Video: Instantaneous Speed

In this video, we will learn how to determine the instantaneous speed of an object by using tangents to find the gradient at a point on the object’s distance-time graph.

12:21

Video Transcript

In this lesson, we’re going to be talking about instantaneous speed. Instantaneous speed refers to the speed that an object is traveling at at a particular instant in time. This makes it a very useful concept when we’re describing the motion of an object whose speed changes over time. In this video, we’re going to focus on how we can calculate the instantaneous speed of an object by looking at a graph of its displacement against time. As we’ll see, the method that we use to calculate the instantaneous speed of an object depends on whether it’s traveling at a constant speed or whether its speed is changing.

Usually, when we calculate an object’s speed, we think of this equation. The speed of an object is given by the distance the object travels divided by the time taken by the object to travel that distance. But in fact, this equation only tells us the average speed that an object was traveling at over the entire time period that we’re looking at. If we were looking at an object that sped up or slowed down, then we might find that its instantaneous speed, that is, its speed at any given moment, is different to its average speed. To illustrate the difference between average speed and instantaneous speed, let’s imagine a car driving down a straight road from point A to point B. And let’s say that along the way, the car stops at a traffic light before driving on to point B.

Let’s also say that the total distance from A to B is 800 meters and that the total time taken for the car’s journey was 80 seconds. Let’s see what happens when we plug these values into our formula for speed. The formula tells us that our speed is given by dividing the distance, that’s 800 meters, by the time taken, which is 80 seconds, which gives us an answer of 10 meters per second. However, when we calculated this value of speed, we included all of the time that the car was stationary at at traffic light. If we looked at the car’s speed at different points in its journey, we’d probably find that, most of the time, its speed was different to 10 meters per second.

For example, we know that while the car was stopped, its speed was zero meters per second. And somewhere else in its journey, the car must’ve been traveling faster than 10 meters per second in order for its speed to average out to 10 meters per second. So it’s really important to note that this equation really just tells us the average speed that an object traveled at based on the overall distances traveled and the total time taken for its journey. If we want to know the speed that the car was traveling at at different points in its journey, that is, its instantaneous speed at various points. Then, we need more information about what exactly the car was doing between point A and point B.

In this lesson, we’re focusing on how we can calculate instantaneous speed from a displacement–time graph. For our car, the displacement–time graph might look like this. This graph shows us that, at a time of zero seconds, the car starts off with a displacement of zero meters. And this increases over the first 20 seconds up to a displacement of 400 meters. So this part of the graph shows the car’s journey from point A to the traffic light. The next part of the graph from 20 seconds to 60 seconds shows that the displacement of the car remains at a constant 400 meters. So this part of the graph corresponds to when the car was stopped at the traffic light. And then, the final part of the graph from 60 seconds to 80 seconds shows the displacement of the car increasing from 400 up to 800 meters, representing the part of the car’s journey after the traffic light as it moved on to point B.

So how do we calculate the instantaneous speed of the car at different times? The answer is that we look at the slope of the graph. Specifically, the instantaneous speed of an object at any time is given by the magnitude of the slope of its displacement–time graph at that time. In other words, the instantaneous speed of our car is given by how steep this graph is at various points. And we can work out the value of its speed at any given time by calculating the slope of the graph at that time. Note that we’re talking about the magnitude of the slope of the graph. This means that regardless of whether the slope is positive or negative, we just take its value with a positive sign as this indicates its magnitude. And that’s because speed can only be positive or zero by definition. It can’t take negative values.

So this displacement–time graph, for example, shows us an object traveling at a relatively low speed because the slope of this line is relatively low. This displacement–time graph, on the other hand, shows us an object traveling at a relatively high speed, which we know because the slope of this line is relatively high. This final displacement–time graph also shows an object traveling at a high speed. The slope of the graph is negative and relatively large. But because speed is just given by the magnitude of the slope of the displacement–time graph, it doesn’t matter that it’s negative. We just take the positive value. A positive value of this slope is approximately the same as the slope of this graph. So actually, both of these graphs show us two objects with roughly the same speed.

If we look back at the graph for our car, we can see that it’s made up of three straight lines. Because a straight line has a constant slope, this tells us that the car is traveling at a constant speed between zero and 20 seconds, and at a different constant speed between 20 and 60 seconds, and at another constant speed between 60 and 80 seconds. Really, this is a simplification, and we’d expect that displacement–time graph for a real car to be curved. But we’ll use this example to show us how we can calculate instantaneous speed from a displacement–time graph which is made up of straight lines.

So let’s say that we want to calculate the instantaneous speed of our car at the time 70 seconds, which is here. In order to do this, we need to calculate the slope of our graph at this point. We can see that the slope of the graph around this point between 60 seconds and 80 seconds is constant. In other words, the graph is a straight line. That means that the magnitude of the slope at the point we’re interested in is the same as the magnitude of the slope of the line between 60 and 80 seconds. So the instantaneous speed at 70 seconds is given by the magnitude of the slope of this line segment.

