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(D) According to the work-energy theorem,the work done by a force is equal to the change in kinetic energy: $dW = F \cdot dx = dK$.Therefore,the force

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At $D$,the particle has maximum acceleration.

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(D) According to the work-energy theorem,the work done by a force is equal to the change in kinetic energy: $dW = F \cdot dx = dK$.Therefore,the force $F$ is given by the slope of the kinetic energy-position graph: $F = \frac{dK}{dx}$.Since $F = ma$,the acceleration $a = \frac{F}{m} = \frac{1}{m} \cdot \frac{dK}{dx}$.This implies that the acceleration is directly proportional to the slope of the $K-x$ graph.At point $A$,the slope is positive,so the force is positive (accelerating).At point $B$,the slope is zero,so the force is zero.At point $C$,the slope is zero,so the force is zero.At point $D$,the slope of the graph is the steepest (maximum),which means the force and consequently the acceleration are at their maximum values.Thus,the correct statement is that at $D$,the particle has maximum acceleration.

For a particle moving under the action of a variable force,the kinetic energy-position $(K-x)$ graph is given. Then:

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