The correct position (x) - time (t) graph for | vedclass

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(A) The equation of motion for a particle with constant acceleration is given by $x(t) = x_0 + v_0 t + \frac{1}{2} a t^2$.If the acceleration $a$ is n

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(A) The equation of motion for a particle with constant acceleration is given by $x(t) = x_0 + v_0 t + \frac{1}{2} a t^2$.If the acceleration $a$ is negative,the graph of $x$ versus $t$ is a parabola opening downwards.In a position-time graph,the slope represents the velocity $(v = \frac{dx}{dt})$.For a downward-opening parabola,the slope starts as positive,decreases to zero at the peak (where the particle momentarily stops),and then becomes negative as the particle moves in the opposite direction.This behavior corresponds to a constant negative acceleration.Therefore,the graph that represents a downward-opening parabola is the correct one,which is shown in option $(A)$.

The correct position $(x)$ - time $(t)$ graph for a particle moving with negative acceleration is

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