1. \hspace{0pt} [In this question \(\mathbf { i }\) and \(\mathbf { j }\) are horizontal unit vectors directed due east and due north respectively and position vectors are given relative to a fixed origin \(O\).]

At 12:00, a ship \(P\) sets sail from a harbour with position vector \(( 15 \mathbf { i } + 36 \mathbf { j } ) \mathrm { km }\). At 12:30, \(P\) is at the point with position vector \(( 20 \mathbf { i } + 34 \mathbf { j } ) \mathrm { km }\). Given that \(P\) moves with constant velocity,

1. show that the velocity of \(P\) is \(( 10 \mathbf { i } - 4 \mathbf { j } ) \mathrm { kmh } ^ { - 1 }\) At time \(t\) hours after 12:00, the position vector of \(P\) is \(\mathbf { p } \mathrm { km }\).
2. Find an expression for \(\mathbf { p }\) in terms of \(\mathbf { i } , \mathbf { j }\) and \(t\). A second ship \(Q\) is also travelling at a constant velocity.
   At time \(t\) hours after 12:00, the position vector of \(Q\) is given by \(\mathbf { q } \mathrm { km }\), where $$\mathbf { q } = ( 42 - 8 t ) \mathbf { i } + ( 9 + 14 t ) \mathbf { j }$$ Ships \(P\) and \(Q\) are modelled as particles.
   If both ships maintained their course,
3. 1. verify that they would collide at 13:30
   2. find the position vector of the point at which the collision would occur. At 12:30 \(Q\) changes speed and direction to avoid the collision.
      Ship \(Q\) now travels due north with a constant speed of \(15 \mathrm { kmh } ^ { - 1 }\)
      Ship \(P\) maintains the same constant velocity throughout.
4. Find the exact distance between \(P\) and \(Q\) at 14:30