AQA M2 2010 January — Question 4

Exam BoardAQA
ModuleM2 (Mechanics 2)
Year2010
SessionJanuary
TopicNon-constant acceleration

4 A particle moves so that at time \(t\) seconds its velocity \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\) is given by $$\mathbf { v } = \left( 4 t ^ { 3 } - 12 t + 3 \right) \mathbf { i } + 5 \mathbf { j } + 8 t \mathbf { k }$$
  1. When \(t = 0\), the position vector of the particle is \(( - 5 \mathbf { i } + 6 \mathbf { k } )\) metres. Find the position vector of the particle at time \(t\).
  2. Find the acceleration of the particle at time \(t\).
  3. Find the magnitude of the acceleration of the particle at time \(t\). Do not simplify your answer.
  4. Hence find the time at which the magnitude of the acceleration is a minimum.
  5. The particle is moving under the action of a single variable force \(\mathbf { F }\) newtons. The mass of the particle is 7 kg . Find the minimum magnitude of \(\mathbf { F }\).