A particle is moving in a straight line and passes through a point O with a velocity of `6ms^(-1)` The particle moves with a constant retardation of `2ms^(-2)` for 4 s and there after moves with constant velocity. How long after leaving O does the particle return to O

Options

A. 3s
B. 8s
C. never
D. 4s

Answer

B

Explanation

Let the particle moves toward right with velocity 6 m/s. Due to retardation after time `t_1` its velocity becomes zero.

From v = v-at `rArr 0 = 6-2 xx t_1 rArr t_1 = 3sec`
But retardation works on it for 4 sec. It means after reaching point A direction of motion get reversed and acceleration works on the particle for next one second.
`S_(OA) = ut_1 - 1/2 at_1^2 = 6 xx 3 - 1/2 (2) (3)^2 = 18-9 = 9m`
`S_(AB) = 1/2 xx 2 xx (1)^2 = 1m`
` therefore S_(BC) = S_(OA) = S_(AB) = 9 -1 = 8m`
Now velocity of the particle at point B in return journey
`v = 0 + 2 xx 1 = 2m//s`
In return journey from B to C, particle moves with constant velocity 2 m/s to cover the distance 8m.
Time taken = 8/2 = 4s`
Total time taken by particle to return at point 0 is
`rArr T = T_(OA) + t_(AB) = t_(BC) = 3 + 1 + 4 = 8 sec`