11.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{3f6f3f19-a1d0-488b-a1a4-302cc4cf5a1e-30_455_997_210_552}
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\caption{Figure 1}
\end{figure}
A tank in the shape of a cuboid is being filled with water.
The base of the tank measures 20 m by 10 m and the height of the tank is 5 m , as shown in Figure 1.
At time \(t\) minutes after water started flowing into the tank the height of the water was \(h \mathrm {~m}\) and the volume of water in the tank was \(V \mathrm {~m} ^ { 3 }\)
In a model of this situation
- the sides of the tank have negligible thickness
- the rate of change of \(V\) is inversely proportional to the square root of \(h\)
- Show that
$$\frac { \mathrm { d } h } { \mathrm {~d} t } = \frac { \lambda } { \sqrt { h } }$$
where \(\lambda\) is a constant.
Given that
- initially the height of the water in the tank was 1.44 m
- exactly 8 minutes after water started flowing into the tank the height of the water was 3.24 m
- use the model to find an equation linking \(h\) with \(t\), giving your answer in the form
$$h ^ { \frac { 3 } { 2 } } = A t + B$$
where \(A\) and \(B\) are constants to be found.
Hence find the time taken, from when water started flowing into the tank, for the tank to be completely full.