6. A particle \(P\) of mass \(m \mathrm {~kg}\) is attached to the end \(A\) of a light elastic string \(A B\), of natural length \(a\) metres and modulus of elasticity 9ma newtons. Initially the particle and the string lie at rest on a smooth horizontal plane with \(A B = a\) metres. At time \(t = 0\) the end \(B\) of the string is set in motion and moves at a constant speed \(U \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in the direction \(A B\). The air resistance acting on \(P\) has magnitude 6mv newtons, where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the speed of \(P\). At time \(t\) seconds, the extension of the string is \(x\) metres and the displacement of \(P\) from its initial position is \(y\) metres. Show that, while the string is taut,

1. \(x + y = U t\)
2. \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 6 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 9 x = 6 \mathrm { U }\) You are given that the general solution of the differential equation in (b) is $$x = ( A + B t ) U e ^ { - 3 t } + \frac { 2 U } { 3 }$$ where \(A\) and \(B\) are arbitrary constants.
3. Find the value of \(A\) and the value of \(B\).
4. Find the speed of \(P\) at time \(t\) seconds.