Riding round and round

The maximum friction force is $F_o$ . The angle between the friction force and the velocity of the motorbike is $\alpha$ . The angle of the arc the motorbike has traveled on the track is $\theta$ .

In the tangent direction:

$m \frac{dv}{dt} = F_o \cos{\alpha}$

In the radial direction:

$m \frac{v^2}{R} = F_o \sin{\alpha} \quad \Rightarrow \quad \frac{2mv}{R}\frac{dv}{dt} = F_o \cos{\alpha} \frac{d\alpha}{dt}$

Divide the second equation by the first equation and we get:

$\frac{2v}{R} = \frac{d\alpha}{dt}$

We can now substitute $v/R=d\theta/dt$ to get $d\theta=d\alpha/2$ .

As the angle $\alpha$ changes from $0$ to $\pi/2$ , the arc the motorbike will travel is $s=\theta R = \pi R/4 = 7.85~\mbox{m}$ .

v~k, v=k s/r,where s is a constant equal to 2 pi*r,

minimizing the value and taking the definite integral we get k=(1/8), since distance traveled is equal to k s,we get (1/8) 62.8=7.85