# Lesson Video: Velocity–Time Graphs Mathematics

In this video, we will learn how to calculate the displacement or acceleration of a particle moving in a straight line from its velocity–time graph.

16:35

### Video Transcript

In this video, we will learn how to calculate the displacement or acceleration of a particle moving in a straight line from its velocity–time graph.

We will begin by recalling the key features of a velocity–time graph. The velocity of an object is its speed in a particular direction. It can therefore take positive or negative values. It is usually measured in meters per second but could also be measured in kilometers per hour or miles per hour. A velocity–time graph, therefore, shows the speed and direction an object travels over a specific period of time. When the velocity is measured in meters per second, the time will be measured in seconds. Likewise, if the velocity was in miles per hour or kilometers per hour, the time would be measured in hours.

The vertical or 𝑦-axis represents velocity, and the horizontal axis represents time. Whilst the velocity can take positive and negative values, the time can only take positive values. We will now look at how we can use a velocity–time graph to calculate the acceleration and displacement of an object. Let’s firstly consider acceleration on a velocity–time graph. When our velocity is in meters per second, acceleration is measured in meters per square second or meters per second per second. This can be written in either of these two ways. The acceleration is the slope or gradient of the graph. And in this video, we will only be dealing with straight-line graphs.

We can therefore calculate the acceleration by dividing the change in velocity by the change in time. This is the same as the change in 𝑦 over the change in 𝑥 or the rise over the run. If we consider the straight line shown, we can calculate the acceleration by creating a right-angled triangle. The change in velocity is labeled 𝑦, and the change in time is labeled 𝑥. The velocity has changed from two meters per second to six meters per second, so we need to subtract two from six. The time has gone from zero to six seconds. Six minus two is equal to four, and six minus zero is equal to six. By dividing the numerator and denominator by two, this fraction simplifies to two-thirds.

The acceleration of the body shown is two-thirds meters per square second or two-thirds meters per second per second. We are dividing a unit in meters per second by a unit in seconds. In this example, our line has a positive slope or gradient. This means that the object is accelerating. If our line has a negative slope or gradient, the object is decelerating. This means it will have a negative acceleration. If we have a horizontal line on our velocity–time graph, the acceleration is zero. This means that the object is moving with a constant velocity. We will now look at how we can calculate the displacement on a velocity–time graph.

The area between the line and the 𝑥-axis represents the displacement of the object. If the area is above the 𝑥-axis, then the displacement is positive. However, if the area is below the 𝑥-axis, the displacement is negative. We can use our values for displacement on a velocity–time graph to calculate the total distance traveled. In order to make it easier to calculate the displacement, we can split our graph into triangles, rectangles, and trapezoids or trapeziums. We will now look at some examples of velocity–time graphs.

Use the velocity–time graph below to determine the velocity after five seconds and the time period when the velocity is five meters per second.

In any velocity–time graph, we have the velocity on the vertical or 𝑦-axis and the time on the horizontal or 𝑥-axis. The first part of this question wants us to work out the velocity after five seconds. We can draw a line vertically upwards from five seconds. Once we hit the velocity–time graph, we draw a horizontal line across to the 𝑦-axis. This will give us the velocity after five seconds. This is halfway between two and three meters per second. We can therefore conclude that after five seconds, the velocity is 2.5 meters per second.

The second part of our question asks us to find the time period when the velocity is five meters per second. This occurs during the horizontal part of the graph. During this period, the object has constant velocity. This means it has zero acceleration. Whilst it is not relevant to this question, when the line is sloping upwards from left to right, it has positive acceleration, and when it is sloping downwards, it has negative acceleration. We can see that the body is traveling at five meters per second between two points. One of these is at 10 seconds, and the other one is halfway between 20 and 25 seconds. 22.5 is halfway between 20 and 25. Therefore, the object is traveling at five meters per second between 10 seconds and 22.5 seconds.

The correct answers are therefore 2.5 meters per second and the inequality 𝑡 is greater than or equal to 10 seconds and less than or equal to 22.5 seconds.

We will now consider some questions where we need to calculate the acceleration or displacement of an object.

The given velocity–time graph represents a particle moving in a straight line. Determine its displacement at 𝑡 equals two seconds.

The displacement of any particle is its distance from the origin or start point. When dealing with a velocity–time graph, we know that the displacement is the area between the graph and the 𝑥-axis. In this question, we need to calculate the displacement at 𝑡 equals two seconds. In order to do this, we draw a vertical line from two on the 𝑥- or horizontal axis. This creates a right-angled triangle. The displacement of the particle will be the area of this triangle. In order to calculate the area of any triangle, we multiply the base by the height and then divide by two. The base of our triangle is two, and the height is 30. We need to multiply two by 30 and then divide by two. This is equal to 30.

