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NEET PHYSICSMOTION IN A STRAIGHT LINEMedium

Question

Velocity-time (v-t) graph for a moving object is shown in the figure. Total displacement of the object during the time interval when there is non-zero acceleration or retardation is:

A

60 m

B

50 m

C

30 m

D

40 m

Step-by-Step Solution

  1. Graph Analysis: In a velocity-time graph, the slope represents acceleration (a=dvdta = \frac{dv}{dt}) . A non-zero slope indicates non-zero acceleration (or retardation). Horizontal lines (zero slope) indicate constant velocity (a=0a=0).
  2. Displacement Principle: The total displacement is given by the area under the velocity-time graph .
  3. Calculation: The question specifically asks for displacement only during the intervals of non-zero acceleration. This requires calculating the area under the inclined parts of the graph (typically the rising and falling segments) and excluding the area under any horizontal (constant velocity) segments.
  4. Result: For this specific PMT 2005 problem, summing the areas of the acceleration and retardation phases (typically triangular regions) yields 50 m.

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 LINEvelocitytimemovingobjectfiguredisplacement

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