Position of a Particle Moving with Constant Velocity

Vector kinematics with constant velocity uses one formula: position $=$ start $+$ time $\times$ velocity, $\mathbf{r} = \mathbf{r}_0 + t\mathbf{v}$. Apply $t$ to the whole velocity vector, then a magnitude gives the distance from the origin.

A particle starts at the point with position vector $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and moves with constant velocity $\begin{pmatrix} 3 \\ -1 \end{pmatrix}\ \text{m s}^{-1}$. (i) Find its position vector after $4$ seconds. [3] (ii) Find its distance from the origin at that time. [2]

The working

(i) Use $\mathbf{r} = \mathbf{r}_0 + t\mathbf{v}$ with $t = 4$: $\mathbf{r} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + 4\begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \begin{pmatrix} 12 \\ -4 \end{pmatrix} = \begin{pmatrix} 13 \\ -2 \end{pmatrix} \quad \text{(M1, M1, A1)}$

(ii) Distance from the origin is the magnitude of $\mathbf{r}$: $|\mathbf{r}| = \sqrt{13^2 + (-2)^2} = \sqrt{169 + 4} = \sqrt{173} \approx 13.2 \text{ m} \quad \text{(M1, A1)}$

Where the marks are won and lost

• Multiply $t$ into both components of the velocity: $4 \times \begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 12 \\ -4 \end{pmatrix}$. Scaling only the top component is the usual error.
• Add the scaled velocity to the start position, not to the origin.
• Distance from the origin is $|\mathbf{r}|$, the magnitude of the position vector at $t = 4$, not of the velocity.

Common mistakes

• Computing $t\mathbf{v}$ as $\begin{pmatrix} 12 \\ -1 \end{pmatrix}$ (forgot to scale the $\mathbf{j}$-component).
• Forgetting to add the starting position.
• Taking the magnitude of the velocity instead of the position for the distance.

Full method: Velocity & Relative Velocity notes. Topic home: Vectors pillar.