Video: Displacement Vectors

In this video we learn what a displacement vector is and how to calculate its magnitude, direction, and components graphically and algebraically.


Video Transcript

In this video, we’re going to learn about displacement vectors. What they are, and how to calculate them both graphically and algebraically. To get started, imagine that you’re taking part in a neighborhood scavenger hunt. Where, given a list of clues, you move from one location to another until you get to the last clue and the last location where the game ends. Each contestant starts at the base of the tree in the southwest corner of town. And clue by clue, you navigate the course until finally at the end, you receive the very last clue and are able to find the last spot. Just as you’re completing the course, a friend of yours tells you that they’re starting out at the course beginning. If you wanted to give them a direction that they could go straight to the end point where you currently are, what would you tell them? It’s displacement vectors that will help us understand this idea more clearly.

The first thing we might want to understand with this topic is, what is a displacement vector? And to figure that out, let’s look at one on a graph. Let’s imagine that this 𝑥𝑦 pair of axes is put onto a table top. And then an ant sits on that tabletop in an initial given position. Based on this frame of reference, the starting position of the ant, which we can call 𝑝 sub one, is given by a vector from the origin to the ant’s location. Say that the ant then begins to move. And after travelling a path for some time ends up in a second position that we can call 𝑝 sub two. If we were to take these two position vectors and subtract the first from the second, then the result would be a vector that we can call 𝑑. This is the displacement of the ant. Graphically, 𝑑 is a vector that begins at the end of our first position vector and ends at the point or the tip of our last position vector.

From this, we understand that displacement vectors connect initial position vectors with final position vectors. It shows the change in position. In this example, we’ve found the displacement vector 𝑑 using a graphical approach. We’ve connected the tip of the initial position vector with the tip of the final position vector. It’s also possible to solve for 𝑑 using an algebraic approach, where 𝑝 sub two and 𝑝 sub one are given in their component form. To get some practice with that method of solving for displacement, let’s look at an example.

The F-35B is a short takeoff and vertical landing fighter jet. An F-35B takes off vertically upward to a height of 20.00 meters, while still facing horizontally. The fighter then follows a flight path that is angled at 30.00 degrees above a line parallel to the ground, 20.00 meters vertically above ground. The fighter flies for a distance of 20.00 kilometers along this trajectory. What is the fighter’s final displacement?

We can call the displacement of the fighter after these maneuvers 𝑑, where we understand that 𝑑 is a vector with both magnitude and direction. To begin on our solution, let’s draw out a sketch of this F-35B’s motion. We’re told that the jet fighter rises vertically a distance we’ve called ℎ of 20.00 meters. And then begins to move at an angle of 30.00 degrees to the horizontal along a path of length 20.00 kilometers. The fighter’s displacement 𝑑 is the vector that connects its initial location on the ground with its final location at the end of its 20.00 kilometer of leg. It’s that vector that we want to solve for based on this given information. We can start by orienting the motion of our fighter to a set of coordinate axes. We’ll position a pair of 𝑥- and 𝑦-coordinate axes, so that their origin, the place where they cross, is the place where our jet fighter began its journey. This means we can write our displacement vector 𝑑 in the following form. A component, we’ll call it 𝑑 sub 𝑥, in the 𝑖-direction plus a component, we’ll call it 𝑑 sub 𝑦, in the 𝑗-direction.

So now, our task is to set out and solve for 𝑑 sub 𝑥 and 𝑑 sub 𝑦. Looking first at 𝑑 sub 𝑥, we see that that value, the 𝑥-component of the jet fighter’s motion, will be equal to 𝑙 times the cos of 30.00 degrees. We know that 𝑙 is given as 20.00 kilometers or 20.00 times 10 to the third meters. When we multiply that by the cos of 30.00 degrees, we find a value of 1.732 times 10 to the fourth meters. That’s 𝑑 sub 𝑥. Now we want to solve for 𝑑 sub 𝑦, the vertical displacement of the jet fighter. Looking at our diagram, we see that 𝑑 sub 𝑦 is equal to ℎ plus 𝑙 times the sin of 30.00 degrees. Plugging in for ℎ and 𝑙, we once again use a value of 20.00 times 10 to the third meters for 𝑙. And ℎ, we’re given as 20.00 meters. With these values entered on our calculator, we find that 𝑑 sub 𝑦, to four significant figures, is 1.002 times 10 to the fourth meters. We insert this value for 𝑑 sub 𝑦 in our expression for displacement, 𝑑. And this gives us the final form for the displacement of the jet fighter, 1.732 times 10 to the fourth meters in the 𝑖-direction and 1.002 times 10 to the fourth meters in the 𝑗-direction.

