A particle has initial velocity `12 m//s`. It moves due to a constant force along the line of velocity which initially produces retardation of `8 m//s^(2)`. Then :

Options

A. the distance travelled in first 3 second is 0 m
B. the distance travelled in first 3 second is `36 m`
C. the distance travelled in first 3 second is `18 m`
D. displacement in first 3 second is 0 m

Answer

C, D

Explanation

`u = 12 m//s`
`a = -8 //s^(2)`
Displacement
`vec(S) = vec(u) t + (1)/(2) vec(a) t^(2)`
`= 12 xx 3 + (1)/(2) xx (-8) xx 3^(2) = 0`
Distance
`v = u + at`
`0 = 12 - 8t`
`t = (3)/(2)s at t = (3)/(2)s` velocity becomes zero & direction of motion reverses
Distance `t = 0 " to " t = (3)/(2)s`
`S = ut + (1)/(2) at^(2) = 9 m`
Distance `t (3)/(2)s "to" 3s`
`S = 9m`
Total distance `= 9 + 9 = 18 m`