Consider a particle moving along a straight line, whose position as a function of time is given by $s(t) = \alpha t^2 - \beta t + \gamma$, where $\alpha = 1\ \text{m s}^{-2}$, $\beta = 6\ \text{m s}^{-1}$ and $\gamma = 5\ \text{m}$. The average speed of the particle, in $\text{m s}^{-1}$, from $t = 0$ to $t = 6$ s is :
Answer: (C) $3$
Concept. Average speed $=$ total distance / time. If the particle reverses, add the distances on each leg.
Step 1. Velocity. $s = t^2 - 6t + 5$, so $v = 2t - 6$, which is zero at $t = 3$ s (the particle turns back).
Step 2. Positions. $s(0) = 5$ m, $s(3) = -4$ m, $s(6) = 5$ m.
Step 3. Distance. $|{-4} - 5| + |5 - (-4)| = 9 + 9 = 18$ m.
Step 4. Average speed. $\dfrac{18}{6} = 3\ \text{m s}^{-1}$. (The average velocity is zero, since the particle returns to the start.)
Answer. $3$ (Option C).
Solution by Sreeraj P, M.Sc Physics