Edexcel M2 2022 October — Question 4

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2022
SessionOctober
TopicNon-constant acceleration

4. At time \(t\) seconds \(( 0 \leqslant t < 5 )\), a particle \(P\) has velocity \(\mathbf { v m s } ^ { - 1 }\), where $$\mathbf { v } = ( \sqrt { 5 - t } ) \mathbf { i } + \left( t ^ { 2 } + 2 t - 3 \right) \mathbf { j }$$ When \(t = \lambda\), particle \(P\) is moving in a direction parallel to the vector \(\mathbf { i }\).
  1. Find the acceleration of \(P\) when \(t = \lambda\) The position vector of \(P\) is measured relative to the fixed point \(O\) When \(t = 1\), the position vector of \(P\) is \(( - 2 \mathbf { i } + \mathbf { j } ) \mathrm { m }\). Given that \(1 \leqslant T < 5\)
  2. find, in terms of \(T\), the position vector of \(P\) when \(t = T\)