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(A) Given the velocity equation: $v^2 = 2 - 3x$. Differentiating both sides with respect to time $t$,we apply the chain rule: $\frac{d}{dt}(v^2) = \fr

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Uniform

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(A) Given the velocity equation: $v^2 = 2 - 3x$. Differentiating both sides with respect to time $t$,we apply the chain rule: $\frac{d}{dt}(v^2) = \frac{d}{dt}(2 - 3x)$ $2v \frac{dv}{dt} = -3 \frac{dx}{dt}$ Since acceleration $a = \frac{dv}{dt}$ and velocity $v = \frac{dx}{dt}$,we substitute these into the equation: $2v \cdot a = -3v$ Assuming $v 
eq 0$,we divide both sides by $2v$: $a = -\frac{3}{2} 	ext{ m/s}^2$ Since the acceleration $a$ is a constant value $(-1.5 	ext{ m/s}^2)$,the acceleration is uniform.

$A$ particle moves in a straight line in such a way that its velocity at any point is given by $v^2 = 2 - 3x$,where $x$ is measured from a fixed point. The acceleration is

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