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 \(t\).

The velocity is the rate of change of position and the acceleration is the rate of change of velocity, so we find them by differentiating with respect to time. \[\begin{align*} \text{Velocity, }\mathbf{v} &= \dfrac{d\mathbf{r}}{dt} = (6t+2)\mathbf{i}-8t\,\mathbf{j} \\ \text{Acceleration, }\mathbf{a} &= \dfrac{d\mathbf{v}}{dt}=6\mathbf{i}-8\mathbf{j}. \end{align*}\]

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.

the direction of the force at angle theta below the horizontal axis

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