The speed v of a particle moving along a stra | vedclass

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(C) Given the equation of motion: $a + b{v^2} = {x^2}$.Differentiating both sides with respect to time $t$:$\frac{d}{dt}(a + b{v^2}) = \frac{d}{dt}({x

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$x/b$

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(C) Given the equation of motion: $a + b{v^2} = {x^2}$.Differentiating both sides with respect to time $t$:$\frac{d}{dt}(a + b{v^2}) = \frac{d}{dt}({x^2})$Since $a$ is a constant,its derivative is $0$. Using the chain rule:$0 + b(2v \cdot \frac{dv}{dt}) = 2x \cdot \frac{dx}{dt}$We know that acceleration $A = \frac{dv}{dt}$ and velocity $v = \frac{dx}{dt}$. Substituting these:$2bv \cdot A = 2xv$Assuming $v 
eq 0$,we can divide both sides by $2v$:$b \cdot A = x$Therefore,the acceleration $A = \frac{x}{b}$.

The speed $v$ of a particle moving along a straight line is given by $a + b{v^2} = {x^2}$ (where $x$ is its distance from the origin). The acceleration of the particle is

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