A particle of mass 1,kg is moving along the x | vedclass

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(B) Given,$v = 2\sqrt{x}$.Acceleration $a = v \frac{dv}{dx}$.First,find $\frac{dv}{dx}$:$\frac{dv}{dx} = \frac{d}{dx}(2x^{1/2}) = 2 \cdot \frac{1}{2}

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$2$

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(B) Given,$v = 2\sqrt{x}$.Acceleration $a = v \frac{dv}{dx}$.First,find $\frac{dv}{dx}$:$\frac{dv}{dx} = \frac{d}{dx}(2x^{1/2}) = 2 \cdot \frac{1}{2} x^{-1/2} = \frac{1}{\sqrt{x}}$.Now,calculate acceleration:$a = (2\sqrt{x}) \cdot \left(\frac{1}{\sqrt{x}}\right) = 2\,m/s^2$.Since the acceleration is constant and independent of time and position,the net force $F$ acting on the particle is given by Newton's second law:$F = m \cdot a = 1\,kg \cdot 2\,m/s^2 = 2\,N$.Thus,the net force at any instant,including at $t = 2\,s$,is $2\,N$.

$A$ particle of mass $1\,kg$ is moving along the $x-$axis,where its velocity depends on the $x-$coordinate as $v = 2\sqrt{x}\,m/s$. Find the net force acting on the particle $2\,s$ after the start of motion.

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