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(B) Given the equation for velocity: ${v^2} = a + bx$. To find the acceleration,we differentiate both sides with respect to time $t$: $\frac{d}{dt}({v

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Uniform

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(B) Given the equation for velocity: ${v^2} = a + bx$. To find the acceleration,we differentiate both sides with respect to time $t$: $\frac{d}{dt}({v^2}) = \frac{d}{dt}(a + bx)$ Using the chain rule: $2v \frac{dv}{dt} = b \frac{dx}{dt}$. Since velocity $v = \frac{dx}{dt}$,we substitute this into the equation: $2v \frac{dv}{dt} = b v$. Assuming $v 
eq 0$,we divide both sides by $2v$: $\frac{dv}{dt} = \frac{b}{2}$. Since $b$ is a constant,the acceleration $\frac{dv}{dt} = \frac{b}{2}$ is also a constant. Therefore,the acceleration is uniform.

$A$ particle moves in a straight line so that its velocity at any point is given by ${v^2} = a + bx$,where $a, b \neq 0$ are constants. The acceleration is

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