A body starting from rest at t = 0 moves along a straight line with a constant acceleration. At t = 2s, the body reverses its direction keeping the acceleration same. The body returns to the initial position at t = `t_(0)`. Then `t_(0)` is

Options

A. 4 s
B. `(4 + 2 sqrt(2) )`s
C. `(2 + 2sqrt(2))`s
D. `(4 + 4 sqrt(2))` s

Answer

B

Explanation

According to the question,

From first equation of the motion,
`v_(1) = u + at_(1) rArr v_(1) = 2a`
Firstly, body decelerate with acceleration to the point C and Then reverse it.s direction and accelerate with acceleration a to the point A. Therefore for distance BC, From first equation of the motion,
`v_(2) = v_(1) - at_(2) rarr 0 = 2a - at_(2)`
or `" "t_(2) = 2`s
Hence, total time taken by body to covered distance AC, t = 2 + 2 = 4s
From second equation one motion,
`s_(1) = AB = ut_(1) + (1)/(2) at_(1)^(2) = 0 + (1)/(2) a xx 2^(2)`
`s_(1)` = 2a `therefore " " s_(1) = s_(2) = 2a`
`therefore " " AC = s_(1) + s_(2) = 4a`
Now, body returns from point C to point A. So, ` u_(1) = 0 , AC = ua`
From second equation of the motion ,
`rArr " "(1)/(2) log_(e) phi (v^(2)) = log_(e) x + log_(e) (sqrt(c)) = log_(e) ( sqrt(c) x)`
`rArr phi (v^(2)) = cx^(2) rArr phi ((y^(2))/(x^(2))) = cx^(2)`
Hence, option (d) is correct.
Hence, option (d) is correct.
S = AC = `u_(1)t + (1)/(2)at^(2) or ua = 0 + (1)/(2) at^(2)`
`rArr t^(2) = 8 rArr t = 2sqrt(2)` s