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) $3$
$s(t) = t^2 - 6t + 5$, so $v = \dfrac{ds}{dt} = 2t - 6$. The particle reverses direction at $t = 3$ s.
Positions: $s(0) = 5$ m, $s(3) = 9 - 18 + 5 = -4$ m, $s(6) = 36 - 36 + 5 = 5$ m.
Distance travelled $= |{-4} - 5| + |5 - (-4)| = 9 + 9 = 18$ m.
$$\text{Average speed} = \frac{18}{6} = 3\ \text{m s}^{-1}$$
(The average velocity is zero because the particle returns to its starting point, but average speed uses total distance.)
Solution by Sreeraj P, M.Sc Physics