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NEET PHYSICSMOTION IN A STRAIGHT LINEMedium

Question

The graph between the displacement xx and time tt for a particle moving in a straight line is shown in the figure. During the intervals OA, AB, BC, and CD, the acceleration of the particle is:

A
  • 0 + +
B

– 0 + 0

C
  • 0 – +
D

– 0 – 0

Step-by-Step Solution

  1. Curvature and Acceleration: In a displacement-time (xtx-t) graph, the acceleration corresponds to the curvature of the graph (the second derivative d2xdt2\frac{d^2x}{dt^2}).
  2. Sign Convention:
  • Concave Upwards: If the graph curves upwards (like a cup), the slope (velocity) is increasing, indicating positive acceleration (a>0a > 0) .
  • Concave Downwards: If the graph curves downwards (like an inverted cup), the slope (velocity) is decreasing, indicating negative acceleration (a<0a < 0) .
  • Straight Line: If the graph is a straight line, the slope is constant (constant velocity), indicating zero acceleration (a=0a = 0).
  1. Analysis of Intervals (based on the correct option):
  • OA: The graph is concave downwards (-).
  • AB: The graph is a straight line (00).
  • BC: The graph is concave upwards (++).
  • CD: The graph is a straight line (00).
  1. Conclusion: The sequence of acceleration signs is – 0 + 0.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOTION IN A STRAIGHT LINE. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOTION IN A STRAIGHT LINEbetweendisplacementparticlemovingstraight

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