Review question

# Given the position of a mass, can we find its velocity and acceleration? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7678

## Solution

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.

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