ m}$.Average speed = $\frac{	ext{Total distance}}{	ext{Total time}} = \frac{18 	ext{ m}}{6 	ext{ s}} = 3 	ext{ ms}^{-1}$.

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