## Calculus (3rd Edition)

We have $$\mathbf{a}(t)=a_{\mathbf{T}}(t) \mathbf{T}(t)+a_{\mathbf{N}}(t) \mathbf{N}(t)$$ where $a_{\mathbf{T}}(t)$ is the tangential component of the accelaration and $a_{\mathbf{N}}(t)$ is the normal one. Now, if the speed is constant, then $v'(t)=0$ and hence $a_{\mathbf{T}}(t) =0$. So $$\mathbf{a}(t)=a_{\mathbf{N}}(t) \mathbf{N}(t)$$ Thus, the acceleration is in the direction of the normal vector, which is orthogonal to the velocity vector. That is, the acceleration and velocity vectors are orthogonal.