Question
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 $\alpha = 1{\text{ }ms}^{- 2},\beta = 6{\text{ }ms}^{- 1}$ and $\gamma = 5\text{ }m$. The average speed of the particle, in ${ms}^{- 1}$ from $t = 0$ to $t = 6\text{ }s$ is:
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Solution
Sol. [Key Concept: Kinematics from a position–time function]
Position (s)
$${\begin{matrix} & & S = \alpha t^{2} - \beta t + \gamma \\ & S = t^{2} - 6t + 5 & \text{(i)} \end{matrix} }{ Velocity\ (\ v}$$ )
$${\begin{array}{r} v = \frac{ds}{dt} = 2t - 6\#(ii) \end{array} } $$When $v = 0 = 2t - 6$
$$t = 3\text{ }s$$
From (ii), graph between $(v/t)$
For distance
$$\begin{matrix} & d = \left( + A_{1} \right) + A_{2} \\ & d = \frac{1}{2} \times 3 \times 6 + \frac{1}{2} \times 3 \times 6 \\ & d = 3 \times 6 = 18 \end{matrix}$$
Average speed $\left( v_{\text{avg}\text{~}} \right) = \frac{\text{~}\text{Distance}\text{~}}{\text{~}\text{Time}\text{~}} = \frac{d}{\Delta t} = \frac{18}{6} = 3{\text{ }ms}^{- 1}$
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