# Maximum acceleration on a turn

The question $$\\$$

Suppose a car enters a turn with an initial speed $$\vec{v_1}$$, and comes out with a final speed $$\vec{v_2}$$ perpendicular to the direction of $$\vec{v_1}$$. The driver does everything in order to minimize the time needed to complete the turn. We're given gravitational acceleration $$g$$, coefficient of friction between wheels and the road $$\mu$$ and the mass of the car $$m$$. What is the acceleration of the car? $$\\$$

The answer, according to the book is $$\vec{a}=\frac{\vec{v_2}-\vec{v_1}}{|\vec{v_2}-\vec{v_2}|}\mu g.$$

My attempt (not getting even close to that) $$\\$$

We know that the acceleration $$\vec{a}$$ of an object is equal to$$\frac{dv}{dt} \hat{\tau}+\frac{v^2}{R} \hat{n},$$ where $$\hat{\tau}$$ and $$\hat{n}$$ are unit vectors in direction tangent to the curve and normal to it, respectively. Now, it's obvious that the car can't go at some arbitrarily huge speed, because it would end up in the ditch. The maximum velocity can be found from the relation $$\frac{mv_{max}^2}{R}=\mu mg \Rightarrow v_{max}=\sqrt{\mu gR}.$$
Likewise, we can find the maximum radial acceleration $$a_{n}=\mu g.$$ Any change of velocity would cause a change in the radial component of acceleration because $$a_n=\frac{v^2}{R}$$. So the velocity must stay constant, and that implies $$a_t=\frac{dv}{dt}=0.$$ Thus we see, that the maximum acceleration is given by $$\vec{a}=\mu g \hat{n}.$$ That is definitely wrong, because $$\hat{n} \neq \frac{\vec{v_2}-\vec{v_1}}{|\vec{v_2}-\vec{v_2}|}$$, and the tangential velocity must change too, because $$v_1 \neq v_2.$$

I'm really frustrated, not only because I can't solve this problem on my own, but I don't even understand the answer. Thanks for help in advance! $$//$$