Lesson Video: Position, Displacement, and Distance | Nagwa Lesson Video: Position, Displacement, and Distance | Nagwa

Lesson Video: Position, Displacement, and Distance Mathematics • Second Year of Secondary School

Join Nagwa Classes

Attend live General Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher!

In this video, we will learn how to differentiate between position, displacement, and distance traveled, including problems that use vector notation.


Video Transcript

In this video, our topic is position, displacement, and distance. These are fundamental qualities for many physical situations. In this video, we’ll learn how to define each of these terms as well as how they relate to one another.

As we get started, say that we have a bee that’s hovering in place at a certain position. We can really only say what that position is if we have a set of coordinate axes defined somewhere. Using this as a reference, we can draw a vector from our origin to the bee’s current position. And further, we can call this vector 𝐫, specifying where the bee is located relative to this frame.

At this point, let’s imagine that our bee starts to move, following some path through three-dimensional space. At some later time, if our bee is positioned here, we can draw a new vector defining that quantity. If we consider the bee’s overall path, we see that its position vector depends on time. This leads us to rewrite our position vectors.

Our initial position vector we’ll say is 𝐫 at a time we’ll call 𝑡 sub i for the initial time. And the final position vector is here, the position of the bee at some time we’ve called 𝑡 sub f. At any point along its journey, we could define the bee’s position by drawing a vector from the origin of our coordinate axes to that particular point on its path. This is the meaning of the bee’s position.

When the bee’s position changes, as we see it does all throughout this path, that relates to the term displacement. In this video, we’ll represent displacement using a lowercase 𝐬. And note that this quantity is a vector. Displacement, like position, is typically a function of time. And it’s defined as the change in an object’s position over time. For example, if we want to calculate the displacement of our bee at the time we’ve called 𝑡 sub f, then we would take the bee’s position at that time and subtract from it the bee’s position at the time we’ve called 𝑡 sub i.

Note that this involves finding the difference between two vectors. If we sketch that difference in, it would look like this. Graphically then, we can see the displacement of the bee over the time interval from 𝑡 sub i to 𝑡 sub f. And notice something interesting about displacement. It only depends on the starting and ending position of our object. It doesn’t take into account any information we have about the path our object followed in traveling between these two positions.

A quantity that does do this though is the one we call distance. We’ll represent distance using a lowercase 𝑑. Like displacement, this is typically a function of time. But unlike displacement, distance is not a vector. If we wanted to determine the distance traveled by our bee over this time interval, we would start at its initial position. And then we would follow the path that it followed all throughout every twist and turn until we reached its end point. And this entire path length is what we call its distance traveled.

Note then that distance is path dependent, but displacement is not. For example, if our bee had followed, say, this path to get from the start to the end point, its displacement will be the same, but its distance traveled will be different. Another thing to note about these two terms is that distance is always at least as great as the magnitude of displacement, if not greater. This is because the shortest distance between a start and an end point follows along the line that’s defined by the relevant displacement vector.

Thanks to the fact that displacement is a vector while distance is a scalar quantity, there’s yet another difference between these two terms. Because an object’s displacement is given by a vector, it’s possible for displacement to be negative. For example, say that we have some particle right here and that over time that particle moves purely in the negative 𝑥-direction according to our coordinate axes. In that case, the particle’s displacement is defined by this vector, which is negative.

Distance, on the other hand, is never negative. Any distance traveled is recorded as a positive value. And so we would say the distance that this particle has traveled is positive. The least distance that an object can travel then is zero, if it doesn’t move at all.

Knowing all this about position, displacement, and distance, let’s get some practice now through a few examples.

Using the given figure, calculate the distance 𝑑 and the displacement 𝑠 of a body that moves from point 𝐴 to point 𝐶 then returns to point 𝐵.

