a body start from a point o with velocity and uniform acceleration a .the direction of acceleration is reversed when the velocity becomes 5v. the velocity of body at point o will be

Initial velocity at O, u = v m/s

Acceleration at O, a = a m/s²

Let the point where direction of acceleration was reversed be Q... then

Velocity at Q, u' = 5v m/s

Acceleration at Q, a' = -a m/s²

Distance covered till Q, s..

(u')² - u² = 2as

(5v)² - v² = 2as

s = 12v²/a

Distance covered till the body stopped, due to negative acceleration, s'..

0² - (u')² = 2a's'

-(5v)² = -2as'

s' = 25v²/2a

total distance covered, d = s+ s' = 49v²/2a

Now, when body is stopped, it will move in reverse direction due to reversed acceleration...

Thus its initial velocity is 0 and acceleration is -a

and the distance it will cover to come back to O is -49v²/2a ( negative, because it is covered in opposite direction)

Let the final velocity at O be v'

(v')² - 0² = 2ad

(v')² = 2(-a)(-49v²/2a)

(v')² = 49v²

v' = 7v

hence final velocity at O is 7v m/s