Scalars and vectors in 1D

What is a physicist’s goal? For the most part, physicists are trying to do the following:

• precisely define the most fundamental, measurable quantities in the universe

• find relationships between those fundamental quantities

To accomplish these goals, physicists need two types of quantities: vectors and scalars. Scalars include magnitude only. Vectors include both magnitude and direction.

Let’s take a closer look at these two types of quantities.

##Scalars

Scalars are quantities with magnitude only. Some examples of scalars in physics are:

• mass
• distance
• speed
• volume
• temperature
• energy

All of these quantities include a magnitude (size), but no direction.

##Vectors

Vectors are quantities with magnitude and direction.

Because vectors include direction, they can be a little more complicated to work with than scalars. Let’s take a closer look, beginning with vector notation.

To indicate that a quantity is a vector, we use vector notation: we draw an arrow over the symbol for that quantity. Some common vector quantities and their symbols in vector notation are shown below.

[[☃ explanation 2]]

When all vector quantities in a given scenario are in the same dimension, their directions can be expressed with a "$+$" or "$-$" sign. In order for the signs to have physical meaning, though, there must be a defined coordinate system stating which direction is assigned as positive and which as negative. A commonly used coordinate system is shown below:
[[☃ image 21]]
In this coordinate system, left is designated as the negative direction and right is designated as the positive direction.

##Drawing vectors

A vector can be visually represented by drawing an arrow. The length of the vector represents the magnitude of the vector quantity. The direction the arrow points represents the direction of the vector quantity.

For example, if we want to represent a velocity of $5,\text{m/s}$ to the right, and another velocity of $10,\text{m/s}$ to the left, we may draw the following vectors:
[[☃ image 34]]
In this example, the grid shows that the $10,\text{m/s}$ vector is twice as long as the $5,\text{m/s}$ vector. But vectors are often drawn by hand, quickly, with no grid available. In that case, the vectors should be drawn reasonably to scale relative to each other, and then labeled with their magnitudes.

The point of the vector arrow is called the tip or head. The other end is called the tail.
[[☃ image 20]]
A vector may be moved anywhere in space and still represent the exact same quantity, as long as its direction and magnitude are not changed. For example, all four vectors below are equivalent.

[[☃ image 32]]

##Example

A one-dimensional coordinate system is shown below, in the horizontal $(x)$ dimension. The coordinate system shows that right is positive and left is negative. A velocity vector is shown above the coordinate system.
[[☃ image 26]]
The same vector represented visually above can be expressed in a variety of ways:

$\vec{v} = 10,\text{m/s}$, left

$\vec{v} = 10,\text{m/s}\leftarrow$

$v_{x} = -10,\text{m/s}$

##Adding vectors visually

Many situations in physics involve adding vectors. First, let's see how to do that visually.

Two position vectors, $\vec{x_{1}}$ and $\vec{x_{2}}$, are modeled below.
[[☃ image 24]]
To add them, we first take vector $\vec{x_{2}}$ and place its tail at the tip of vector $\vec{x_{1}}$.
[[☃ image 35]]
Then, we draw a new vector from the tail of the first vector that we are adding to the tip of the last vector. This new vector is the sum, or resultant, of $\vec{x_{1}}$ and $\vec{x_{2}}$.
[[☃ image 37]]
The order in which you add the vectors doesn’t matter. You get the same resultant either way:
[[☃ image 39]]
This method of adding vectors “tip-to-tail” works for any number of vectors in one, two, or three dimensions!

##Subtracting vectors visually

Subtracting a vector is equivalent to adding the negative of that vector. The negative of a vector is a vector of the same magnitude, pointing in the opposite direction.

Let's try an example by returning to our position vectors, $\vec{x_{1}}$ and $\vec{x_{2}}$:
[[☃ image 28]]
How can we subtract $\vec{x_{2}}$ from $\vec{x_{1}}$? We need to add $-\vec{x_{2}}$ to $\vec{x_{1}}$. So, the first step is to draw$-\vec{x_{2}}$.
[[☃ image 33]]
Then we place the vectors $\vec{x_{1}}$ and $-\vec{x_{2}}$ tip-to-tail, just like before.
[[☃ image 35]]
Finally, we draw a new vector from the tail of $\vec{x_{1}}$ to the tip of $-\vec{x_{2}}$.
[[☃ image 36]]

##Adding (or subtracting) vectors numerically

These visual representations above help us see what the sum of two or more vectors will look like relative to the originals. But sometimes, we need to know the magnitude of the resultant vector. We can do this numerically, without drawings.

Let's say we have two vectors, $\vec{x_{1}}$ and $\vec{x_{2}}.$ $\vec{x_{1}}$ is $5,\text{m}$ to the right, and $\vec{x_{2}}$ is $8,\text{m}$ to the left. How can we find the resultant, $\vec{R},$ of these two vectors?

First, we need to define a coordinate system. Let’s use the same one as before, in which left is the negative direction and right is the positive direction.
[[☃ image 33]]
Now, we can assign a positive or negative sign to each vector.

• $\vec{x_{1}}$ points right, so $\vec{x_{1}}=5,\text{m}$

• $\vec{x_{2}}$ points left, so $\vec{x_{2}}=-8,\text{m}$

So:

$\vec{R}=\vec{x_{1}}+\vec{x_{2}}=5,\text{m}+(-8,\text{m})=\boxed{-3,\text{m}}$

##Try it!

