= Solution
Choose the barn entrance as $x=0$, its exit as $x=3$, and the positive $x$ direction as the javelin's motion. Put $t=0$ at the trailing end entering the barn. Write $s=ct$ and $s'=ct'$ in metres. In the javelin frame let the trailing end have $x'=0$, the leading end $x'=4$, and use the same origin event. The <Lorentz transformation> is
$$
\boxed{s'=\gamma(s-\beta x),\qquad x'=\gamma(x-\beta s),\quad\beta=12/13,\quad\gamma=13/5.}
$$
In the barn frame, the four <world lines> are
$$
x_{\rm entrance}=0,\qquad x_{\rm exit}=3,\qquad
x_{\rm trailing}=\frac{12}{13}s,\qquad x_{\rm leading}=\frac{12}{13}s+\frac{20}{13}.
$$
In the javelin frame they are
$$
x'_{\rm trailing}=0,\qquad x'_{\rm leading}=4,\qquad
x'_{\rm entrance}=-\frac{12}{13}s',\qquad x'_{\rm exit}=\frac{15}{13}-\frac{12}{13}s'.
$$
The two <spacetime diagrams> below use vertical time coordinates and horizontal spatial coordinates. Event labels refer to the same physical events in both panels; dashed lines show light-ray directions.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/ib/paper-4-javelin.png]
{title=Javelin and barn world lines in the barn and javelin rest frames, with entry event A and exit event B}
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