Average speed is defined as the total distance traveled divided by the total time taken. Note that for motion in a straight line, if the particle changes direction, distance is not equal to the magnitude of displacement.
The velocity $v(t)$ is the derivative of position $s(t)$ with respect to time: $$v(t) = \frac{ds}{dt} = 2\alpha t - \beta$$ Substituting $\alpha = 1$ and $\beta = 6$: $$v(t) = 2t - 6$$
The particle changes direction when $v(t) = 0$: $$2t - 6 = 0 \implies t = 3 \text{ s}$$ Since $t=3$ lies within the interval $[0, 6]$, the particle moves in one direction until $t=3$ and reverses thereafter.
Calculate positions at $t=0, 3, 6$: $$s(0) = 1(0)^2 - 6(0) + 5 = 5 \text{ m}$$ $$s(3) = 1(3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4 \text{ m}$$ $$s(6) = 1(6)^2 - 6(6) + 5 = 36 - 36 + 5 = 5 \text{ m}$$ Distance from $t=0$ to $t=3$: $|s(3) - s(0)| = |-4 - 5| = 9 \text{ m}$ Distance from $t=3$ to $t=6$: $|s(6) - s(3)| = |5 - (-4)| = 9 \text{ m}$ Total distance = $9 + 9 = 18 \text{ m}$
$$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{18 \text{ m}}{6 \text{ s}} = 3 \text{ ms}^{-1}$$
Final Answer: 3
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