A mass of \(\quantity{3}{kg}\) has position vector \(\mathbf{r} = (3t^2+2t)\mathbf{i}-4t^2\mathbf{j}\), where \(\mathbf{r}\) is measured in metres and \(t\) is the time in seconds.

Find its velocity and acceleration vectors in terms of \(t\).

Show that it is subject to a force of magnitude \(\quantity{30}{N}\), and find the direction of the force.

Newton’s second law tells us that the force, \(\mathbf{F}=m\,\mathbf{a}\). The acceleration and therefore the force are constant. So we have (in \(\mathrm{N}\)) \[\mathbf{F} = 3 \times (6\mathbf{i}-8\mathbf{j}) = 6(3\mathbf{i}-4\mathbf{j}).\]

So the magnitude of the force is \[6\sqrt{3^2+4^2} = \quantity{30}{N}\] as required.

From the diagram, \(\tan\theta=\frac{8}{6}\) so the direction of the force is \(53.1^\circ\) below the \(\mathbf{i}\)-direction.