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(D) Given the velocity $v = k\sqrt{x}$.We know that acceleration $a = v \frac{dv}{dx}$.First,find $\frac{dv}{dx}$ by differentiating $v$ with respect

Vedclass · Accepted Answer

$a =$ constant

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(D) Given the velocity $v = k\sqrt{x}$.We know that acceleration $a = v \frac{dv}{dx}$.First,find $\frac{dv}{dx}$ by differentiating $v$ with respect to $x$: $\frac{dv}{dx} = \frac{d}{dx}(k x^{1/2}) = k \cdot \frac{1}{2} x^{-1/2} = \frac{k}{2\sqrt{x}}$.Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula: $a = (k\sqrt{x}) \cdot \left(\frac{k}{2\sqrt{x}}\right)$.Simplifying this,we get $a = \frac{k^2}{2}$.Since $k$ is a constant,$a$ is also a constant.

If the velocity of a particle moving along the $x-$ axis is given by $v = k\sqrt{x}$,then which of the following is true? ($a$ is acceleration)

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