Power from force and derived speed (non-equilibrium)

2. A particle of mass 0.5 kg is moving under the action of a single force \(\mathbf { F N }\) so that its velocity \(\mathrm { v } \mathrm { ms } ^ { - 1 }\) at time \(t\) seconds is given by $$\mathbf { v } = 3 t ^ { 2 } \mathbf { i } - 8 t \mathbf { j } + 2 \mathrm { e } ^ { - t } \mathbf { k }$$

1. Find an expression for the acceleration of the particle at time \(t \mathrm {~s}\).
2. Determine an expression for F.v at time \(t \mathrm {~s}\).
3. Find the kinetic energy of the particle at time \(t \mathrm {~s}\).
4. Describe the relationship between the kinetic energy of a particle and the rate of working of the force acting on the particle. Verify this relationship using your answers to part (b) and part (c).

1 A particle, \(P\), of mass 2 kg moves in two dimensions. Its initial velocity is \(\binom { - 19.5 } { - 60 } \mathrm {~ms} ^ { - 1 }\).

1. Calculate the initial kinetic energy of \(P\). For \(t \geqslant 0 , P\) is acted upon only by a variable force \(\mathbf { F } = \binom { 4 t } { - 2 } \mathrm {~N}\), where \(t\) is the time in seconds.
2. Find

1 A particle \(Q\) of mass \(m \mathrm {~kg}\) is acted on by a single force so that it moves with constant acceleration \(\mathbf { a } = \binom { 1 } { 2 } \mathrm {~ms} ^ { - 2 }\). Initially \(Q\) is at the point \(O\) and is moving with velocity \(\mathbf { u } = \binom { 2 } { - 5 } \mathrm {~ms} ^ { - 1 }\). After \(Q\) has been moving for 5 seconds it reaches the point \(A\).

1. Use the equation \(\mathbf { v } . \mathbf { v } = \mathbf { u } . \mathbf { u } + 2 \mathbf { a } . \mathbf { x }\) to show that at \(A\) the kinetic energy of \(Q\) is 37 m J .
2. 1. Show that the power initially generated by the force is - 8 mW W.
   2. The power in part (b)(i) is negative. Explain what this means about the initial motion of \(Q\).
3. 1. Find the time at which the power generated by the force is instantaneously zero.
   2. Find the minimum kinetic energy of \(Q\) in terms of \(m\).

A car of mass 1200 kg is travelling along a straight horizontal road \(AB\). There is a constant resistance force of magnitude 500 N. When the car passes point \(A\), it has a speed of \(15 \text{ m s}^{-1}\) and an acceleration of \(0.8 \text{ m s}^{-2}\).

1. Find the power of the car's engine at the point \(A\). [3]

The car continues to work with this power as it travels from \(A\) to \(B\). The car takes 53 seconds to travel from \(A\) to \(B\) and the speed of the car at \(B\) is \(32 \text{ m s}^{-1}\).

1. Show that the distance \(AB\) is 1362.6 m. [3]

Relative to a fixed origin \(O\), the position vector \(\mathbf{r}\) m at time \(t\) s of a particle \(P\), of mass 0.4 kg, is given by $$\mathbf{r} = e^{2t}\mathbf{i} + \sin(2t)\mathbf{j} + \cos(2t)\mathbf{k}.$$

1. Show that the velocity vector \(\mathbf{v}\) and the position vector \(\mathbf{r}\) are never perpendicular to each other. [6]
2. Given that the speed of \(P\) at time \(t\) is \(v\) ms\(^{-1}\), show that $$v^2 = 4e^{4t} + 4.$$ [2]
3. Find the kinetic energy of \(P\) at time \(t\). [1]
4. Calculate the work done by the force acting on \(P\) in the interval \(0 < t < 1\). [2]
5. Determine an expression for the rate at which the force acting on \(P\) is working at time \(t\). [2]