Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 79

Show that $\displaystyle a(t) = v(t) \frac{dV}{ds}$ of a particle that moves along a straight line with displacement $s(t)$, velocity $v(t)$ and the acceleration $a(t)$. Explain the difference between the meanings of the derivatives $\displaystyle \frac{dv}{dt}$ and $\displaystyle \frac{dv}{ds}$

Recall that $a(t) \displaystyle = \frac{dV}{dt}$ and $v(t) = \displaystyle \frac{ds}{dt}$


$
\begin{equation}
\begin{aligned}

a(t) =& \frac{dV}{dt} = \frac{dV}{ds} \left( \frac{ds}{dt} \right) = \frac{dV}{ds} v
\\
\\
& \text{So,}
\\
\\
a(t) =& v \frac{dV}{ds}
\\
\\


\end{aligned}
\end{equation}
$


$\displaystyle \frac{dV}{dt}$ represents how the velocity changes with respect to time. On the other hand, $\displaystyle \frac{dV}{ds}$ represents how the velocity changes with respect to displacement of a certain particle.