A car at a steady 20 m/s around a bend is accelerating, and this is not a trick of wording. Velocity carries a direction; change the direction and you have changed the velocity, whatever the speedometer says. The change points at the centre of the circle, and its size is $a_c = \dfrac{v^2}{r}$ — a-c equals v squared over r — where $a_c$ is the centripetal acceleration in $\mathrm{m/s^2}$, $v$ is the speed along the circle in m/s, and $r$ is the radius in metres, measured to the CENTRE of the turn. Multiply by the mass and you have the force that must be supplied: $F_c = \dfrac{mv^2}{r}$, in newtons.

Say the next part out loud, because it is where the marks are: centripetal force is not a new force. It is a JOB, and some ordinary force has to take it — friction on a flat road, the track's normal force on a banked one, tension in a string, gravity for a satellite. Nothing pushes outward. The shove you feel against the car door is your own inertia going straight while the door turns into you.

Two standard designs follow. On a FLAT curve friction does the whole job, and friction has a ceiling: set $\mu_s mg = \dfrac{mv^2}{r}$, cancel the mass, and the top speed is $v_{\max} = \sqrt{\mu_s g r}$ — a property of the road and the weather, identical for a loaded truck and an empty hatchback. On a BANKED curve the track is tilted so its own perpendicular push leans inward, and the angle that does it with no friction at all is $\theta = \tan^{-1}\!\left(\dfrac{v^2}{rg}\right)$, in degrees, for one chosen design speed.

And keep one number in view above all others: the $v^2$. Double the speed and the demand does not double, it QUADRUPLES. That single exponent is why highway curves carry advisory speeds, why the last 10 km/h costs more grip than the first fifty, and why wet asphalt rewrites the arithmetic before you have finished reading the sign.