5. A passenger train always moves at a constant speed. When it passes a bridge that is $450 \mathrm{~m}$ long, it takes $33 \mathrm{~s}$ from the moment the front of the train enters the bridge until the rear of the train leaves the bridge; when it passes through a tunnel that is $760 \mathrm{~m}$ long, the entire train is inside the tunnel for $22 \mathrm{~s}$. A freight train of length $a \mathrm{~m}$ and speed $v \mathrm{~m} / \mathrm{s}$ is coming from the opposite direction. The two trains pass each other, from the moment the fronts of the trains meet until the rears of the trains separate, it takes $t \mathrm{~s}$.
(1) Write the function expression for $t$ in terms of $a$ and $v$;
(2) If the speed of the freight train is no less than $12 \mathrm{~m} / \mathrm{s}$ and less than $15 \mathrm{~m} / \mathrm{s}$, and its length is $324 \mathrm{~m}$, find the range of the time it takes for the two trains to pass each other.
(2004, Hebei Province Junior High School Mathematics Innovation and Knowledge Application Competition Final)
(Tip: (1) Let the speed of the passenger train be $x \mathrm{~m} / \mathrm{s}$, and the length of the passenger train be $y \mathrm{~m}$. According to the problem, we have $450+y=33 x, 760-y=22 x$. Solving these equations yields $x=22, y=276$. Therefore, $t=\frac{a+276}{v+22}(v>0, a>0)$. (2) When $a=324, 12 \leqslant v<15$, from (1) we get $t=\frac{600}{v+22}$. Also, $34 \leqslant$ $v+22<37$, so $\frac{600}{37}<\frac{600}{v+22} \leqslant \frac{300}{17}$.)
