Question 47: Motion in 1D

Question

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?

Accepted Answer

$${X_P} = \alpha t + \beta {t^2}$$
$${X_Q} = ft - {t^2}$$
$$	herefore$$ $${V_P} = \alpha  + 2\beta t$$
$${V_Q} = f - 2t$$
$$\because$$ $${V_P} = {V_Q}$$
$$ \Rightarrow \alpha  + 2\beta t = f - 2t$$
$$ \Rightarrow t = {{f - \alpha } \over {2(1 + \beta )}}$$

Answer

$${{\alpha  - f} \over {1 + \beta }}$$

Answer

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

Answer

$${{\alpha  + f} \over {2(1 + \beta )}}$$

Answer

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

Motion in 1D

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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