6 A particle \(P\) moves in a straight line starting from a point \(O\) and comes to rest 14 s later. At time \(t \mathrm {~s}\) after leaving \(O\), the velocity \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) of \(P\) is given by $$\begin{array} { l l } v = p t ^ { 2 } - q t & 0 \leqslant t \leqslant 6
v = 63 - 4.5 t & 6 \leqslant t \leqslant 14 \end{array}$$ where \(p\) and \(q\) are positive constants.
The acceleration of \(P\) is zero when \(t = 2\).

1. Given that there are no instantaneous changes in velocity, find \(p\) and \(q\).
2. Sketch the velocity-time graph.
3. Find the total distance travelled by \(P\) during the 14 s .
   \includegraphics[max width=\textwidth, alt={}, center]{e1b91e54-a3ae-436c-a4f7-7095891f7034-10_326_1109_255_520} Two particles \(A\) and \(B\) of masses 2 kg and 3 kg respectively are connected by a light inextensible string. Particle \(B\) is on a smooth fixed plane which is at an angle of \(18 ^ { \circ }\) to horizontal ground. The string passes over a fixed smooth pulley at the top of the plane. Particle \(A\) hangs vertically below the pulley and is 0.45 m above the ground (see diagram). The system is released from rest with the string taut. When \(A\) reaches the ground, the string breaks. Find the total distance travelled by \(B\) before coming to instantaneous rest. You may assume that \(B\) does not reach the pulley.
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