Fast Track

For more than 300 years physicists and mathematicians (Bernoulli, Fermat, Newton and many others) have been discussing the ways to find the curve that minimizes the travel time of an object subject to a force field, see here . The most commonly discussed case is about the motion in uniform gravitational field. The solution is the so-called Brachistochrone curve, a cycloid (see, for example this link ):

$x=R(\theta-\sin\theta)$

$y=R(1-\cos\theta)$

if the gravitation pulls in the $y$ direction and the $x$ direction is perpendicular to that. Interestingly, this curve does not depend on the mass or on the acceleration of gravity. According to the result, the skier has to start in the direction of the steepest slope (y direction), and gently turn towards the target initially.

The target coordinates determine $R$ and another parameter $\theta_0$ :

$x_0=R(\theta-\sin\theta_0)$

$y_0=R(1-\cos\theta_0)$

By dividing the two equations we get $\frac{x_0}{y_0}(1-\cos \theta_0)=\theta_0- \sin\theta_0$ . We can solve this graphically to get $\theta_0=5.12 rad$ . Using this result, we have $R=\frac{y_0}{1-\cos \theta_0} = 16.57m$ . The time of the travel is (again, see this link )

$T=\sqrt{\frac{R}{g'}} \theta_0$

where $g'$ is the acceleration of the body if it is left to move freely. In our case $g'=g \sin \alpha = 4.23m/s^2$ and the time is

$T=10.13s$ .

Notice that the skier has to travel well below the target (a distance $2R= 33m$ downhill relative to the starting point), before turning uphill and arriving to the target from below. At the lowest point she will travel with a velocity of $v_{max}= \sqrt{4gR\sin\alpha}=16.7m/s$ , about twice the speed of a 100m sprint runner. Of course, this is true only as long as friction is negligible. In reality the skier should probably take a shorter path, to minimize losses due to friction and avoid a possible embarrassment if the skis stop a few maters below the target.

For the motion along the straight line the acceleration is $g''=g' \frac{y_0}{\sqrt{x_0^2+y_0^2}}$ and the distance to travel is $s=\sqrt{x_0^2+y_0^2}$ . The time is

$T'= \sqrt{\frac{2(x_0^2+y_0^2)}{g y_0 \sin \alpha}}=21.86s$

more than 2 times longer.