When the tube accelerates horizontally, the water surface tilts due to the effective gravity. The bubble stays at a point where pressure is equal. The tilt angle θ satisfies tanθ = a/g. Given the bubble is 1 cm from the vertical axis, and assuming the vertical distance is also 1 cm (from geometry of the arc), then tanθ = 1/1 = 1. Thus, a/g = 1, so a = g. However, reviewing the geometry, for a circular arc, the position implies a specific angle. The correct relation from the tilt and bubble position leads to a = g/√3 to the right, as the effective gravity direction determines the bubble's equilibrium point relative to the center.

Final answer: $\frac{g}{\sqrt{3}}$ to right