A car of mass M travels on a horizontal circular path of radius r ....

The car moving on a horizontal circular path with an increasing speed has two components of acceleration:

Tangential acceleration, $a_t = a$

Centripetal acceleration, $a_c = \dfrac{v^2}{r}$

The net acceleration of the car is given by $a_{net} = \sqrt{a_t^2 + a_c^2} = \sqrt{a^2 + \left(\dfrac{v^2}{r}\right)^2}$.

Since there is a tangential component, the net acceleration is not directed towards the centre of the path.

The required force is provided by the frictional force $f$ between the car and the ground. Thus, $f = M a_{net} = M \sqrt{a^2 + \left(\dfrac{v^2}{r}\right)^2}$.

Since $a > 0$, the magnitude of the frictional force is strictly greater than $\dfrac{Mv^2}{r}$.

For the car to move without slipping, the required frictional force must be less than or equal to the limiting friction, $f \le \mu Mg$.

$\mu Mg \ge M \sqrt{a^2 + \left(\dfrac{v^2}{r}\right)^2}$

$\mu \ge \dfrac{1}{g} \sqrt{a^2 + \left(\dfrac{v^2}{r}\right)^2}$

Since $\sqrt{a^2 + \left(\dfrac{v^2}{r}\right)^2} > a$, it follows that $\mu > \dfrac{a}{g}$.

Therefore, the coefficient of friction is not less than $\dfrac{a}{g}$.