If the speed is constant there is only normal acceleration (from the motorist to the center of the circunference).
To get both accelerations you can make:
1) Get your position vector in polar coordinates:
$$
\vec{r} = x\vec{i}+y\vec{j} =r \cos \theta \; \hat{i} + r \sin \theta \; \hat{j} = r \hat{r}.
$$
Now you can get the speed vector making the derivative with respect to time:
$$
\vec{v} = \dfrac{d\vec{r}}{dt}=\omega r \left( - \sin \theta \; \hat{i} + \cos \theta \; \hat{j}\right) = \omega r \hat{v},
$$
wehre
$$
\omega = \dfrac{d\theta}{dt}.
$$
Finally you get the acceleration
$$
\vec{a}=\dfrac{d\vec{v}}{dt}=
\omega^2 r
\left(
-\cos \theta \; \hat{i} - \sin \theta \; \hat{j}
\right)
+
r\alpha
\left( - \sin \theta \; \hat{i} + \cos \theta \; \hat{j}\right),
$$
where
$$
\alpha = \dfrac{d\omega}{dt}.
$$
Now notice that you can rewrite the acceleration in terms of the position and speed vectors:
$$
\vec{a}=
-\omega^2 r \hat{r}
+
r\alpha
\hat{v},
$$
and that's it. One of it is the normal acceleration and the other is the tangential. Who to rembember it? Easy: position vector goes from the center of the circle, so
$
-\omega^2 r \hat{r}
$
is an acceleration that goes to the center of the circle: we call it normal or radial acceleration $a_n=\omega ^2 r $. The other acceleration goes in the speed vector, and the speed is alway tangential to the circle in circular motion. Then we call ir tangential acceleration $a_t = \alpha r$.
If you go back to the original coordinates, you get what you said:
$$
a_n = \dfrac{v^2}{r}, \qquad a_t = \dfrac{dv}{dt}.
$$
In your homework notice that you hace a 70 km/h speed with no acceleration. Then one of the circular accelerations is zero. Guess what? Notice also that in a circular motion theres is ALWAYS one acceleration that MUST be non-zero.
