Determining particle speed from two position vectors

Key concept: This is a constant-velocity vector question. The speed comes from displacement divided by time, followed by the magnitude of the velocity vector.

Step by step

What the data tells you

The particle moves with constant velocity, so the velocity vector is the same throughout the motion. That means we can use the two position vectors to find the displacement over the time interval, then divide by the time elapsed.

The position at $t=4$ is

$\begin{pmatrix} 0.2 \\ 0.8 \end{pmatrix}$

and the position at $t=10$ is

$\begin{pmatrix} -1.6 \\ 3.2 \end{pmatrix}.$

Step-by-step calculation

First find the displacement:

$\Delta \mathbf{r} = \begin{pmatrix} -1.6 \\ 3.2 \end{pmatrix} - \begin{pmatrix} 0.2 \\ 0.8 \end{pmatrix} = \begin{pmatrix} -1.8 \\ 2.4 \end{pmatrix}.$

The time interval is

$10 - 4 = 6\text{ s}.$

So the velocity vector is

$\mathbf{v} = \frac{1}{6}\begin{pmatrix} -1.8 \\ 2.4 \end{pmatrix} = \begin{pmatrix} -0.3 \\ 0.4 \end{pmatrix}\text{ m/s}.$

Now find the speed, which is the magnitude of the velocity:

$|\mathbf{v}| = \sqrt{(-0.3)^2 + (0.4)^2} = \sqrt{0.09 + 0.16} = \sqrt{0.25} = 0.5.$

Why this method is reliable

With constant velocity, the average velocity over any interval equals the actual velocity. That is why using the two positions is enough. You do not need calculus here; a simple displacement-over-time calculation gives the exact speed.

Final answer

$\boxed{0.5\text{ m/s}}$

Pitfall alert

A common mistake is to calculate the distance between the two position vectors and then divide by time without first checking whether the question asks for speed or velocity. In this constant-velocity setting, that approach happens to give the same scalar speed only if you use the straight-line displacement magnitude, not the path length. Another error is reversing the subtraction order for the position vectors, which flips the velocity direction. The speed would still be positive, but the intermediate vector would be wrong and could cause trouble in later parts of the question. Keep the time interval as $10-4$, not $4-10$.

Try different conditions

If the question instead asked for the velocity vector, the answer would be $\begin{pmatrix} -0.3 \\ 0.4 \end{pmatrix}$ m/s rather than just the speed. If the times were $t=2$ and $t=8$ but the same position vectors were given, the method would be identical, except the time interval would still be 6 s. For a non-constant velocity problem, you would need more information, such as an acceleration function, because two positions alone would not determine the instantaneous speed at a later time.

Further reading

displacement vector, constant velocity, vector magnitude