= Tangential and normal acceleration
{title2=$a=\dot v\,T+\kappa v^2N$}
For a twice continuously differentiable <regular curve> describing motion, let $v=\|r'(t)\|>0$, $T=r'/v$, and let $N$ be the <principal normal vector> where the <curvature> $\kappa$ is positive. Differentiating the <velocity vector> $r'=vT$ gives
$$
\boxed{a=r''=\dot v\,T+\kappa v^2N.}
$$
The <tangential component> changes the <speed>; the <normal component> changes the direction of the <velocity vector>. A particle can therefore have nonzero <acceleration> at constant <speed>, as in <circular motion> with constant <speed>. The formula follows from $dT/dt=(dT/ds)(ds/dt)=\kappa vN$ and the <product rule>. When $\kappa=0$, the normal contribution vanishes even though the <principal normal vector> is undefined there. https://ocw.mit.edu/courses/16-07-dynamics-fall-2009/84852a46fa77de9a750245ceb761255a_MIT16_07F09_Lec06.pdf[Widnall and Peraire, MIT 16.07 Lecture 6: intrinsic coordinates, pp. 1–4]{ref}
In dimensionless coordinates, take $r(t)=(t,t^2)$. At $t=1$,
$$
v=\sqrt5,\qquad
\dot v=\frac4{\sqrt5},\qquad
\kappa=\frac{2}{5\sqrt5},\qquad
T=\frac{(1,2)^T}{\sqrt5},\qquad
N=\frac{(-2,1)^T}{\sqrt5}.
$$
The resulting <vector> components are
$$
a_T=\left(\frac45,\frac85\right)^T,\qquad
a_N=\left(-\frac45,\frac25\right)^T,\qquad
\boxed{a_T+a_N=(0,2)^T=r''(1).}
$$
Both increasing <speed> and turning contribute to the same <acceleration>.
\Image[/tangent-normal-acceleration-on-parabola.png]
{title=Acceleration on a parabola has both tangential and normal parts}
{description=At $t=1$, the <acceleration> $(0,2)^T$ is the sum of $(4/5,8/5)^T$ tangent to the path and $(-4/5,2/5)^T$ normal to it. This is a dimensionless mathematical trajectory.}
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