The displacement (in meters) of a particle moving in a straight line is given by the equation of motion $s=\frac{8}{t^{2}}$, where $t$ is measured in seconds. Find the velocity (in $\mathrm{m} / \mathrm{s}$ ) of the particle at times $t=a, t=1, t=2$, and $t=3$. \[ \begin{array}{llr} t=a & v= & \mathrm{m} / \mathrm{s} \\ t=1 & v=\square & \mathrm{m} / \mathrm{s} \\ t=2 & v=\square & \mathrm{m} / \mathrm{s} \\ t=3 & v= & \mathrm{m} / \mathrm{s} \end{array} \]

Step 1 ：Given the displacement function: \(s = \frac{8}{t^2}\)

Step 2 ：Find the derivative with respect to time: \(v = \frac{ds}{dt} = -\frac{16}{t^3}\)

Step 3 ：At \(t = a\), the velocity is: \(v = -\frac{16}{a^3} \, \mathrm{m} / \mathrm{s}\)

Step 4 ：At \(t = 1\), the velocity is: \(v = -16 \, \mathrm{m} / \mathrm{s}\)

Step 5 ：At \(t = 2\), the velocity is: \(v = -2 \, \mathrm{m} / \mathrm{s}\)

Step 6 ：At \(t = 3\), the velocity is: \(v = -\frac{16}{27} \, \mathrm{m} / \mathrm{s}\)

Step 7 ：\(\boxed{\begin{array}{llr} t=a & v=-\frac{16}{a^3} & \mathrm{m} / \mathrm{s} \\ t=1 & v=-16 & \mathrm{m} / \mathrm{s} \\ t=2 & v=-2 & \mathrm{m} / \mathrm{s} \\ t=3 & v=-\frac{16}{27} & \mathrm{m} / \mathrm{s} \end{array}}\)