A hare is 160 m away from a straight railway line at point $A$. The perpendicular projection of point $A$ onto the railway line is $T$. A train is approaching point $T$ at a speed of $30 \mathrm{~m} / \mathrm{s}$, and the initial distance of the front of the train from point $T$ is $300 \mathrm{~m}$. The hare can run at a speed of $15 \mathrm{~m} / \mathrm{s}$. Can the hare cross the railway line in front of the train in some direction?
Solution. If the rabbit ran perpendicular to the tracks, it would cover the $160 \mathrm{~m}$ distance in $\frac{160}{15}=\frac{32}{3} \mathrm{sec}$, while the train would reach the point in $\frac{300}{30}=10 \mathrm{sec}$. Thus, the rabbit would not make it across the tracks before the train. Therefore, the rabbit must start in some diagonal direction. Let $E$ be the point where it reaches the tracks. Let the distance $T E$ be $x$. In the right triangle $A T E$, $A E=\sqrt{160^{2}+x^{2}}$, which is the distance the rabbit covers.

![](https://cdn.mathpix.com/cropped/2024_05_02_d3a0a9601a514e23e200g-1.jpg?height=263&width=593&top_left_y=405&top_left_x=755)

The train is $300 \mathrm{~m}$ away from point $T$ at point $B$. The train's distance to point $E$ is $300+x$ meters. The rabbit will make it across the tracks before the train if it covers its distance faster than the train, that is, if

$$
\frac{300+x}{30}>\frac{\sqrt{160^{2}+x^{2}}}{15}
$$

Let's solve the inequality:

$$
300^{2}+600 x+x^{2}>2^{2} \cdot 160^{2}+2^{2} x^{2}
$$

Rearranging:

$$
3 x^{2}-600 x+12400<0
$$

From this, $23.41<x<176.59$.

The rabbit can make it across the tracks before the train if it reaches the tracks at least 23.41 meters and at most 176.59 meters from point $T$ $\left(8.32^{\circ}<\angle T A E<47.82^{\circ}\right)$.