If t = frac{v^2}{2},then left( - frac{df}{dt} | vedclass

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(B) Given $t = \frac{v^2}{2}$.Differentiating with respect to $t$,we get $\frac{dt}{dt} = \frac{1}{2} 	imes 2v \frac{dv}{dt}$.Since $f = \frac{dv}{dt

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$f^3$

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(B) Given $t = \frac{v^2}{2}$.Differentiating with respect to $t$,we get $\frac{dt}{dt} = \frac{1}{2} 	imes 2v \frac{dv}{dt}$.Since $f = \frac{dv}{dt}$,we have $1 = v \cdot f$,which implies $f = \frac{1}{v}$ or $v = f^{-1}$.Now,differentiate $f = v^{-1}$ with respect to $t$:$\frac{df}{dt} = -v^{-2} \frac{dv}{dt}$.Substitute $\frac{dv}{dt} = f$ and $v = \frac{1}{f}$:$\frac{df}{dt} = -\left( \frac{1}{f} ight)^{-2} \cdot f = -f^2 \cdot f = -f^3$.Therefore,$-\frac{df}{dt} = -(-f^3) = f^3$.

If $t = \frac{v^2}{2}$,then $\left( - \frac{df}{dt} \right)$ is equal to,(where $f$ is acceleration)

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