"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?"

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Accepted Answer

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${X_P} = \alpha t + \beta {t^2}$

${X_Q} = ft - {t^2}$

$\therefore$ ${V_P} = \alpha + 2\beta t$

${V_Q} = f - 2t$

$\because$ ${V_P} = {V_Q}$

$\Rightarrow \alpha + 2\beta t = f - 2t$

$\Rightarrow t = {{f - \alpha } \over {2(1 + \beta )}}$

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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?

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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?

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