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(A) Given: Initial velocity $u = 0$, acceleration $a = 8 	ext{ m/s}^2$.Using the equation of motion $S = ut + \frac{1}{2}at^2$, since $u = 0$, we hav

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$\frac{\sqrt{2}-1}{2} 	ext{ s}$

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(A) Given: Initial velocity $u = 0$, acceleration $a = 8 	ext{ m/s}^2$.Using the equation of motion $S = ut + \frac{1}{2}at^2$, since $u = 0$, we have $S = \frac{1}{2}at^2$.$1$. Time taken to cover the first metre $(S_1 = 1 	ext{ m})$:$1 = \frac{1}{2} 	imes 8 	imes t_1^2 \implies 1 = 4t_1^2 \implies t_1^2 = \frac{1}{4} \implies t_1 = \frac{1}{2} 	ext{ s}$.$2$. Time taken to cover the first two metres $(S_2 = 2 	ext{ m})$:$2 = \frac{1}{2} 	imes 8 	imes t_2^2 \implies 2 = 4t_2^2 \implies t_2^2 = \frac{1}{2} \implies t_2 = \frac{1}{\sqrt{2}} 	ext{ s}$.$3$. Time taken to move the second metre is the difference between the time taken to cover $2 	ext{ m}$ and $1 	ext{ m}$:$\Delta t = t_2 - t_1 = \frac{1}{\sqrt{2}} - \frac{1}{2} = \frac{\sqrt{2}-1}{2} 	ext{ s}$.

The acceleration of a particle starting from rest and moving in a straight line with uniform acceleration is $8 \text{ m/s}^2$. The time taken by the particle to move the second metre is

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