Let’s recall that the slope of a straight line is given by the vertical difference between two points on that line divided by the horizontal difference between the same two points. So let’s use the end points of this line segment as the two points that we’ll use to calculate the slope. This point is at 60 seconds on the time axis and 400 meters on the displacement axis. So the coordinates of this point are 60, 400. This second point is at 80 seconds on the time axis and 800 meters on the displacement axis. So its coordinates are 80, 800. We can now use these coordinates to work out the slope of this line segment. The vertical difference between the two points we’ve chosen is given by the difference between their 𝑦-coordinates which give their vertical position on the graph. So the vertical difference in this case is 800 minus 400.

Similarly, the horizontal difference between these two points is given by 80 minus 60. 800 minus 400 is 400, and 80 minus 60 is 20. And 400 divided by 20 is 20. So the slope of the line at 70 seconds is 20. Now, we know that the instantaneous speed is given by the magnitude of the slope of the displacement–time graph. Since the slope that we calculated is positive, we already have its magnitude, 20. And the units for the instantaneous speed which we’ve calculated are just given by whatever units are used for our axes. So in this case, the vertical difference that we calculated is measured in meters, and the horizontal difference that we calculated is measured in seconds. That means that the instantaneous speed of the car at this point was 20 meters per second.

If we want to find out the instantaneous speed of the car at any other instant, we just need to work out the slope of the graph at that instant. For example, at a time of 10 seconds, we can see that the slope of the graph is just given by the slope of this line segment. And in this particular instance, that’s actually the same as the slope of this line segment. Although, of course, that won’t always be true. So the instantaneous speed at 10 seconds is also 20 meters per second. If we look for the instantaneous speed at, say, 50 seconds, we can see that the slope of the graph at this point is given by the slope of this line segment. Because this line segment is horizontal, we can say that its slope is zero, which means that the instantaneous speed at this point is zero meters per second. And that’s true for anywhere within this line segment.

So we can see that between 20 seconds and 60 seconds, the instantaneous speed of the car is zero. But for the time shown on the graph before 20 seconds or after 60 seconds, the instantaneous speed of the car is 20 meters per second. Now that we’ve looked at how we can calculate the instantaneous speed of an object from a straight-line, displacement–time graph, let’s have a look at how we can calculate the instantaneous speed of an object with a curved displacement–time graph.

As we mentioned before, the straight-line graph that we’ve just been looking at was really a simplification of how the displacement of a car might change over time. The displacement–time graph for a real car making the journey described would probably look more like this. Instead of being made out of straight lines which have constant slopes and therefore represent constant speeds, this displacement–time graph is curved. If we look at this section of the graph, we can see how it goes from being relatively steep to slightly less steep until it’s horizontal. This gradual change of slope represents the fact that the instantaneous speed of the car, when it’s slowing down, gradually decreases to zero.

Similarly, if we look at this part of the graph, we can see that the slope gradually and continuously increases from zero to some positive value to a higher positive value, representing the fact that the car gradually speeds up from a standstill. But how can we calculate the instantaneous speed at these points? We can do this by drawing tangents to the graph. A tangent is a straight line that touches a curve in such a way that both the curve and the tangent have the same slope at the point where they touch. For example, we could draw a tangent to the curve at this point. Here, the curve is sloped like this, so the tangent to the curve at this point looks like this. Drawing a tangent gives us a way of calculating the slope of a curve at a given point.

This tangent touches our graph at a time of about 18 seconds. So by calculating the slope of this tangent, we would find the slope of the graph at 18 seconds. And therefore, we would find the instantaneous speed of the car at 18 seconds. Since the tangent is a straight line, we can find its slope by picking two points on the line and dividing the vertical difference between those points by the horizontal difference. Since we just drew the tangent by eye, that means that our answer will really be an estimate of the slope. But we can minimize the error in our measured gradient and make our answer as accurate as possible by choosing our two points so they’re near the ends of the line.

Looking at this point to start with, we can see that it’s at a time of approximately eight seconds, and it’s at a displacement of zero meters. So the coordinates of this point are eight, zero. Now, looking at this point, we can see that it’s at a time of approximately 37 seconds and a displacement of 800 meters. So the coordinates of this point are 37, 800. Once again, the slope is given by the vertical difference divided by the horizontal difference between these points. The vertical difference between the points is given by the difference between their vertical coordinates, in other words, 800 minus zero. And the horizontal difference is 37 minus eight. 800 minus zero is, of course, 800. And 37 minus eight is 29. 800 divided by 29 is 27.586. So to one decimal place, that’s 27.6.

So if the tangent to our graph at the time 18 seconds has a slope of 27.6, that means that the graph itself has a slope of 27.6 at 18 seconds. The instantaneous speed at this time is given by the magnitude of 27.6, which is just 27.6 because it’s already positive. And once again, the units for our instantaneous speed are given by the units used on the vertical axis divided by the units used on the horizontal axis. So that’s meters per second. So the instantaneous speed of the car at the time 18 seconds is 27.6 meters per second.

If we wanted to find the instantaneous speed of the car at some other time, for example, at 65 seconds. We would once again draw a tangent to the curve at that time, calculate the slope of that tangent using two points near the end, and then take the magnitude of that slope. Which means if we calculate a negative slope, we just take the positive version of that number.

So let’s just summarize what we’ve learned about calculating instantaneous speed. Instantaneous speed is the speed of an object at a certain instant in time. It’s given by the magnitude of the slope of a displacement–time graph at that certain instant in time. And for a curved displacement–time graph, we can draw a tangent to the graph to help us calculate the slope at a certain point.

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