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(B) Let the total distance covered be $d$. For the total journey: Initial velocity $= u$,final velocity $= 0$,acceleration $= -a$,time $= T$. Using $v

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$T\left(1-\frac{1}{\sqrt{2}}\right)$

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(B) Let the total distance covered be $d$. For the total journey: Initial velocity $= u$,final velocity $= 0$,acceleration $= -a$,time $= T$. Using $v = u + at$: $0 = u - aT \Rightarrow u = aT$ Total distance $d = uT - \frac{1}{2}aT^2 = (aT)T - \frac{1}{2}aT^2 = \frac{1}{2}aT^2$. For the first half of the journey: Distance $= \frac{d}{2} = \frac{1}{4}aT^2$. Let the time taken be $t$. Using $s = ut - \frac{1}{2}at^2$: $\frac{1}{4}aT^2 = (aT)t - \frac{1}{2}at^2$ Dividing by $a$: $\frac{T^2}{4} = Tt - \frac{t^2}{2}$ $T^2 = 4Tt - 2t^2 \Rightarrow 2t^2 - 4Tt + T^2 = 0$. Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $t = \frac{4T \pm \sqrt{16T^2 - 8T^2}}{4} = \frac{4T \pm \sqrt{8T^2}}{4} = \frac{4T \pm 2\sqrt{2}T}{4} = T \pm \frac{T}{\sqrt{2}}$. Since $t Thus,the correct option is $(b)$.

$A$ particle starts with an initial speed $u$ and retardation $a$ to come to rest in time $T$. The time taken to cover the first half of the total path travelled is .......

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