An open rectangular tank is fixed on a truck.

Step 1. The tilt (a). Each bit of water of mass $dm$ feels $g\,dm$ downwards and $a\,dm$ backwards. The free surface is at right angles to the combined pull, so it makes an angle $\theta$ with the horizontal where

It slopes up towards the rear of the truck.

Step 2. Depths (b). The amount of water is fixed, so the surface turns about the middle of the tank. Over half the length, $2.0\ \text{m}$, it rises by $2.0\ \text{m} \times 0.1 = 0.2\ \text{m}$. So the depth is $1.4\ \text{m}$ at the rear and $1.0\ \text{m}$ at the front.

Step 3. Pressures (c). Vertically nothing has changed: the pressure grows by $\rho g$ per metre of depth below the surface above that point. At the rear, $1000 \times 10 \times 1.4\ \text{Pa} = 14.0\ \text{kPa}$; at the front, $10.0\ \text{kPa}$.

Step 4. Forces on the end walls (d). On a vertical wall wetted to a depth $h$, the pressure grows steadily from zero at the top to $\rho g h$ at the bottom, so the force is $\tfrac12\rho g h^2$ times the width:

Step 5. The limit (e). Water spills when the rear depth reaches the top of the wall, a rise of $0.3\ \text{m}$ over half the length: $\dfrac{a}{g} = \dfrac{0.3}{2.0}$, so $a = 1.50\ \text{m/s}^2$.