A stone is held on the surface of a pond and released. The stone falls vertically through the water and the distance, \(x\) metres, that the stone has fallen in time \(t\) seconds is measured. It is given that \(x=0\) and \(\dfrac{\mathrm dx}{\mathrm dt}=0\) when \(t=0\).
The motion of the stone is modelled by the differential equation \[\frac{\mathrm d^2x}{\mathrm dt^2}+2\frac{\mathrm dx}{\mathrm dt}=10.\]
By substituting \(y=\dfrac{\mathrm dx}{\mathrm dt}\), show that the differential equation can be written as \(\dfrac{\mathrm dy}{\mathrm dt}=10-2y\).[1]
Find \(y\) in terms of \(t\) and hence find \(x\) in terms of \(t\).[6]
A second model for the motion of the stone is suggested, given by the differential equation \[\frac{\mathrm d^2x}{\mathrm dt^2}=10-5\sin\frac12t.\] Find \(x\) in terms of \(t\) for this model.[3]
The pond is \(5\) metres deep. For each of these models, find the time the stone takes to reach the bottom of the pond, giving your answers correct to \(2\) decimal places.[2]