Answer
$3.7\times 10^{-37}m$
Work Step by Step
The uncertainty in the velocity is given in the problem. Calculate the lowest uncertainty (maximum precision) in the position by using equation 28–1.
$$\Delta x \geq \frac{\hbar}{\Delta p}=\frac{\hbar}{m \Delta v}$$
$$=\frac{1.055\times 10^{-34}J \cdot s}{(1300kg)(0.22m/s)}$$
$$=3.7\times 10^{-37}m $$