Two vectors, $\vec{A}$ and $\vec{B},$ have the following values:

$\vec{A}=0.5,\text{m}$, west

$\vec{B}=0.3,\text{m}$, east

What is the resultant (sum) of these two vectors?
Use a coordinate system in which west is negative and east is positive.

[[☃ graded-group 1]]

[[☃ explanation 1]]

##Problem-solving tips

• Be careful to not add vectors that represent different types of quantities. This seems obvious when talking about scalar quantities like temperature and volume. For example, you can't add $25\degree\text{C}$ to $2,\text{L}.$ Similarly, you can't add a displacement vector of $25,\text{m}$ to a velocity vector of $-2,\text{m/s}.$ The result would be meaningless. As always in science, paying careful attention to units can help avoid mistakes.

• Remember to clearly define your coordinate system before doing any calculations with vectors. It's easy to forget that you're working with vectors, especially since positive vector values "look like" scalars with no sign. But direction and sign are very important with vectors, even if the direction is an implied "$+$."

An object's position is its location relative to a chosen reference frame. Position is a vector with both magnitude and direction. The variable $\vec{x}$ is often used to represent horizontal position. The variable $\vec{y}$ is often used to represent vertical position.

If an object moves relative to its reference frame, then the object’s position changes. This change in position is known as displacement.

##Calculating displacement

Displacement is defined mathematically as:

$\Delta \vec{x}=\vec{x}-\vec{x}_0$

where $\Delta \vec{x}$ represents the displacement, $\vec{x}$ represents the final (or current) position, and $\vec{x}_0$ represents the initial position. This equation says that displacement is the change in an object's position, or equivalently, the difference between the object's final position and initial position.

[[☃ explanation 1]]

Since position is a vector, displacement is also a vector with both magnitude and direction. When working in one dimension, the sign $(+/-)$ of position and displacement describe their direction. So, we can simplify our equation by removing the vector notation, while carefully remembering that these are vector quantities:

$\Delta x=x-x_0$

##Modeling displacement visually

Displacement can be modeled as an arrow that points from an object's initial position to its final position. For example, consider a professor who walks from one position to another as shown below.

[[☃ image 3]]

The professor’s initial position is $x_0=1.5\text{ m}.$ Her final position is $x=3.5\text{ m}.$ Thus, her displacement is:

$\Delta x=x−x_0=3.5 \text{ m}−1.5\text{ m}=2.0 \text{ m}$

Next, consider an airplane passenger who walks from his seat to the back of the plane.

[[☃ image 4]]

The passenger’s initial position in the airplane's reference frame is $x_0=6.0\text{ m}.$ His final position is $x=2.0\text{ m}.$ So, his displacement is:

$\Delta x=x−x_0=2.0 \text{ m}−6.0\text{ m}=-4.0 \text{ m}$

His displacement is negative because his motion is toward the rear of the plane, which is the negative direction in our chosen coordinate system.

##Displacement and distance travelled

When an object moves, there are two different ways we can describe "how far" it moved. One way is by calculating its displacement, which is its change in position (a vector). Another way is by calculating the distance travelled, which is the total path length (a scalar). At first, these two quantities may sound like the same thing. But they are NOT the same.

Let's return to our example of the professor walking. When she walks from an initial position of $x_0=1.5\text{ m}$ to a final position of $x=3.5\text{ m},$ her displacement is $\Delta x=2.0 \text{ m}$ (to the right, as implied by the positive value). Her distance travelled is also $2.0 \text{ m}$ (no direction included or implied). Displacement and distance travelled have the same magnitude in this case.

However, imagine the professor didn't stop when she reached $x=3.5\text{ m}.$ Instead, she turned around and walked back to a final position of $x=2.5\text{ m}.$ In this case, her displacement is:

$\Delta x=x−x_0=2.5\text{ m}−1.5\text{ m}=1.0 \text{ m}$ (to the right of where she started)

In terms of displacement, she travelled "less far" in this scenario because she ended closer to her initial position. But what about her distance travelled? She walked $2.0\text{ m}$ forward $(\text{from}\space x=1.5\text{ m}\space\text{to}\space x=3.5\text{ m})$ and then another $1.0\text{ m}$ back $(\text{from}\space x=3.5\text{ m}\space\text{to}\space x=2.5\text{ m}).$ So, the professor's distance travelled is:

$\text{distance travelled}=2.0\text{ m} + 1.0\text{ m}=3.0\text{ m}$

In terms of distance, she travelled "farther" this time, since she covered a longer path length. So, we must be careful about which value we're interested in and what we're calculating.

##Try it!

Four objects move according to the paths shown below. Assume the units of the horizontal scale are given in meters.

[[☃ image 1]]

Which objects had positive displacements?

[[☃ explanation 4]]

Which objects had a displacement that was equal in magnitude to distance travelled?

[[☃ explanation 8]]

What was the displacement of object D?

[[☃ explanation 7]]

##Common mistakes and misconceptions

• Swapping the initial and final positions in the displacement equation. It's easy to get mixed up and put the initial position first in the displacement equation to get $-\Delta x$ instead of $\Delta x$. When calculating displacement, make sure to start with the final position and then subtract the initial position.

• Thinking that distance and displacement are just different names for the same quantity. However, distance and displacement are different concepts. If an object changes direction in its journey, the total distance traveled will be greater than the displacement between those two points.

[[☃ image 1]]

• Confusion about choosing the “correct” coordinate system. All coordinate systems are equally valid, so we can choose any system. The choice will affect the numbers in the calculation, but as long as we use the coordinate system consistently, the physical meaning of the answer won't change.