Displacement

Have you ever walked literally anywhere? Then guess what, you're making use of the measurement we know as displacement. Displacement is used everywhere in the field of physics: if something is moving, you need to find its displacement to know everything else about it. It's a variable that we simply couldn't live without! But what is displacement, and how do we solve for it? Let's find out.

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Jetzt kostenlos anmeldenHave you ever walked literally anywhere? Then guess what, you're making use of the measurement we know as displacement. Displacement is used everywhere in the field of physics: if something is moving, you need to find its displacement to know everything else about it. It's a variable that we simply couldn't live without! But what is displacement, and how do we solve for it? Let's find out.

Suppose an object changes position: it goes from position \(A\) to position \(B\).

The object's **displacement** is the vector that points from position \(A\) to position \(B\): it is the difference between these positions.

If something started in an initial position, moved in any direction, for any length of time, and in a variety of different ways, and ended in a final position, a line could be drawn from the initial to the final position. If we make this line into an arrow pointing toward the final position, we would have a graphic representation of the displacement vector.

Displacement is a vector quantity. As a vector, displacement has both a magnitude and a direction. From the definition being a difference in positions, we see that displacement has units of meters.

Displacement, as we know, is a vector. This means we have both a magnitude and a direction. If we take away the displacement and keep only the magnitude, we would have the distance from one point to another instead, turning our vector displacement into the scalar distance.

The **distance** between positions \(A\) and position \(B\) is the magnitude of the displacement between these two positions.

As you may know, a direct line from a starting position to a final position is not the only way to measure a length. What if the person traveling between those points took a less direct journey? If you’re measuring the entire journey from point \(A\) to point \(B\), ignoring direction, you would be measuring the distance traveled instead. The distance is a scalar, which unlike a vector does not take into account direction, meaning it can’t be negative. For example, if someone traveled left for \(9\,\mathrm{ft}\), their displacement would be \(-9\,\mathrm{ft}\) if we choose left to be the negative direction. However, this person's distance to their starting point would be \(9\,\mathrm{ft}\), as the direction they traveled in does not matter at all to the distance. An easy way to understand it is that if you took your displacement and threw away the information on the direction, you would be left with only information about the distance.

As previously stated, displacement is the vector going from an initial position \(x_\text{i}\) to a final position \(x_\text{f}\). Therefore, the equation to calculate the displacement \(\Delta x\) looks like this:

\[\Delta\vec{x}=\vec{x}_\text{f}-\vec{x}_\text{i}.\]

It is important to know that when it comes to displacement, the value can be negative depending on the direction of the displacement. If we choose upwards to be positive, then the displacement of a skydiver between jumping and landing is negative. However, if we choose upwards to be negative, then their displacement is positive! Meanwhile, the distance between their jumping and landing will be positive in both cases.

Here are a few examples we can use to practice how displacement can be used to solve problems.

James moves \(26\,\mathrm{ft}\) east across a football stadium, before moving \(7\,\mathrm{ft}\) west. He then moves another \(6\,\mathrm{ft}\) west, before traveling back \(15\,\mathrm{ft}\) east. What is James’ displacement after he travels the described journey? What is the distance to his initial position?

First, we decide for ourselves to make east the positive direction. James moves \(26\,\mathrm{ft}\) east, so after this step, James' displacement is \(26\,\mathrm{ft}\) to the east. Next, he moves \(7\,\mathrm{ft}\) west, which is the same as \(-7\,\mathrm{ft}\) east. This means that we subtract \(7\) from \(26\), giving us a total displacement of \(19\,\mathrm{ft}\) to the east now. Next, James moves another \(6\,\mathrm{ft}\) west, giving us a displacement of \(19\,\mathrm{ft}-6\,\mathrm{ft}=13\,\mathrm{ft}\) to the east. Finally, James moves \(15\,\mathrm{ft}\) east, making the final total displacement \(28\,\mathrm{ft}\) to the east.

The distance between his final position and his initial position is \(28\,\mathrm{ft}\).

Sofia walks north up the street for \(50\,\mathrm{ft}\). She then travels \(20\,\mathrm{ft}\) west across the street, then another \(25\,\mathrm{ft}\) north. What will her two-dimensional displacement be when she has arrived at her destination?

