= Solution
The upward jump at $-1/2$ creates a <rarefaction wave> with $u=(x+1/2)/t$, bounded by the characteristic lines $x=-1/2$ and $x=t-1/2$. The downward jump at zero joins the states $1,0$, so part (a) gives the shock position $s(t)=t/2$. The intervening plateau exists exactly while $t-1/2<t/2$, or $t<1$. The fan satisfies the <Inviscid Burgers equation> classically and matches the constant branches continuously; the shock satisfies the <Rankine-Hugoniot condition> and is causal because $1>1/2>0$. Its initial trace is the prescribed pulse.
At $t=1$, the fan reaches the shock at $x=1/2$. Subsequently the states immediately to its left and right are $(s(t)+1/2)/t$ and zero, so
$$
\dot s=\frac{s+1/2}{2t},\qquad s(1)=\frac12.
$$
Solving gives $s(t)+1/2=\sqrt t$. Thus the continuation is
$$
\boxed{u(x,t)=\begin{cases}(x+1/2)/t,&-1/2<x<\sqrt t-1/2,\\0,&\text{otherwise},\end{cases}\qquad t\geq1,\qquad s(t)=\sqrt t-\frac12}.
$$
Values assigned precisely at the shock do not affect the <weak solution>. Its speed $1/(2\sqrt t)$ lies strictly between the left state $1/\sqrt t$ and zero, so it remains a causal <entropy shock>. The area under the triangular profile is $\tfrac12(\sqrt t)(1/\sqrt t)=1/2$, checking conservation of the initial mass. This is the <rectangular pulse solution of the inviscid Burgers equation>.
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