The acceleration of a particle starting from rest moving in a straight line with uniform acceleration is \(8 \mathrm{~m} / \mathrm{sec}^2\). The time taken by the particle to move the second metre is

• Correct answer
  A

  \(\frac{\sqrt{2}-1}{2} \mathrm{sec}\)
• B

  \(\frac{\sqrt{2}+1}{2} \mathrm{sec}\)
• C

  \((1+\sqrt{2}) \mathrm{sec}\)
• D

  \((\sqrt{2}-1) \mathrm{sec}\)

Solution

Hint:
\(\begin{array}{l|l} \mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2 & \mathrm{~S}=\mathrm{uT}+\frac{1}{2} \mathrm{aT}^2 \\ 1=\frac{1}{2} \cdot 8 \cdot \mathrm{t}^2 & 2=\frac{1}{2} 8 \cdot \mathrm{T}^2 \\ \frac{1}{4}=\mathrm{t}^2 & \mathrm{~T}^2=\frac{1}{2} \\ \mathrm{t}=\frac{1}{2} & \mathrm{~T}=\frac{1}{\sqrt{2}} \end{array}\) \(\text {Time }=\frac{1}{\sqrt{2}}-\frac{1}{2}=\frac{2}{2}=\frac{\sqrt{2}-1}{2}\)

Hence, the correct answer is option: 'A'.