Calculating Displacement from Velocity-Time Graphs - Online Tutor, Practice Problems & Exam Prep

In a velocity-time graph, displacement (ΔX) is determined by calculating the area under the curve between two time points. This area can be broken down into simpler shapes like rectangles and triangles, using the formulas: rectangle area = base × height and triangle area = ½ × base × height. For example, if the area above the time axis is positive and below is negative, total displacement is the sum of these areas. Understanding this concept is crucial for analyzing motion and applying the principles of kinematics effectively.

Calculating Displacement from Velocity-Time Graphs

Hey, guys. So sometimes you'll have to calculate the displacement by using the velocity-time graph rather than using your constant velocity equations. So that's what we're going to cover in this video, but we're going to see that all of these problems will just break down to just using really simple equations for really simple shapes like rectangles and triangles. So let's get to it.

Guys, in a velocity-time graph, the displacement, the change in the position, that's Δx between two points, is just going to be the area that is underneath the curve. This is an expression that your textbooks and your professors are going to use. And what under the curve really just means, is it's the area that's between the values of where you are on the velocity-time graph and the time axis itself. So under the curve just means all of the area inside of the shape that I've made between the red lines. Guys, that's all there is to it. We just have to figure out the area of this shape.

So let's just go ahead and take a look at this example. We've got a velocity-time graph for this moving object. We're going to calculate the displacement for the first 4 seconds. So what you do is you just highlight 0 and 4 seconds. And then you calculate or you just figure out what the area underneath that curve is. So that's all we have to do. But I might not know what the area formula is for this complicated shape here. I've got some straight lines, some slanted lines. I'm not really sure what that is. So what I can do is I can just break it up into smaller shapes that I do know the area for. So I can break this up and now I've just made a square and a triangle. And guys, I actually know what the formulas for those shapes are. A rectangle is just base times height, and a triangle is just one-half base times height.

So for this first part here, if I want to calculate the displacement, I'm going to call that Δx1. This is just going to be the area of the square. So I'm going to call this Δx2, and I'm going to call this Δx3. So this Δx1 is just going to be Δx2 plus Δx3. So now I just have to figure out those areas. Well, Δx2 is just a rectangle, so this is going to be base times height. The base of this rectangle is 2 and the height is 2, so it's just going to be 2 times 2 and that's 4. So I've got 4 meters here. And then this triangle here, I knew the equation for that is 12 of base times height. So now I've got 12, and now the base is 2 and the height is 2 also. So I've just got 12×2×2. The 12 and the 2 will cancel, and you'll just get 2 meters. So I'm just going to add those 4 and 2 together, and I get a total of 6 meters. Now, another way you can think about this 6 meters is just if you count up all the little boxes that are inside of this curve. I've got 1, 2, 3, 4, 5, and then these are both one-half, like, this is like one-half and one-half. Altogether, that makes 6. That's another way you can visualize that.

Alright, guys. So let's move on to part B. Now we're going to calculate the displacement for the entire motion. So the entire motion is actually from 0 all the way to 6. So we have to figure out the area that's underneath the curve for all of that. It's going to include the area that we have for part A. So if I want the total displacement,Δxtotal, that's going to be Δx1 that I just found out over here, which I already know the answer for, plus now the area that's under the curve from 4 to 6. So I'm going to call thisΔx4. I've got Δx1 plus Δx4. All of that added together is going to be my total entire displacement. So let's just go ahead and do that.

So I know Δx1 is 6, and then I have to figure out this displacement, and that's going to be my total. So let's just go ahead to the graph, Δx4 is just going to be one-half of base times height because it's a triangle again, the base is 2 and the height is 2. So this is just going to be 12×2×2. So again, we're going to cancel and you're just going to get 2 out of this. But what you have to remember is that areas above the time axis are going to be positive displacements because your velocity's positive. You're moving forward. So these displacements that we got over here, 4 and 2, are positive, whereas the areas below the time axis are going to be negative displacements because you're moving backward. Your velocity is negative there. So that means this is not +2, this is actually -2. So this is our Δx4. So we've got 6 +(-2), and you add that together, and your total displacement is just going to be positive 4 meters.

And that's really all there is to it, guys. Let's get some more practice problems, and let me know if you have any other questions!

A moving box's motion is described by the velocity graph below. The box's initial position is x = 0.
(a) How far has the box moved at t = 4?
(b) What is the displacement of the box from t=4 to t=10?

• A. (a) 16 m;
  (b) 0 m
• B. (a) 16 m;
  (b) 24 m
• C. (a) 24 m;
  (b) 0 m
• D. (a) 24 m;
  (b) 24 m
• E. (a) 32 m;
  (b) 0 m
• F. (a) 32 m;
  (b) 24 m

Displacement of a Car

Hey, guys. Let's work out this example together. We're shown the velocity-time diagram. So that's \( v \) versus \( t \) or the velocity-time graph of a moving car, and we're told that the initial position is negative 21, we're supposed to find out what's the car's final position at \( t = 5 \) seconds, so final position. So the variable that we are looking for is going to be \( x \) or \( x_{\text{final}} \). So that's going to be what we're looking for here. So how do we find the final position? Well, we find position by finding the displacements. Remember \( \Delta x \). \( \Delta x \) can be written as \( x_{\text{final}} - x_{\text{initial}} \). But when we have a velocity-time graph, \( \Delta x \), the displacement, is also the area that is under the curve. And so, what's the variable I'm actually looking for here? I'm looking for the final position. So I can rearrange this formula really quickly and say that the final position is going to be the initial position plus \( \Delta x \). In other words, plus the area that's under the curve. So if this is what I'm looking for, I already have the initial position, which is -21 meters. So now all I have to do is I just have to figure out the area that's under the curve. I have to figure out what displacement is. So let's go ahead and do that over here.