Looking at this figure, we’re told to imagine a body that starts here at point 𝐴, moves all the way from point 𝐴 to point 𝐶, and then doubles back, ending up at point 𝐵. Knowing that the distance this object travels is represented by 𝑑 while its displacement is represented by 𝑠, we can recall that the distance an object travels includes the total path length that it follows. That means that this body’s distance will include the distance from 𝐴 to 𝐵 to 𝐶 and then back to 𝐵. We see that that’s equal to 28 centimeters plus 24 centimeters then plus 24 centimeters again. That comes out to 76 centimeters in total.

The displacement of our body, on the other hand, only takes into account its starting and ending positions. We know that our body starts at point 𝐴 and it ends up at point 𝐵. That means that its displacement includes only this length of 28 centimeters. Note that if our body had traveled just from point 𝐴 directly to point 𝐵, then its distance and displacement would’ve been the same. We see through the actual scenario though how these two terms are different.

Let’s now look at an example where our object of interest moves in two dimensions.

According to the figure, a body moved from 𝐴 to 𝐵 along the line segment 𝐴𝐵, and then it moved to 𝐶 along 𝐵𝐶. Finally, it moved to 𝐷 along 𝐶𝐷 and stopped there. Find the distance covered by the body 𝑑 one and the magnitude of its displacement 𝑑 two.

Looking at the figure, we’re told that a body begins at point 𝐴, right here, and then follows the line segment 𝐴𝐵 to point 𝐵 then moves to point 𝐶 following this path, finally moving along line segment 𝐶𝐷 to end up at point 𝐷. Given this motion, we want to calculate the distance the body has covered, we call that 𝑑 one, and the magnitude of its displacement 𝑑 two.

Clearing some space, let’s first work on solving for the distance our body travels 𝑑 one. We can recall that distance in general is equal to the total path length followed by some body as it moves from one location to another. In our case, our body moved from point 𝐴 to point 𝐷 along the path shown in orange. We see that that involves 6.6 centimeters of travel from point 𝐴 to point 𝐵, 8.8 centimeters from 𝐵 to 𝐸, 16.4 centimeters from 𝐸 to 𝐶, and on the last leg of the journey 12.3 centimeters. 𝑑 one then equals the sum of these four distances. Adding them all up gives a result of 44.1 centimeters. This is the distance our body traveled.

And now let’s consider its displacement magnitude 𝑑 two. Displacement is different from distance in that displacement only takes into account the start point and end point for some body. In our situation, our body begins at point 𝐴 and it ends up at point 𝐷. So this straight pink line connecting these two points represents the displacement magnitude 𝑑 two. We see that the length of this line segment can be divided up into two parts: one part right here and then the second part here. Each of these is a hypotenuse of a right triangle. This means we can use the Pythagorean theorem to solve for these lengths.

If we call the length of the first hypotenuse 𝑙 one and that of the second 𝑙 two, then we can say that the displacement magnitude 𝑑 two is equal to their sum and that 𝑙 one and 𝑙 two are defined this way. 𝑙 one equals the square root of 6.6 centimeters quantity squared plus 8.8 centimeters quantity squared. And 𝑙 two equals the square root of 16.4 centimeters quantity squared plus 12.3 centimeters quantity squared. When we enter these expressions on our calculator, we find that 𝑙 one is equal to exactly 11 centimeters, while 𝑙 two is 20.5 centimeters. Adding these together gives us a result of 31.5 centimeters. This is the displacement magnitude of our body.

Now let’s look at calculating displacement based on particle position as a function of time.

A particle started moving in a straight line. After 𝑡 seconds, its position relative to a fixed point is given by 𝑟 equals quantity 𝑡 squared minus four 𝑡 plus seven meters when 𝑡 is greater than or equal to zero. Find the displacement of the particle during the first five seconds.

Okay, so here we have a particle’s position, given as a function of time when 𝑡 is greater than or equal to zero. And based on this, we want to solve for the particle’s displacement during the first five seconds, that is, from 𝑡 equals zero up to 𝑡 equals five. To do this, we can recall that the displacement of a particle — we can represent that displacement with a lowercase 𝑠 — at some time we’ll call 𝑡 sub f, is given by the change in the particle’s position from some initial time 𝑡 sub i to this final time 𝑡 sub f.

