Two cars leave the same point with 1 minute gap and move with an acceleration of0.2ms−2 . The time after the departure of the second car at which the distance between them becomes three times of its initial value is

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Two cars leave the same point with 1 minute gap and move with an acceleration of $0 .2 {ms}^{- 2}$ . The time after the departure of the second car at which the distance between them becomes three times of its initial value is

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$\frac{3}{2}$ minute

$\frac{1}{2}$ minute

1 minute

2 minute

answer is C.

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Detailed Solution

$S_{1} = 1 / 2 \times 0.2 \times (60)^{2}$

$\begin{array}{r} S_{1} = \frac{1}{2} \times \frac{2}{10} \times 60 \times 60 = 360 m \\ S_{2} = \frac{1}{2} \times 0 .2 \times (t + 60)^{2} \\ S_{2} - S_{1} = 3 s_{1} \\ \frac{1}{2} \times (0 .2) (t + 60)^{2} - \frac{1}{2} \times (0 .2) (t^{2}) = 3 \times 360 \\ \Rightarrow \frac{1}{2} \times 0 .2 [(t + 60)^{2} - t^{2}] = 1080 \\ \Rightarrow 0 .1 [t^{2} + 3600 + 120 t - t^{2}] = 1080 \\ \Rightarrow [3600 + 120 t] = 10800 \\ \Rightarrow 3600 + 120 t = 10800 \\ \Rightarrow 120 t = 10800 - 3600 \\ 120 t = 7200 \\ t = 60 s \\ ∴ t = 1 \min \end{array}$