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.
Let's look at the general acceleration vector in polar coordinates$^*$:
$$\mathbf a=\left(\ddot r-r\dot\theta^2\right)\hat r+\left(r\ddot\theta+2\dot r\dot\theta\right)\hat\theta$$
If we want our object to remain on the same circle, which I'm assuming this is what you are interested in, we must have $\dot r=0$ and $\ddot r=0$. This means our acceleration must have the form:
$$\mathbf a=-r\dot\theta^2\hat r+r\ddot\theta\hat\theta$$
Since, by Newton's laws, the acceleration vector is proportional to the force via the mass $m$ of our particle, we see that we need a radial force magnitude$^{**}$ of
$$F_r=-mr\dot\theta^2$$
and a tangential force magnitude of
$$F_\theta=mr\ddot\theta$$
Because we are confined to move around a single circle we can determine the role of each force. The radial force must only be responsible for changing the direction of the velocity, since if it could effect the speed then this means the object would have to change its $r$ coordinate, hence knocking us off the circle. Similarly, the tangential force must only be responsible for changing the speed of the particle as it moves around the circle, since if it could effect the direction of the velocity then it would do so by knocking us off the circle.
You can probably tell by now that if we want to stay on the circle, these forces are required to be "linked", in a sense. Indeed, if you take the time derivative of $F_r$ and use the requirement that $\dot r=0$, you will find that
$$\frac{\text dF_r}{\text d t}=-2\dot\theta F_\theta$$
showing that the presence of a tangential force requires a change in the magnitude of the radial force in order for the particle to remain on the circle (or, on the flip side, a change in the magnitude of the radial force must be accompanied by a tangential force).
What if this condition is not met? Well, looking at the derivative of $F_r$ without the condition that $\dot r=0$, we must have
$$-m\dot r\dot\theta^2\neq 0$$
This means that $\dot r\neq0$, which means we are no longer engaging in circular motion. Therefore, we need these two force components to be linked in this way to keep the motion circular.
A subtle point remains to be cleared up (as realized in comments to other answers). This does not necessarily mean that these forces are physically linked in general. This answer assumes we have an object undergoing non-uniform circular motion, and then investigates what must be true about the forces acting on the object. However, there could be situations where the radial and tangential force are not physically linked, at which point to achieve non-uniform circular motion you would have to make these forces act in such a way so that non-uniform circular motion is achieved.
$^*$ dots represent a rate of change with respect to time if you are not familiar with calculus. for example, $\dot r$ is the rate of change of the variable $r$ with respect to time. The derivation of this equation can be found here.
$^{**}$ Note that this is what you usually encounter in your introductory physics classes as $F_{r}=mv^2/r$, since for motion along a circle of radius $r$, $\dot\theta=v/r$. The negative sign in this answer is to keep track of the direction of increasing/decreasing $r$, but if you are working a question where you only care about the magnitude of the radial force then this is irrelevant, hence why you usually don't see the negative sign.
Best Answer
The centripetal acceleration is $r\omega^2$ provided by a force $mr\omega^2$.
If the trajectory is to be a circle the radius $r$ must stay constant whilst $\omega$ is changing.
If it were a satellite orbiting the Earth such a change could not happen as for the same radius of orbit (equal to the separation between satellite and Earth) the gravitational force of attraction between the Earth and the satellite would need to change.
In terms of a rocket attached to a merry-go-round it can happen because in such a case the force applied to the rocket by the merry-go-round could increase in order to compensate for the increasing speed of the rocket (and merry-go-round).
As $a=\frac{v^2} {r}$ the rate of change of centripetal acceleration is $\dot a = \frac{2v}{r}\frac {dv}{dt}$.