A particle starts moving from rest in a straight line with constant acceleration. After time $t_{0}$ , acceleration changes its sign (just opposite to the initial direction), remaining the same in magnitude. Determine the time from the beginning of motion in which the particle returns to the initial postion

2 t_{0}

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(2 + \sqrt{2}) t_{0}

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3 t_{0}

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(2 - \sqrt{2} t_{0})

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Solution

The correct option is B $(2 + \sqrt{2}) t_{0}$
Writing, s = ut + $\frac{1}{2} a t^{2}$ , from point B onwards we have,
$- \frac{1}{2} a t_{0}^{2} = (a t_{0}) t - \frac{1}{2} a t^{2}$

$∴ t^{2} - 2 t_{0} t - t_{0}^{2} = 0$
$t = \frac{2 t_{0} \pm \sqrt{4 t_{0}^{2} + 4 t_{0}^{2}}}{2}$
$= (t_{0} + \sqrt{2} t_{0})$
$∴$ Total time $= t_{A B} + t = (2 + \sqrt{2}) t_{0}$

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The correct option is B (2+√2)t0Writing, s = ut + 12at2, from point B onwards we have, −12at20=(at0)t−12at2 ∴t2−2t0t−t20=0 t=2t0±√4t20+4t202 =(t0+√2t0) ∴ Total time =tAB+t=(2+√2)t0