The diagram shows a V-shaped tank with dimensions $L = 4\text{ m}$, $W = 0.6\text{ m}$ and $H = 0.7\text{ m}$. The tank is initially empty. Water is pumped into the tank at a rate of $0.0025\text{ m}^3/\text{s}$. At any instant from the start of water flowing into the tank, the water in the tank has a depth of $y\text{ m}$ and a surface width of $x\text{ m}$.

1. Use similar triangles to express the surface width $x$ in terms of the depth $y$.

By similar triangles on the V-shaped cross-section:
$\frac{x}{y} = \frac{W}{H} = \frac{0.6}{0.7} = \frac{6}{7}$
$x = \frac{6}{7}y$

2. Express the volume $V$ of the water in terms of $y$.

The volume of the water is the cross-sectional triangular area multiplied by the length $L$:
$V = \frac{1}{2} x y L$
$V = \frac{1}{2} \left(\frac{6}{7}y\right) (y) (4)$
$V = \frac{12}{7}y^2$

3. Differentiate $V$ with respect to $y$.

$\frac{\mathrm{d}V}{\mathrm{d}y} = \frac{24}{7}y$

4. Apply the chain rule to find the rate of change of depth $\frac{\mathrm{d}y}{\mathrm{d}t}$ when $y = 0.4$.

Using the chain rule, $\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}y} \times \frac{\mathrm{d}y}{\mathrm{d}t}$:
$0.0025 = \left(\frac{24}{7}y\right) \frac{\mathrm{d}y}{\mathrm{d}t}$
Substitute $y = 0.4$:
$0.0025 = \left(\frac{24}{7}(0.4)\right) \frac{\mathrm{d}y}{\mathrm{d}t}$
$0.0025 = \frac{9.6}{7} \frac{\mathrm{d}y}{\mathrm{d}t}$
$\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{0.0025 \times 7}{9.6} = \frac{7}{3840} \approx 0.0018229\dots$
$\frac{\mathrm{d}y}{\mathrm{d}t} = 0.0018\text{ m/s (to 4 d.p.)}$

1. Calculate the maximum volume of the tank.

The tank is a triangular prism. Its total volume $V_{\text{full}}$ is the area of the V-shaped cross-section multiplied by its length $L$:
$V_{\text{full}} = \frac{1}{2} \times W \times H \times L = \frac{1}{2} \times 0.6 \times 0.7 \times 4 = 0.84 \text{ m}^3$

2. Express the volume of water $V$ in terms of $y$, and calculate the volume when $y = 0.4$.

Using similar triangles on the cross-section, the surface width $x$ is proportional to the depth $y$:
$\frac{x}{0.6} = \frac{y}{0.7} \implies x = \frac{6}{7}y$
The volume of water at depth $y$ is:
$V = \frac{1}{2}xyL = 2\left(\frac{6}{7}y\right)y = \frac{12}{7}y^2$
When $y = 0.4$:
$V_{0.4} = \frac{12}{7}(0.4)^2 = \frac{1.92}{7} \text{ m}^3$

3. Determine the remaining volume and the time taken to fill it.

Remaining volume to be filled:
$V_{\text{rem}} = V_{\text{full}} - V_{0.4} = 0.84 - \frac{1.92}{7} = \frac{5.88 - 1.92}{7} = \frac{3.96}{7} \text{ m}^3$
Since water is pumped in at a constant rate of $0.0025 \text{ m}^3/\text{s}$, the time taken is:
$\text{Time} = \frac{V_{\text{rem}}}{0.0025} = \frac{3.96/7}{0.0025} \approx 226.29 \text{ s}$
Rounding to the nearest second gives $226$ seconds.