Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

The graph between the displacement $x$ and time $t$ 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:

0 + +

– 0 + 0

0 – +

– 0 – 0

Step-by-Step Solution

Curvature and Acceleration: In a displacement-time ( $x-t$ ) graph, the acceleration corresponds to the curvature of the graph (the second derivative $\frac{d^2x}{dt^2}$ ).
Sign Convention:

Concave Upwards: If the graph curves upwards (like a cup), the slope (velocity) is increasing, indicating positive acceleration ( $a > 0$ ) .
Concave Downwards: If the graph curves downwards (like an inverted cup), the slope (velocity) is decreasing, indicating negative acceleration ( $a < 0$ ) .
Straight Line: If the graph is a straight line, the slope is constant (constant velocity), indicating zero acceleration ( $a = 0$ ).

Analysis of Intervals (based on the correct option):

OA: The graph is concave downwards ( $-$ ).
AB: The graph is a straight line ( $0$ ).
BC: The graph is concave upwards ( $+$ ).
CD: The graph is a straight line ( $0$ ).

Conclusion: The sequence of acceleration signs is – 0 + 0.

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