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(C) Given that acceleration $a = kx$. We know that acceleration can be expressed as $a = v \frac{dv}{dx}$. Substituting the given expression for $a$:

Vedclass · Accepted Answer

$v^2=u^2+kx^2$

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(C) Given that acceleration $a = kx$. We know that acceleration can be expressed as $a = v \frac{dv}{dx}$. Substituting the given expression for $a$: $v \frac{dv}{dx} = kx$ Rearranging the terms to integrate: $v \, dv = kx \, dx$ Integrating both sides with initial conditions $v=u$ at $x=0$ and final velocity $v$ at position $x$: $\int_{u}^{v} v \, dv = \int_{0}^{x} kx \, dx$ Evaluating the integrals: $\left[ \frac{v^2}{2} \right]_{u}^{v} = k \left[ \frac{x^2}{2} \right]_{0}^{x}$ $\frac{v^2 - u^2}{2} = \frac{kx^2}{2}$ Multiplying by $2$ on both sides: $v^2 - u^2 = kx^2$ Therefore,$v^2 = u^2 + kx^2$.

The initial velocity of a particle moving along the $x$-axis is $u$ (at $t=0$ and $x=0$) and its acceleration $a$ is given by $a=kx$. Which of the following equations is correct for its velocity $(v)$ and position $(x)$?

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