Since this is a calculation of two-dimensional displacement, we choose the east and north directions to be positive. We consider Sofia to start out at a displacement of \((0,0)\,\mathrm{ft}\) east and north, respectively. First, she travels north for \(50\,\mathrm{ft}\), and since north-south displacement goes last in our coordinates, we call her displacement after this move \((0,50)\,\mathrm{ft}\). Next, \(20\,\mathrm{ft}\) west gives us a negative value on our east-west displacement, making the total displacement equal to \((-20,50)\,\mathrm{ft}\). Finally, she moves \(25\,\mathrm{ft}\) north. Adding that to our north-south displacement gives us our final displacement of \((-20,75)\,\mathrm{ft}\) in our coordinates. To answer the question, we translate our coordinates back to reality and conclude that Sofia's displacement is \(75\,\mathrm{ft}\) to the north and \(20\,\mathrm{ft}\) to the west.

The distance from her starting point to her destination can be calculated using the Pythagorean Theorem.

We've looked at displacement and we know that it is a vector, meaning that displacement has both a magnitude and a direction when we describe it. The vector that we call displacement can be given in one, two, or three dimensions. We've looked at displacement in two dimensions already, but what if we added a third? We live our lives in three-dimensional space, so it is important to know how displacement is used in three dimensions.

In three dimensions, a vector is shown in a matrix like so: \(\begin{pmatrix}i\\ j\\ k\end{pmatrix}\). Here, the \(i\) represents the displacement in the \(x\) direction, \(j\) represents the displacement in the \(y\) direction, and \(k\) represents the displacement in the \(z\) direction.

In terms of addition and subtraction in vectors, it's quite simple. All you need to do is take the \(i\), \(j\), and \(k\) values of one vector and add or subtract them from the corresponding values of the other vector. This is useful in displacement as the displacement between two positions is equal to the difference between the positions.

Suppose you climbed the highest point in the United States, Denali, and you want to know your displacement between the start of the climb (at coordinates \((62.966284,\,-151.156684)\,\text{deg}\) and elevation \(7500\,\mathrm{ft}\)) and the top (at coordinates \((63.069042,\,-151.006347)\,\text{deg}\) and elevation \(20310\,\mathrm{ft}\)). What you do is calculate the difference between these two vectors to get the displacement vector \(\Delta\vec{x}\):

\[\Delta\vec{x}=\begin{pmatrix}63.069042\,\mathrm{deg} - 62.966284\,\mathrm{deg} \\ -151.006347\,\mathrm{deg}+151.156684\,\mathrm{deg} \\ 20310\,\mathrm{ft}-7500\,\mathrm{ft}\end{pmatrix} =\begin{pmatrix}0.102758\,\mathrm{deg} \\ 0.150337\,\mathrm{deg} \\ 12810\,\mathrm{ft} \end{pmatrix}.\]

Of course, it is convenient to convert this to meters, and we get

\[\Delta\vec{x}=\begin{pmatrix} 11.5 \\ 7.6 \\ 3.9 \end{pmatrix}\,\mathrm{km}.\]

We now have the displacement as a vector, so we can take it apart and conclude that your displacement was \(11.5\,\mathrm{km}\) to the north, \(7.6\,\mathrm{km}\) to the east, and \(3.9\,\mathrm{km}\) up.

We can calculate the total distance \(d\) between your starting point and the top of Denali as follows:

\[d=\sqrt{\Delta x_1^2 +\Delta x_2^2 +\Delta x_3^2}=\sqrt{(11.5\,\mathrm{km})^2+(7.6\,\mathrm{km})^2+(3.9\,\mathrm{km})^2}=14.3\,\mathrm{km}.\]

Displacement is a vector describing the difference between a starting position and an ending position.

The formula for displacement is \(\Delta\vec{x}=\vec{x}_\text{f}-\vec{x}_\text{i}\).

Distance is the length, or magnitude, of the displacement vector.

Displacement and distance differ based on the fact that they are a vector and a scalar, respectively.

Distance cannot be negative.

The formula for displacement is the initial position subtracted from the final position.

Is displacement a vector or a scalar quantity?

Vector.

Briefly describe what displacement is.

Displacement is the difference between two locations in space, taking into account both the magnitude and the direction of the location difference/change.

What is the scalar equivalent of displacement?

Distance.

What is the difference between displacement and distance?

Displacement is a vector and distance is a scalar. Distance is the length of the displacement vector.

What is the equation used to calculate displacement?

\(\Delta x=x_f-x_i\).

If we know the displacement, we always know the distance.

True.

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