The velocity v of a particle as a function of | vedclass

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(B) Given the velocity function: $v = \sqrt{c_1 - c_2 x}$.Squaring both sides,we get: $v^2 = c_1 - c_2 x$.Differentiating both sides with respect to t

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$-\frac{c_2}{2}$

Vedclass · Answer

(B) Given the velocity function: $v = \sqrt{c_1 - c_2 x}$.Squaring both sides,we get: $v^2 = c_1 - c_2 x$.Differentiating both sides with respect to time $t$:$\frac{d}{dt}(v^2) = \frac{d}{dt}(c_1 - c_2 x)$.Using the chain rule: $2v \frac{dv}{dt} = -c_2 \frac{dx}{dt}$.Since acceleration $a = \frac{dv}{dt}$ and velocity $v = \frac{dx}{dt}$,we have:$2v a = -c_2 v$.Assuming $v 
eq 0$,we divide both sides by $2v$:$a = -\frac{c_2}{2}$.Thus,the acceleration of the particle is constant and equal to $-\frac{c_2}{2}$.

The velocity $v$ of a particle as a function of its position $x$ is expressed as $v = \sqrt{c_1 - c_2 x}$,where $c_1$ and $c_2$ are positive constants. The acceleration of the particle is

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