Now let’s look at another displacement vector example, where we’ll find displacement magnitude as well as direction.

A delivery man starts at the post office, drives 40 kilometers north, then 20 kilometers west, then 60 kilometers northeast, and finally, 50 kilometers north to stop for lunch. Find the magnitude of the delivery man’s net displacement vector. Give your answer to a precision of three significant figures. Find the angle north of east made by the delivery man’s net displacement vector. Give your answer to a precision of two significant figures.

In this two-part exercise, we want to solve first for the magnitude of the net displacement vector of the delivery man. We’ll call that magnitude of 𝑑. We also want to solve for the angle north of east made by this net displacement vector. We’ll label that angle 𝜃. Let’s start out by drawing a sketch of the delivery man’s motion. Beginning with our four compass directions — north, south, east, and west — we can let the delivery man begin at the intersection of these two axes. We’re told the first leg of the delivery man’s journey is 40 kilometers north, then 20 kilometers west, then 60 kilometers northeast. And finally, 50 kilometers north to the end point of the delivery man’s journey for that morning.

With the journey laid out graphically like this, we can draw in our displacement vector 𝑑 that shows from the start to the end of these four journey legs. We can see that this vector 𝑑 has both a component to the north, we can call it 𝑑 sub 𝑁 in the northern direction. And a component to the east, we can call that 𝑑 sub 𝐸 in the eastern direction. In order to solve for the magnitude of the displacement vector, we’ll want to solve for the displacement vector itself first. And to do that, we’ll solve for 𝑑 sub 𝑁 and 𝑑 sub 𝐸. Starting with 𝑑 sub 𝑁, 𝑑 sub 𝑁 is equal to the sum of all the northern components of each leg of our journey. Noting that the third leg of our journey, where we travel 60 kilometers northeast, involves motion 45 degrees above the horizontal. We can write out 𝑑 sub 𝑁 as 40 kilometers plus 60 kilometers times the sin of 45 degrees plus 50 kilometers. When we add these three terms together, we find a value, to four significant figures, of 132.4 kilometers.

Now we move on to solving for the eastern component of displacement 𝑑 sub 𝐸. Looking at our diagram of the delivery man’s motion, we can write the terms of 𝑑 sub 𝐸 as negative 20 kilometers, that’s when we’re moving to the west, plus 60 kilometers times the cos of 45 degrees. Adding these terms together, to four significant figures, they’re equal to 22.46 kilometers. We now have our displacement vector 𝑑, written out in terms of its northern and eastern components. To find the magnitude of that displacement vector, we’ll take the square root of 𝑑 sub 𝑁 squared plus 𝑑 sub 𝐸 squared. Entering these values on our calculator, to three significant figures, the magnitude of 𝑑 is 134 kilometers. That’s the net displacement of the delivery man.

We next want to solve for 𝜃 which, looking at our diagram, is the angle north of east at which the displacement vector points. The tangent of that angle 𝜃 is equal to 𝑑 sub 𝑁, the northern component of the displacement, divided by 𝑑 sub 𝐸, the eastern component. If we take the arc tangent of both sides of the equation, plug in values for 𝑑 sub 𝑁 and 𝑑 sub 𝐸, and then enter this expression on our calculator. We find that, to two significant figures, 𝜃 is 80 degrees. That’s the angle north of east in which the displacement vector points.

Let’s now summarize what we’ve learnt about displacement vectors. Displacement vectors have a length or magnitude equal to the shortest distance between a start point and an end point. Adding vectors to get a net displacement requires separating them by dimensions, 𝑥 and 𝑦. And displacement vectors, like all vectors, have both a magnitude, or length, and a direction. So the next time you’re looking for the shortest way to get from 𝐴 to 𝐵, consider using a displacement vector.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.