question_answer

A point mass starts moving in straight line with constant acceleration a from rest at \[t=0.\] At time \[t=2s\], the acceleration changes the sign, remaining the same in magnitude. The mass returns to the initial position at time \[t={{t}_{0}}\] after start of motion. Here \[{{t}_{0}}\] is:

A) 4s

B) \[(4+2\sqrt{2})s\]

C) \[(2+2\sqrt{2})s\]

D) \[(4+4\sqrt{2})s\]

Correct Answer: B

Solution :

[b] Let \[{{t}_{0}}=(2+t)\] second

\[-2a=(2a)t-\frac{1}{2}a{{t}^{2}}\]

\[\therefore \]    \[{{t}^{2}}-4t-4=0\]

\[t=\frac{4\pm \sqrt{16+16}}{2}=2+2\sqrt{2}\]

\[\therefore \]            \[{{t}_{0}}=2+t=(4+2\sqrt{2})s\]