The displacement, \( \Delta x \), is just going to be the area that's under the curve between the two times I'm looking for. It says what's the car's final position at \( t = 5 \) seconds, and we're also told that the initial position is at \( t = 0 \). So what this really means is we're looking for the areas under the curve of this entire graph over here. So the area that's under the curve is just going to be all of this stuff that's highlighted in yellow. So let me see if I could do that real quick. Yeah. It's pretty good. Alright. So how do we find now the area of this really complicated shape? Remember, anytime we do this, we just have to break it up into a bunch of little smaller shapes that are easy, that are a little bit more manageable. So, I always like to break it up into the smallest number of shapes. I see a big triangle over here, and if I keep this line going, I actually have a smaller triangle and then a big rectangle. So it's only 3 shapes I need to worry about and that's good. So if I want to figure out the total displacement over here, now I've got these 3 smaller shapes, so I'm just gonna label them. \( \Delta x_1 \). Let's call this guy \( \Delta x_2 \), and then we'll call the smaller one \( \Delta x_3 \). Okay? So let's just get to it. And the total displacement is just gonna be by adding up all of those areas or all of those smaller displacements up. And so \( \Delta x_1 \) is going to be a triangle, so we're going to use \(\frac{1}{2}\) base times height. Now we just have to figure out what the base and height is. And so we're going from 0 all the way to 3, so the base here is 3, and the height is going to be going from 2 all the way up to 10. So that means that this height of this triangle is going to be 8. So it means we have \(\frac{1}{2} \times 3 \times 8\) and that's 12. Let's move on to part, the second one. The second one is going to be from 0 all the way to 5. It's a rectangle, which means the equation we're going to use is base times height. And we have that the base is 5 and the height is going to be 2, right here. So we have just displacement is 10. So we've already got these first two. Now, we just have to look at the last one, \( \Delta x_3 \). It's a small triangle, so we're still going to use the same equation, \(\frac{1}{2}\), \(\frac{1}{2}\) wow. Okay. \(\frac{1}{2}\) base times height. So we've got \(\frac{1}{2}\), the base is going to be 2 because going from 3 to 5, and the height is also going to be from 2 because we're going from 2 up to 4. So we have \(\frac{1}{2} \times 2 \times 2\). \( \frac{1}{2} \) and 2 will cancel, just leaving one factor of 2 there. And so now we just put everything together, all of these areas, and then add everything up. The \( \Delta x \) is going to be \( 12 + 10 + 2 \), and so that's just going to be 24. So now the question is, are we done? Is this our answer? No, it's not. It's not our answer. Remember, this just represents the displacement. We actually have to take this number and we have to plug it back into this formula over here to figure out the final position. So you can't forget that last step there. So we're going to take this number over here, and we're going to pop it into this equation. And so, for the final answer, the final position is going to be the initial position of negative 21 plus 24, which is the displacement. And so what we get is positive 3 meters. That is our final position. Alright, guys. Let me know if you have any questions, and I'll see you in the next session.

Here’s what students ask on this topic:

How do you calculate displacement from a velocity-time graph?

To calculate displacement from a velocity-time graph, you need to find the area under the curve between two time points. This area represents the displacement (ΔX). You can break the area into simpler shapes like rectangles and triangles. For a rectangle, use the formula: $\mathrm{area}=\mathrm{base}\times \mathrm{height}$. For a triangle, use: $\mathrm{area}=\frac{1}{2}\mathrm{base}\times \mathrm{height}$. Sum these areas to get the total displacement. Remember, areas above the time axis are positive, and areas below are negative.

What does the area under a velocity-time graph represent?

The area under a velocity-time graph represents the displacement (ΔX) of an object over a given time interval. This is because displacement is the integral of velocity with respect to time. By calculating the area under the curve, you are effectively summing up the product of velocity and time, which gives you the total change in position. Positive areas indicate forward motion, while negative areas indicate backward motion.

How do you find the displacement if the velocity-time graph has both positive and negative areas?

To find the displacement when a velocity-time graph has both positive and negative areas, calculate the area of each section separately. Positive areas (above the time axis) represent forward displacement, while negative areas (below the time axis) represent backward displacement. Sum these areas, taking care to add positive areas and subtract negative areas. The result is the net displacement. For example, if you have +6 m and -2 m, the total displacement is 4 m.

What is the formula for the area of a triangle on a velocity-time graph?

The formula for the area of a triangle on a velocity-time graph is: $\mathrm{area}=\frac{1}{2}\mathrm{base}\times \mathrm{height}$. Here, the base is the time interval, and the height is the velocity. This formula helps you calculate the area under the curve when the section of the graph forms a triangular shape, contributing to the total displacement.

Why is it important to understand displacement from velocity-time graphs?

Understanding displacement from velocity-time graphs is crucial because it allows you to analyze and interpret the motion of objects accurately. This skill is essential in kinematics, helping you solve problems related to motion in physics. By mastering this concept, you can determine how far an object has traveled over a period, which is fundamental for understanding real-world applications like vehicle movement, projectile motion, and more.