Calculating the viscosity of water using particles There are two particles with radius 0.212 μm. The particles are in a thin liquid film and the microscope can only detect movement in the $x$- and $y$-axis. The temperature is 25 °C.
$$
\begin{array}{c|rrr}
\hline
t/s &amp; 30 &amp; 60 &amp; 90 &amp; 120 \\
\hline
x_1 &amp; 4.4 &amp; 10.7 &amp; 11.0 &amp; 12.4\\
y_1 &amp; -3.1 &amp; -5.2 &amp; -10.8 &amp; -9.3\\
x_2 &amp; 3.9 &amp; 2.0 &amp; 8.3 &amp; 5.8 \\
y_2 &amp; 3.5 &amp; 6.4 &amp; 14.1 &amp; 12.0 \\
\hline
\end{array}
$$
Use this information to estimate the viscosity of water.

I know that I should use the equations:
$$x^2 = 2Dt \tag{1}$$
$$D = \frac{kT}{6πηa} \tag{2}$$
Where if I plot $x^2$ against $t,$ I can calculate the viscosity from the slope $\displaystyle\frac{2kT}{6πηa}.$ However, what I don't understand is how to calculate $x^2.$
I tried taking the average of the two particles in direction $x$ and squaring the result, but that doesn't give me the right answer. I also tried taking the average of $x^2$ of the two particles, but once again I get the wrong answer.
How should I calculate $x^2?$
The basic idea here is that the spherical particles, each of radius a, are being bounced around via Brownian motion. So the two relevant equations are
$$r(t)^2 = 2Dt \tag{1}$$
$$D = \frac{kT}{6πηa} \tag{2}$$
Note that the first equation uses r(t) rather than x, to avoid needless confusion with the x and y cartesian coordinates of the spherical particles.
Now consider particle 1. At t = 0 s, it is at the defined location 0,0, where all distances are in $\mu m$. At four subsequent times, the x and y coordinates are as given in the OP's table. The following figure shows particle 1's location at t = 0s and at t = 30 s.

At any given time,
$$r_1(t)^2 = x_1(t)^2 + y_1(t)^2 \tag{3}$$
To estimate D, an ordinary least squares fit is performed on the tabulated data for particle 1. The result is shown in the next figure:

Note that the origin is used: at t = 0 s, then, by definition, $x_1(0) = 0 \space µm$ and $y_1(0) = 0 \space µm$. So $D = 1.1486 \times 10^{-12} m^2 s^{-1}$. Rounding off to two significant figures will be performed at the end.
Solving equation (2) for $\eta $ yields:
$$\eta = \frac{kT}{6πDa} \tag{4}$$
where T = 298.15 K, $a = 2.12 \times 10^{-7} m$ and $k = 1.38064852 \times 10^{-23} J K^{-1} $. The result for $\eta $ is
$$\eta = 8.97 \times 10^{-4} Nt \space m^{-2} \space s \tag{5}$$
where the fact that $1 J = 1 Nt \times 1 m$ has been used. N.B. 1 Joule equals 1 Newton times 1 meter.
Other units for $\eta $: Since 1 Pascal of pressure is 1 Newton of force per square meter, i.e., $1 Pa = 1 Nt / m^2$, and since 1 centipoise (symbol: cP) is 1 milliPascal times 1 s, i.e., $1 cP = 1 mPa \space s$, then
$$\eta = 8.97 \times 10^{-4} Pa \space s = 0.897 \space mPa \space s = 0.897 cP\tag{6}$$
Finally, rounding to two significant figures, $\eta = $0.90 cP. The viscosity of water at 25 °C is reported to be 0.91 cP.