JEE MAIN - Physics (2022 - 25th June Evening Shift - No. 2)
Two buses P and 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?
$${{\alpha - f} \over {1 + \beta }}$$
$${{\alpha + f} \over {2(\beta - 1)}}$$
$${{\alpha + f} \over {2(1 + \beta )}}$$
$${{f - \alpha } \over {2(1 + \beta )}}$$
Explanation
$${X_P} = \alpha t + \beta {t^2}$$
$${X_Q} = ft - {t^2}$$
$$\therefore$$ $${V_P} = \alpha + 2\beta t$$
$${V_Q} = f - 2t$$
$$\because$$ $${V_P} = {V_Q}$$
$$ \Rightarrow \alpha + 2\beta t = f - 2t$$
$$ \Rightarrow t = {{f - \alpha } \over {2(1 + \beta )}}$$
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