= Solution
Sampling at $0.2$ Hz means $\Delta t=5$ s. The target times are therefore \b[$5$, $10$ and $15$ s], all within the supplied observation range. For bracketing observations, <linear interpolation> gives $d(t)=d_i+(d_{i+1}-d_i)(t-t_i)/(t_{i+1}-t_i)$. Thus
$$
\begin{aligned}
d(5)&=5.2+(7.8-5.2)\frac{5-1.2}{6.7-1.2}=6.99636,\\
d(10)&=7.8+(15.5-7.8)\frac{10-6.7}{11.6-6.7}=12.98571,\\
d(15)&=15.5+(8.4-15.5)\frac{15-11.6}{15.3-11.6}=8.97568.
\end{aligned}
$$
The estimated displacements are \b[$7.00$, $12.99$ and $8.98\,\mu\mathrm m$]. Interpolation assumes approximately linear behavior within each interval; it does not recover any unobserved high-frequency motion.
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