A model boat with a mass of $m=0.5 \mathrm{~kg}$ is launched with an initial speed of $v=10 \mathrm{~m} / \mathrm{s}$. The water's drag force is proportional to the speed of the model boat ( $F=-k v, k=0.5 \mathrm{~kg} / \mathrm{s}$ ). Calculate the distance traveled by the small boat

a) during the time it takes for its speed to halve,

b) until it comes to a complete stop! (The equation $\frac{d f(x)}{d x}=a f(x)$ (where $a$ is a constant) is only satisfied by functions of the form $f(x)=b \cdot e^{a x}$, where $b$ is a constant and $e$ is the base of the natural logarithm.)
The boat is only affected by a frictional force of magnitude $k v$, so the equation of motion is:

$$
m a=-k v
$$

Acceleration is the time derivative of velocity, so the velocity as a function of time satisfies the following equation:

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

Based on the guidance provided in the problem, we can write the general solution to this equation as:

$$
v(t)=b e^{-(k / m) t}
$$

Since initially (at $t=0$) the boat's velocity was $v_{0}$, in this case $b=v_{0}$, so

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

The distance $x(t)$ traveled in time $t$ can be obtained by integrating $v(t)$:

$$
x(t)=\int_{0}^{t} v\left(t^{\prime}\right) \mathrm{d} t=\left(v_{0} m / k\right)\left[1-e^{-(k / m) t}\right]
$$

Let $t_{1 / 2}$ denote the time at which the boat's velocity is halved, then

$$
v_{0} / 2=v_{0} e^{-(k / m) t_{1 / 2}}
$$

From this,

$$
x\left(t_{1 / 2}\right)=\left(v_{0} m / k\right)[1-(1 / 2)]=v_{0} m / 2 k
$$

Given the data, $x\left(t_{1 / 2}\right)=5 \mathrm{~m}$.

From the formula for $v(t)$, it is clear that as $t \rightarrow \infty$, the velocity approaches zero, but it remains positive for any finite time. The total distance traveled can therefore be obtained from $x(t)$ by calculating its limit as $t \rightarrow \infty$, thus

$$
x_{\text {total }}=v_{0} m / k
$$

which is twice the value of $x\left(t_{1 / 2}\right)$.

Thus, at any finite time $t$, the boat has a positive velocity, but for large values of $t$, the velocity is so small that the total distance traveled approaches $x_{\text {total }}$ over a very long time, but never reaches it.

Therefore, under the influence of such a force, the body would never come to a complete stop, but the limit of the distance traveled would be finite. This force well approximates the drag force in fluids, but reality deviates from this approximation in that the force expression $F=-k v$ includes another term whose effect results in the body coming to a stop in a finite time.

Lajos Polacsek (Jászberény, Lehel Vezér Gymnasium, 3rd year student)