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(A) Given the velocity $v = 5 \sqrt{x}$.We know that acceleration $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$.First,find

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$12.5$

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(A) Given the velocity $v = 5 \sqrt{x}$.We know that acceleration $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$.First,find $\frac{dv}{dx}$ by differentiating $v = 5x^{1/2}$ with respect to $x$:$\frac{dv}{dx} = 5 \cdot \frac{1}{2} x^{-1/2} = \frac{2.5}{\sqrt{x}}$.Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula:$a = (5 \sqrt{x}) \cdot \left( \frac{2.5}{\sqrt{x}} \right) = 5 \cdot 2.5 = 12.5 \, m/s^2$.Thus,the acceleration of the particle is $12.5 \, m/s^2$.

The velocity of a particle moving in the positive direction of the $x$-axis varies as $v = 5 \sqrt{x}$. Assuming that at $t = 0$,the particle was at $x = 0$,what is the acceleration of the particle in $m/s^2$?

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