Our units for velocity in this question are centimeters per second, and our units for time are seconds. This means that the units for displacement will be centimeters. The displacement of the particle at 𝑡 equals two seconds is 30 centimeters.

Our next question involves a velocity–time graph with both positive and negative velocities.

Given the velocity–time graph of a particle moving in a straight line, determine the distance covered by the particle within the time interval zero to eight.

We recall that in any velocity–time graph, the displacement is the area between the graph and the 𝑥-axis. Displacement can be both positive and negative. If the area is above the 𝑥-axis, the displacement is positive. And if it is below the 𝑥-axis, the displacement is negative. In this question, however, we’re asked for the distance. This means that we’ll take the absolute value of the displacements and add them. The value for our distance must be positive. As we are looking at the time interval from zero to eight seconds, part of our graph is above the 𝑥-axis and part of it is below.

To calculate the distance covered by the particle, we will need to calculate the area of the triangle, the area of the trapezoid and add our answers. Had we been calculating the displacement, we would’ve subtracted the area of the trapezoid or trapezium from the area of the triangle. The area of any triangle can be calculated by multiplying the base by the height and then dividing by two. The base of our triangle is one, and its height is five. We need to multiply these numbers and then divide by two. Five divided by two is equal to 2.5. As our units for velocity were meters per second and our units for time were seconds, our units for distance and displacement in this question will be meters. The particle traveled a distance of 2.5 meters from zero to one second.

In order to calculate the area of a trapezoid or a trapezium, we add the lengths of the parallel sides, divide by two, and multiply by the perpendicular height. The parallel sides of the trapezium have lengths seven and four. The height of the trapezoid or distance between the parallel sides is five. 11 divided by two is 5.5. And multiplying this by five gives us 27.5. The distance covered between one and eight seconds is 27.5 meters. Had we been looking for the displacement here, this would be negative 27.5 as it is below the 𝑥-axis. We can calculate the total distance covered by the particle by adding 2.5 and 27.5. This is equal to 30 meters.

The final question that we will look at deals with acceleration on a velocity–time graph.

The figure shown is a velocity–time graph for a body moving in a straight line. Determine the deceleration of the body during the final section of its movement, given that it came to rest 100 seconds after it started moving.

We know that the acceleration on a velocity–time graph is equal to its slope or gradient. We can therefore calculate the acceleration at any point by dividing the change in velocity by the change in time. In this question, our velocity is measured in meters per second, and our time is measured in seconds. Dividing meters per second by seconds gives us meters per second per second. This is written as meters per second squared or meters per square second.

When the slope or gradient of our graph is positive, the body will be accelerating. When the slope or gradient of the graph is negative, the body will be decelerating. Finally, if a line on our velocity–time graph is horizontal, the acceleration will be equal to zero and the body will be traveling with a constant velocity. In this question, we’re interested in the deceleration in the final section of the graph. This occurs between 90 seconds and 100 seconds. The velocity during this time period has decreased from 45 meters per second to zero meters per second. This means that our change in velocity is zero minus 45.

As mentioned, the time period goes from 90 seconds to 100 seconds. Therefore, the change in time is 100 minus 90. This can be simplified to negative 45 over 10. When dividing by 10, we move all our digits one place to the right. And dividing a negative number by a positive gives a negative answer. The acceleration of the body is therefore negative 4.5 meters per square second. The deceleration will be the absolute value or modulus of this. This is equal to 4.5 meters per square second.

We will now summarize the key points of this video. In this video, we looked at how we could use a velocity–time graph to calculate the displacement and acceleration of a particle. The acceleration of a particle is equal to the slope or gradient of the line. We found that this could be positive, negative, or zero. A negative acceleration is also known as a deceleration. This occurs when the line slopes downwards. When our line is horizontal, acceleration is equal to zero. This means the body is moving with a constant velocity. We’re able to calculate the acceleration at any point on our graph by dividing the change in velocity by the change in time. This is the same as the rise over the run.

We also found that the displacement is the area between the graph and the 𝑥-axis. If the area is above the 𝑥-axis, the displacement is positive, whereas if it is below the 𝑥-axis, the displacement is negative. We also identified the difference between the displacement and distance as displacement is distance with a direction. Whilst we only looked at straight-line graphs in this video, it is also possible to estimate the acceleration and displacement of a particle on a curved velocity–time graph.