= Solution
{c}
Here $f=0$ and $ct=1$, so <D'Alembert's formula> gives
$$
u\left(\frac32,\frac1c\right)
=\frac1{2c}\int_{1/2}^{5/2}g(s)\,ds.
$$
Only the overlap $[1/2,1]$ with the <support> of the initial velocity contributes. Therefore
$$
\begin{aligned}
u\left(\frac32,\frac1c\right)
&=\frac1{2c}\int_{1/2}^{1}(1-s^2)\,ds\\
&=\frac1{2c}\left[s-\frac{s^3}{3}\right]_{1/2}^{1}
=\frac1{2c}\left(\frac23-\frac{11}{24}\right)
=\boxed{\frac5{48c}}.
\end{aligned}
$$
The relevant interval is determined by the two <characteristic lines for the one-dimensional wave equation> through the observation point.
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