The graph between the displacement `x` and time `t` for a particle moving in a straight line is shown in the figure. During the interval `OA, AB, BC` and `CD` the acceleration of the particle is
`OA, AB, BC, CD`

Options

A. `{:(OA,AB,BC,CD),(+,0,+,+):}`
B. `{:(OA,AB,BC,CD),(-,0,+,0):}`
C. `{:(OA,AB,BC,CD),(+,0,-,+):}`
D. `{:(OA,AB,BC,CD),(-,0,-,0):}`

Answer

B

Explanation

In portion OA, slope of graph is decreaging,so velocity is decreasing and acceleration is negative. In portion AB, displacement is constant, v= 0 and a = 0. In portion BC, slope of x-t is increasing, hence v is increasing and accelerations is positive. In portion CD, slope is constant, v=const. and a=0.