#### Answer

See details below.

#### Work Step by Step

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.