This question is intereting. We are asked to consider a motorcycle travelling on flat ground. The coefficient of friction between the motorcycle and the ground is
$\mu$, and the motorcycle begin moving from rest. If the motorcycle wihes to travel in a circle of radius $R$, we must determine the least distance that the motorcycle
must travel to accelerate to its maximum allowable speed.
In order to achieve maximum acceleration, the motorcycle must use the entirety of the friction avaliable to it,
in order travel forwards. Thus, we know that the force of friction will be given by $F_f \ = \ mg \mu$. We let the radial componenet of friction be $F_r$, and the tangent componenet be $F_t$.
We thus have:
$$F_f \ = \ \sqrt{F_r^2 \ + \ F_t^2} \ = \ mg \mu$$
At some given instance in time, the radial forces acting on the motorcycle must
be balanced, or else the motorcycle will skid outwards. Thus, it mut be true that at any given time $t$:
$$F_r(t) \ = \ \frac{mv(t)^2}{R}$$
It is also true that we will have:
$$ma \ = \ F_t(t) \ = \ \sqrt{m^2 g^2 \mu^2 \ - \ F_r(t)^2}$$
Thus, we can relate the velocity and the acceleration of the motorcycle:
$$ma \ = \ \sqrt{m^2 g^2 \mu^2 \ - \ \frac{m^2 v(t)^4}{R^2}}$$
Our goal is to now find the distance travelled by our motorcycle as a function of the velocity. This is due to the fact that we can easily find the final velocity of our motorcycle:
at maximum $v$, the motorcycle will not be accelerating tangentially anymore, and all the friction will be acting radially, the prevent the motorcycle from skidding outwards. We thus
have:
$$mg \mu \ = \ \frac{mv_f^2}{R} \ \Rightarrow \ v_f \ = \ \sqrt{R g \mu}$$
Going back to the velocity-acceleration equation, we know that $a \ = \ v dv/dx$, so we have:
$$v \frac{dv}{dx} \ = \ \sqrt{m^2 g^2 \mu^2 \ - \ \frac{m^2 v(t)^4}{R^2}} \ \Rightarrow \ \frac{dv}{dx} \ = \ \frac{\sqrt{ g^2 \mu^2 \ - \ \frac{ v(t)^4}{R^2}}}{v}$$
$$\Rightarrow \ \displaystyle\int \ dx \ = \ x(v) \ = \ \displaystyle\int \ \frac{v}{\sqrt{ g^2 \mu^2 \ - \ \frac{v^4}{R^2}}} dv \ = \ R \displaystyle\int \frac{v}{\sqrt{ R^2 g^2 \mu^2 \ - \ v^4}} dv$$
So we will have:
$$x(v_f) \ = \ R \displaystyle\int_{0}^{\sqrt{R g \mu}} \ \frac{v}{\sqrt{ R^2 g^2 \mu^2 \ - \ v^4}} dv \ = \ R \displaystyle\int_{0}^{\alpha} \ \frac{v}{\sqrt{ \alpha^4 \ - \ v^4}} dv$$
Where $\alpha \ = \ R^2 g^2 \mu^2$. Now, we perform a $u$-substitution. We let $\beta \ = \ v^2$. Then $d\beta \ = \ 2v dv$, so $dv \ = \ \frac{d \beta}{2}$. We thus rewrite our integral:
$$x(v_f) \ = \ \frac{R}{2} \displaystyle\int_{0}^{\alpha^2} \ \frac{1}{\sqrt{a^4 \ - \ \beta^2}} \ d\beta \ = \
\frac{R}{2} \displaystyle\int_{0}^{\alpha^2} \ \frac{1}{\alpha^2} \frac{1}{\sqrt{1 \ - \ \Big(\frac{\beta}{\alpha^2}\Big)^2}} \ d\beta \ = \ \frac{R}{2} \sin^{-1}\Big(\frac{\beta}{\alpha^2}\Big) \biggr\rvert_{0}^{a^2} \ = \ \frac{\pi R}{4}$$
Small confession: I solved this integral numerically the first time I did the problem, but whenI was typing this solution, I felt I owed it to the (likely non-existent) readers to solve it
exactly.
Now, let's move onto the second method that can be used to solve this question. Morin dubs this the "slick method", and boy is he right. We set $\beta(t)$ to be the angle that the force of
friction makes with the radial direction. We can set up the radial and tangential equations:
$$ma \ = \ m g \mu \sin \beta(t)$$
$$\frac{mv^2}{R} \ = \ mg \mu \cos \beta(t)$$
We take the derivative of the radial equation with respect to $x$, getting:
$$\frac{2v}{R} \frac{dv}{dx} \ = \ -g \mu \sin \beta \frac{d \beta}{dx}$$
We know that $v \frac{dv}{dx} \ = \ a$, thus we get:
$$\frac{2}{R} a \ = \ \frac{2}{R} g \mu \sin \beta \ = \ -g \mu \sin \beta \frac{d \beta}{dx} \ \Rightarrow \ -\frac{2}{R} \ = \ \frac{d \beta}{dx}$$
Now, when the motorcycle begins moving, it's velocity will be $0$, thus the centripetal force will be $0$, and no friction is needed to balance it. This means that $\beta(0) \ = \ \frac{\pi}{2}$.
In addition, as we discussed earlier, at maximum velocity, the motorcycle will not be accelerating and all the friction will be used to balance the centripetal force, thus $\beta(t_f) \ = \ 0$.
We thus have:
$$-\frac{2}{R} \displaystyle\int \ dx \ = \ -\frac{2x}{R} \ = \ \displaystyle\int_{\frac{\pi}{2}}^{0} \ d\beta \ = \ -\frac{\pi}{2} \ \Rightarrow \ x \ = \ \frac{\pi R}{4}$$
The two solutions agree with each other, thus I am fairly certain that this is in fact the correct answer.