5 A particle \(P\) of mass \(m \mathrm {~kg}\) is projected vertically upwards through a liquid. Student \(A\) measures \(P\) 's initial speed as \(( 8.5 \pm 0.25 ) \mathrm { m } \mathrm { s } ^ { - 1 }\) and they also record the time for \(P\) to attain its greatest height above the initial point of projection as over 3 seconds.
In an attempt to model the motion of \(P\) student \(B\) determines that \(t\) seconds after projection the only forces acting on \(P\) are its weight and the resistance from the liquid. Student \(B\) models the resistance from the liquid to be of magnitude \(m v ^ { 2 } - 6 m v\), where \(v\) is the speed of the particle.

1. (a) Show that \(\frac { \mathrm { d } t } { \mathrm {~d} v } = - \frac { 1 } { ( v - 3 ) ^ { 2 } + 0.8 }\).
   (b) Determine whether student \(B\) 's model is consistent with the time recorded by \(A\) for \(P\) to attain its greatest height. After attaining its greatest height \(P\) now falls through the liquid. Student \(C\) claims that the time taken for \(P\) to achieve a speed of \(1 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) when falling through the liquid is given by $$- \int _ { 0 } ^ { 1 } \frac { 1 } { ( v - 3 ) ^ { 2 } + 0.8 } \mathrm {~d} v$$
2. Explain why student \(C\) 's claim is incorrect and write down the integral which would give the correct time for \(P\) to achieve a speed of \(1 \mathrm {~ms} ^ { - 1 }\) when falling through the liquid.