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Two buses P & Q start from a point at the same time and move in a straight line and their positions are represented by $X_{p}(t) = \alpha t + \beta t^{2}$ and $X_{Q}(t) = ft - t^{2}$. At what time, both the buses have same velocity?
Explanation
For bus P $\therefore$ Distance = $[\alpha[4-1]/2 + \beta[8-1]/3] = 3\alpha/2 + 7\beta/3$ $x_{p}(t) = \alpha t + \beta t^{2}$ $V_{p}(t) = \alpha + 2\beta t [ \because V_{p} = dx_{p}/dt]$ For bus Q $x_{q}(t) = ft - t^{2}$ $V_{q}(t) = f - 2t [ \because V_{q} = dx_{q}/dt]$ As, $V_{p}(t) = V_{q}(t) \Rightarrow \alpha + 2\beta t = f - 2t$ $\Rightarrow \alpha - f = -2\beta t - 2t \Rightarrow f - \alpha = 2\beta t + 2t$ $\Rightarrow t = (f - \alpha)/(2\beta + 2)$