## Statement

Assuming that the following vector magnitudes refer to rectilinear motion, give their corresponding scalar representation:

- $\overrightarrow{r}=3\xb7\overrightarrow{i}m$
- $\u2206\overrightarrow{r}=-3\xb7\overrightarrow{j}m$
- $\overrightarrow{v}=4\xb7(-\overrightarrow{j})m/s$
- $\overrightarrow{v}=3\xb7t\xb7\overrightarrow{i}m/s$

## Solution

**Resolution**

$\overrightarrow{r}=3\xb7\overrightarrow{i}m$

It represents the position vector of the motion. Remember that a vector has magnitude and direction. In this case:

- Magnitude: 3
- Direction: same as the unit vector $\overrightarrow{i}$

Movements that take place in the direction of the unit vector $\overrightarrow{i}$ are those associated with the *x axis*, therefore:

$$\overline{)\overrightarrow{r}=3\xb7\overrightarrow{i}m\Rightarrow x=3m}$$

$\u2206\overrightarrow{r}=-3\xb7\overrightarrow{j}m$

It is the displacement vector of the motion:

- Magnitude: 3 (magnitude is never negative)
- Direction: That given by the unit vector $-\overrightarrow{j}$

Movements that take place in the direction of the unit vector $\overrightarrow{j}$ are those associated to the *y-axis*, therefore:

$$\overline{)\u2206\overrightarrow{r}=-3\xb7\overrightarrow{j}m\Rightarrow \u2206y=-3m}$$

$\overrightarrow{v}=4\xb7(-\overrightarrow{j})m/s$

It is normally written $\overrightarrow{v}=4\xb7(-\overrightarrow{j})=-4\xb7\overrightarrow{j}$ .

Which is the velocity vector of the motion. Following a similar reasoning to the previous one:

$$\overline{)\overrightarrow{v}=4\xb7(-\overrightarrow{j})m/s=-4\xb7\overrightarrow{j}m/s\Rightarrow {v}_{y}=-4m/s}$$

$\overrightarrow{v}=3\xb7t\xb7\overrightarrow{i}m/s$

In this case the velocity vector of the motion depends on time. We can write:

- Magnitude 3·
*t* - Direction: given by the unit vector $\overrightarrow{i}$

The vector is associated to *x-axis*, therefore:

$$\overline{)\overrightarrow{v}=3\xb7t\xb7\overrightarrow{i}m/s\Rightarrow {v}_{x}=3\xb7tm/s}$$