If x is the displacement,v is the velocity,an | vedclass

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(B) We know that velocity $v = \frac{dx}{dt}$,so $\frac{dt}{dx} = \frac{1}{v}$.Now,differentiate with respect to $x$:$\frac{d^2t}{dx^2} = \frac{d}{dx}

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$f = -v^3 \frac{d^2t}{dx^2}$

Vedclass · Answer

(B) We know that velocity $v = \frac{dx}{dt}$,so $\frac{dt}{dx} = \frac{1}{v}$.Now,differentiate with respect to $x$:$\frac{d^2t}{dx^2} = \frac{d}{dx} \left( \frac{1}{v} \right) = -\frac{1}{v^2} \frac{dv}{dx}$.Using the chain rule,$\frac{dv}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx} = f \cdot \frac{1}{v} = \frac{f}{v}$.Substituting this back into the expression:$\frac{d^2t}{dx^2} = -\frac{1}{v^2} \left( \frac{f}{v} \right) = -\frac{f}{v^3}$.Rearranging the terms,we get $f = -v^3 \frac{d^2t}{dx^2}$.

If $x$ is the displacement,$v$ is the velocity,and $f$ is the acceleration of a particle at time $t$,which of the following is true?

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