A $70 \mathrm{~kg}$ water skier is pulled by a motorboat at a constant speed of $40 \mathrm{~km} / \mathrm{h}$. The tow rope is then tensioned by a force of $400 \mathrm{~N}$. If the rope is released, how far will the water skier glide, assuming the minimum speed required to remain on the surface is $10 \mathrm{~km} / \mathrm{h}$? (Assume that the braking force is proportional to the square of the speed.)
I. solution. The drag force is proportional to the square of the velocity:

$$
F_{k}=-k v^{2}
$$

We know that at a speed of $v_{0}=40 \mathrm{~km} / \mathrm{h}$, the rope is stretched by a force of $F_{0}=-F_{k}\left(v_{0}\right)=400 \mathrm{~N}$. From this,

$$
k=F_{0} / v_{0}^{2}=3.24 \mathrm{~kg} / \mathrm{m}
$$

After the rope is released, only the drag force acts on the water skier; the equation of motion is:

$$
\begin{aligned}
& m a=F_{k}=-k v^{2} \\
& m \cdot(\mathrm{d} v / \mathrm{d} t)=k v^{2}
\end{aligned}
$$

The obtained relationship can also be written as:

$$
\frac{\mathrm{d}}{\mathrm{d} t}(-1 / v)=-k / m
$$

thus, integrating:

$$
-1 / v=-(k / m) t+C_{1}
$$

where $C_{1}$ is a constant. The value of the constant $C_{1}$ can be determined from the condition that at $t=0$ the speed is $v_{0}$:

$$
C_{1}=-1 / v_{0}
$$

Thus, the time dependence of the speed is:

$$
v(t)=\frac{m v_{0}}{k v_{0} t+m}
$$

The water skier slides on the water for $t_{1}$ time after the rope is released, until their speed is greater than $v_{\min}=10 \mathrm{~km} / \mathrm{h}$. Accordingly,

$$
v_{\min }=\frac{m v_{0}}{k v_{0} t_{1}+m}, \quad \text { hence } \quad t_{1} \frac{m}{k v_{0}} \cdot \frac{v_{0}-v_{\min }}{v_{\min }}=5.83 \mathrm{~s}
$$

Knowing the equation (6), the distance traveled in the time interval from 0 to $t_{1}$ is:

$$
s=\int_{0}^{t_{1}} v(t) \mathrm{d} t=\int_{0}^{t_{1}} \frac{m v_{0}}{k v_{0} t+m} \mathrm{~d} t=\left[\frac{m}{k} \ln \left(k v_{0} t+m\right)\right]_{0}^{t_{1}}=\frac{m}{k} \ln \frac{k v_{0} t+m}{m}
$$

Substituting the values of $t_{1}$ and $k$ into the required distance:

$$
s=\left(m v_{0}^{2} / F_{0}\right) \ln \left(v_{0} / v_{\min }\right)=30 \mathrm{~m}
$$

Zoltán Szalontai (Törökszentmiklós, Bercsényi M. Gymn., III. grade)

II. solution. The solution is essentially simpler if we examine the speed as a function of the distance traveled (after the rope is released). The work done by the drag force over a distance $\Delta s$ reduces the kinetic energy of the water skier:

$$
\begin{gathered}
-\left|F_{k}\right| \Delta s=\Delta E_{k i n^{1}} \\
-k v^{2} \Delta s=(1 / 2) m \Delta v^{2} \\
\Delta v^{2} / \Delta s=-(2 k / m) v^{2}
\end{gathered}
$$

From this, taking the limit as $\Delta s \rightarrow 0$, we get:

$$
d v^{2} / \mathrm{d} s=-(2 k / m) v^{2}
$$

The obtained equation can also be written as:

$$
\frac{\mathrm{d}}{\mathrm{d} s}\left(\ln v^{2}\right)=-2 k / m
$$

after integration, we get:

$$
v^{2}=v_{0}^{2} e^{-(2 k / m) s}
$$

where we have taken into account that at the moment the rope is released $(s=0)$, the speed was $v_{0}$.

Taking the square root:

$$
v=v_{0} \mathrm{e}^{-(k / m) s}
$$

The skier remains above the water as long as $v$ reaches $v_{\min }$:

$$
s=-(m / k) \ln \left(v_{\min } / v_{0}\right)=\left(m v_{0}^{2} / F_{0}\right) \ln \left(v_{0} / v_{\min }\right)=30 \mathrm{~m}
$$

Remark. The speed-distance relationship can also be obtained directly from the equation of motion:

$$
\begin{gathered}
m a=-k v^{2} \\
m(\mathrm{~d} v / \mathrm{d} t)=m(\mathrm{~d} v / \mathrm{d} s)(\mathrm{d} s / \mathrm{d} t)=m(\mathrm{~d} v / \mathrm{d} s) v=-k v^{2} \\
\text { thus } \mathrm{d} v / \mathrm{d} s=-(k / m) v, \quad \text { hence } \quad v=v_{0} \mathrm{e}^{-(k / m) s}
\end{gathered}
$$

Bettina Keszthelyi (Bp., Apáczai Csere J. Gymn., IV. grade)