Question

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

• A. + 0 + +
• B. - 0 + 0
• C. + 0 - +
• D. - 0 - 0

Answer 1

The correct option is B - 0 + 0

Slope of the displacement-time graph gives the instantaneous velocity of the particle.

Region OA : Slope is (+) and decreasing i.e $v$ decreases in positive direction.

$⟹$ The particle is retarding i.e $a$ is negative (-)

Region AB : Slope is zero i.e $v=constant$ .

$⟹$ The particle is moving with zero acceleration i.e $a=0$

Region BC : Slope is (+) and increasing i.e $v$ increases in positive direction.

$⟹$ The particle is accelerating i.e $a$ is positive (+)

Region CD : Slope is positive but constant i.e $v=constant$ .

$⟹$ The particle is moving with zero acceleration i.e $a=0$