Acceleration in curvilinear motion in terms of arc length

Question

The following expressions for velocity$(v)$ and acceleration$(a)$ in curvilinear motion of a particle are from the book Engineering Dynamics - A Primer by Oliver M. O'Reilly. $r$ is a vector function, $s$ is the arc length parameter and t is the time,

$\LARGE v=\frac{dr(t)}{dt}=\frac{dr}{ds}\frac{ds}{dt}$

$\LARGE a=\LARGE\frac{dv(t)}{d(t)}=\frac{d^2r}{ds^2}$$\LARGE(\LARGE\frac{ds}{dt})^2$+$\LARGE\frac{dr}{ds}\frac{d^2s}{dt^2}$

I understand the velocity but how do you derive the expression for acceleration?

Answer 1

Given the velocity equation

$v = \dfrac{dr}{dt} = \dfrac{dr}{ds} \dfrac{ds}{dt}, \tag 1$

we obtain acceleration by taking yet another $t$-derivative:

$a = \dfrac{dv}{dt} = \dfrac{d}{dt} \left ( \dfrac{dr}{ds} \dfrac{ds}{dt} \right ) = \dfrac{ds}{dt} \dfrac{d}{ds} \left ( \dfrac{dr}{ds} \dfrac{ds}{dt} \right )$ $= \dfrac{d^2r}{ds^2} \left ( \dfrac{ds}{dt} \right )^2 + \dfrac{dr}{ds}\dfrac{ds}{dt} \dfrac{d}{ds } \left( \dfrac{ds}{dt} \right )$ $= \dfrac{d^2r}{ds^2} \left ( \dfrac{ds}{dt} \right )^2 + \dfrac{dr}{ds} \dfrac{d}{dt} \left( \dfrac{ds}{dt} \right ) = \dfrac{d^2r}{ds^2} \left ( \dfrac{ds}{dt} \right )^2 + \dfrac{dr}{ds} \dfrac{d^2s}{dt^2}. \tag 2$

Answer 2

hint

The acceleration is the derivative with respect to $ t $ of the velocity $ v$.

$v $ is a product of the form $ A.B $

thus, the acceleration will be

$$a= \frac{dA}{dt}B + A \frac{dB}{dt}$$

but, the derivative wrt the time $ t $ is the product of the derivative wrt $ s $ by the deivative of $ s $ wrt $ t $. $$\frac {d}{dt} = \frac{d}{ds}.\frac{ds}{dt}$$