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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 $\alpha=1$ ms$^{-2}$, $\beta=6$ ms$^{-1}$ and $\gamma=5$ m. The average speed of the particle, in ms$^{-1}$ from t = 0 to t = 6 s is:
Detailed Solution
$s=t^2-6t+5 \Rightarrow v=\frac{ds}{dt}=(2t-6)$ m/s. The velocity is -6 m/s at t = 0, zero at t = 3 s and +6 m/s at t = 6 s. Distance = $|A_1|+|A_2|=2\left[\frac{1}{2}\times3\times6\right]=18$ m. Average speed $=\frac{18}{6}=3$ m/s.