The graph between the displacement x and time | vedclass

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(B) The acceleration of a particle is given by the second derivative of displacement with respect to time,$a = \frac{d^2x}{dt^2}$,which corresponds to

Vedclass · Accepted Answer

$-, 0, +, 0$

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(B) The acceleration of a particle is given by the second derivative of displacement with respect to time,$a = \frac{d^2x}{dt^2}$,which corresponds to the concavity of the $x-t$ graph.$1$. Region $OA$: The graph is concave downwards (bending towards the time axis). The slope (velocity) is decreasing,so the acceleration is negative $(-)$.$2$. Region $AB$: The graph is a straight line parallel to the time axis. The slope (velocity) is constant and zero. Therefore,the acceleration is zero $(0)$.$3$. Region $BC$: The graph is concave upwards (bending towards the displacement axis). The slope (velocity) is increasing,so the acceleration is positive $(+)$.$4$. Region $CD$: The graph is a straight line with a constant non-zero slope. Since the velocity is constant,the acceleration is zero $(0)$.Thus,the signs of acceleration for $OA, AB, BC, CD$ are $-, 0, +, 0$ respectively. The correct option is $(b)$.

The graph between the displacement $x$ and time $t$ for a particle moving in a straight line is shown in the figure. Determine the sign of the acceleration of the particle for the intervals $OA$,$AB$,$BC$,and $CD$ respectively.

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