3D vector motion problems

6 A rock of mass 8 kg is acted on by just the two forces \(- 80 \mathbf { k } \mathrm {~N}\) and \(( - \mathbf { i } + 16 \mathbf { j } + 72 \mathbf { k } ) \mathrm { N }\), where \(\mathbf { i }\) and \(\mathbf { j }\) are perpendicular unit vectors in a horizontal plane and \(\mathbf { k }\) is a unit vector vertically upward.

1. Show that the acceleration of the rock is \(\left( - \frac { 1 } { 8 } \mathbf { i } + 2 \mathbf { j } - \mathbf { k } \right) \mathrm { ms } ^ { - 2 }\). The rock passes through the origin of position vectors, O , with velocity \(( \mathbf { i } - 4 \mathbf { j } + 3 \mathbf { k } ) \mathrm { m } \mathrm { s } ^ { - 1 }\) and 4 seconds later passes through the point A .
2. Find the position vector of A .
3. Find the distance OA .
4. Find the angle that OA makes with the horizontal. Section B (36 marks)

2 A particle, Q , moves so that its velocity, \(\mathbf { v }\), at time \(t\) is given by \(\mathbf { v } = ( 6 t - 6 ) \mathbf { i } + \left( 3 - 2 t + t ^ { 2 } \right) \mathbf { j } + 4 \mathbf { k }\), where \(0 \leqslant t \leqslant 6\).

1. Explain how you know that Q is never stationary. When Q is at a point A the direction of the acceleration of Q is parallel to the \(\mathbf { i }\) direction. When Q is at a point B the direction of the acceleration of Q makes an angle of \(45 ^ { \circ }\) with the \(\mathbf { i }\) direction.
2. Determine the straight-line distance AB .

5. A particle of mass 2 kg is moving under the action of a force \(\mathbf { F N }\) which, at time \(t\) seconds, is given by $$\mathbf { F } = 4 t \mathbf { i } - \sqrt { t } \mathbf { j } + 6 \mathbf { k }$$ When \(t = 1\), the velocity of the particle is \(\left( 3 \mathbf { i } - \frac { 1 } { 3 } \mathbf { j } - \mathbf { k } \right) \mathrm { ms } ^ { - 1 }\).

1. Find an expression for the velocity vector of the particle at time \(t \mathrm {~s}\).
2. Determine the values of \(t\) when the particle is moving in a direction perpendicular to the vector \(( - \mathbf { i } + 3 \mathbf { k } )\).

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 \(\mathbf { F }\).