Question

The displacement of a particle as a function of time is shown in figure. It indicates

• A. The particle starts with a certain velocity, but the motion is retarded and finally the particle stops.
• B. The velocity of the particle decreases.
• C. The acceleration of the particle is in opposite direction to the velocity.
• D. The particle starts with a constant velocity, the motion is accelerated and finally the particle moves with another constant velocity.

Answer 1

Answer: (A)

The particle starts with a certain velocity, but the motion is retarded and finally the particle stops.

Answer 2

The given graph shows displacement (s) as a function of time (t).

1. Velocity from s-t graph: The slope of the displacement-time (s-t) graph represents the instantaneous velocity of the particle ($v = \frac{ds}{dt}$).

   • At $t=0$, the curve has a positive and non-zero slope, indicating that the particle starts with a certain positive velocity.
   • As time increases, the slope of the curve gradually decreases. This means the velocity of the particle is decreasing.
   • As $t$ approaches a large value (e.g., beyond $t=4$), the curve becomes horizontal, meaning the slope approaches zero. This indicates that the velocity of the particle approaches zero, i.e., the particle eventually stops.

2. Acceleration from s-t graph: The rate of change of velocity is acceleration. Since the velocity (slope) is continuously decreasing while remaining positive (as displacement is always increasing), the acceleration must be negative.

   • If velocity is positive and acceleration is negative, they are in opposite directions. This condition describes retarded motion (or deceleration).

Now let's evaluate each option:

• A. The particle starts with a certain velocity, but the motion is retarded and finally the particle stops.

  • "The particle starts with a certain velocity": Correct, as the initial slope is non-zero.
  • "but the motion is retarded": Correct, as the velocity is decreasing (slope is decreasing), and since the velocity is positive, the acceleration must be in the opposite direction, causing retardation.
  • "and finally the particle stops": Correct, as the slope approaches zero, indicating the velocity becomes zero. This option provides a complete and accurate description of the entire motion.

• B. The velocity of the particle decreases.

  • Correct, as observed from the decreasing slope of the graph. However, this is only a partial description of the motion.

• C. The acceleration of the particle is in opposite direction to the velocity.

  • Correct. Since the velocity is positive (displacement increases) and it is decreasing, the acceleration must be negative (opposite to the positive velocity). This is also a partial description and a consequence of the velocity decreasing.

• D. The particle starts with a constant velocity, the motion is accelerated and finally the particle moves with another constant velocity.

  • "starts with a constant velocity": Incorrect, the slope is not constant initially.
  • "the motion is accelerated": Incorrect, the motion is retarded (decelerated) as velocity is decreasing.
  • "finally the particle moves with another constant velocity": Incorrect, the particle finally stops (velocity becomes zero), not moves with a constant non-zero velocity. This option is entirely incorrect.

Comparing options A, B, and C, option A provides the most comprehensive and accurate description of the particle's motion from start to finish, incorporating the initial state, the nature of the motion (retarded), and the final state. Options B and C are true statements but describe only aspects of the motion.