This definition tells us that to calculate the displacement of our particle over the first five seconds, we’ll need to know its position at 𝑡 equals zero and 𝑡 equals five seconds. We can write that like this. Displacement at 𝑡 equals five seconds equals position at 𝑡 equals five seconds minus position at 𝑡 equals zero seconds.

Our given equation for particle position is valid for both of these times. So we’ll start by solving for the particle’s position at 𝑡 equals five. This is equal to five squared minus four times five plus seven meters. And we subtract from this the particle’s position at 𝑡 equals zero. That’s equal to zero squared minus four times zero plus seven meters. We know that zero squared is zero, and so is negative four times zero. So our particle’s position at time 𝑡 equals zero simplifies to seven meters. At five seconds, its position is five squared, which is 25, minus four times five, which is 20, added to seven meters. This adds up to 12 meters. So 12 meters minus seven meters gives us the displacement we want to solve for. It’s five meters. This is the particle’s displacement during the first five seconds.

Let’s look now at calculating displacement from a two-dimensional position vector.

If the position vector of a particle is expressed as 𝐫 equals two 𝑡 plus three 𝐢 plus five 𝑡 minus two 𝐣, then the magnitude of the displacement at 𝑡 = 2 seconds equals blank length units.

Here then we want to fill in this blank by solving for the magnitude of this particle’s displacement at 𝑡 equals two seconds. We can begin to do this by recalling that the displacement vector of an object at a time we’ll call 𝑡 sub f is equal to that object’s position vector at that same time minus the object’s position vector at some initial time we’ll call 𝑡 sub i.

In our case, as we want to solve for displacement magnitude at a time of 𝑡 equals two seconds, we can start by figuring out the displacement vector at this time. According to our definition, this displacement vector is equal to the position vector at that same time, two seconds, minus the position vector at a time 𝑡 equals zero. Knowing this, we can then use our position vector, which we see is given to us as a function of time, to let us solve for particle displacement at two seconds.

If we plug in 𝑡 equals two to our equation for the position vector, we get two times two plus three 𝐢, that’s seven 𝐢, plus five times two minus two 𝐣. That’s equal to eight 𝐣. We can then replace this vector with the expression we’ve just solved for. Next, we want to solve for the particle’s position at time 𝑡 equals zero seconds. That’s equal to two times zero plus three 𝐢, that simplifies to three 𝐢, plus five times zero minus two 𝐣. The vector overall then is three 𝐢 minus two 𝐣. And we substitute this in for 𝐫 of zero.

We’re now ready to calculate the displacement at a time 𝑡 equals two seconds. And we get a result of four 𝐢 plus 10𝐣. This, however, is not our final answer because we want to solve for the magnitude of this displacement. Let’s recall that if we have a two-dimensional vector, we’ll call it 𝐕, then the magnitude of that vector is equal to the square root of the sum of the squares of its 𝑥- and 𝑦-components. This means that the magnitude of our particle’s displacement at 𝑡 equals two seconds equals the square root of four squared plus 10 squared. Four squared is 16, and 10 squared is 100. So we’re taking the square root of 116.

116 though is divisible by four. It’s equal to four times 29. And then, knowing that four is equal to two squared, we can move it outside the square root sign, which makes our final answer two times the square root of 29. This is the magnitude of our particle’s displacement at 𝑡 equals two seconds.

Let’s finish up now by reviewing some key points about position, displacement, and distance. In this video, we saw that a particle’s position is a vector quantity that depends on the coordinate system being used to define that position. We then saw that displacement is also a vector quantity, and it’s equal to a difference in position. And lastly, distance is a scalar quantity that’s equal to the length of an object’s path, and distance cannot be negative.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

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