1. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

A particle is moving along a straight line.
At time t seconds, \(\mathrm { t } > 0\), the velocity of the particle is \(\mathrm { Vms } ^ { - 1 }\), where $$v = 2 t - 7 \sqrt { t } + 6$$

1. Find the acceleration of the particle when \(t = 4\) When \(\mathrm { t } = 1\) the particle is at the point X .
   When \(\mathrm { t } = 2\) the particle is at the point Y . Given that the particle does not come to instantaneous rest in the interval \(1 < \mathrm { t } < 2\)
2. show that \(X Y = \frac { 1 } { 3 } ( 41 - 28 \sqrt { 2 } )\) metres.