# Video: Acceleration over Distance and Time

The graph shows the change in velocity of an object with time. What is the object’s velocity at 𝑡 = 0? For how long does the object accelerate? What is the object’s velocity after it has accelerated? What is the object’s acceleration? What is the object’s displacement?

09:44

### Video Transcript

The graph shows the change in velocity of an object over time.

Now we can see that we’ve been given a graph here with velocity on the vertical axis in meters per second and time on the horizontal axis given in second. We can also see that at the time of zero seconds, the object starts out with a velocity of 20 meters per second, whereas at a time of 15 seconds, the velocity of the object has increased to 50 meters per second. And we can see that this increase in velocity happens at a steady rate because the slope of this graph is constant. In other words, the blue line is a straight line. And so, the velocity of the object increases by the same amount in every unit of time.

Now the first question is asking us: What is the velocity of the object at 𝑡 is equal to zero?

In other words, we’re being asked to find the velocity of the object at the time 𝑡 is equal to zero right at the beginning of its trajectory, or at least at the beginning of the part that’s plotted on this graph. So when the time is 𝑡 is equal to zero, we can see that the velocity of the object is 20 meters per second. And so, that is our answer to the first question. Let’s write down this information up here and let’s look at the next question then.

For how long does the object accelerate?

Now this question is asking us to find the amount of time for which the object is accelerating. And an acceleration involves a change in velocity, whether that’s an increase in velocity or decrease in velocity. Now, in this case, we can see that the velocity of the object is increasing as time passes. But more to the point, the velocity of the object is increasing over the entire range of the graph. In other words, for the full 15 seconds that we’ve plotted the graph, the velocity is changing and therefore the object is accelerating. And so our answer to this part of the question is that the object accelerates for 15 seconds. Let’s write down that piece of information up here and let’s move on to the next question.

What is the object’s velocity after it has accelerated?

Okay, so we know that the object starts accelerating at this point in time. And we’ve said that it finishes accelerating at this point in time when we finish plotting graph. We need to find out what the object’s velocity is at that point, i.e., after it has finished accelerating. And so in order to find out this quantity, we simply need to go across to the vertical axis and see that the velocity value is 50 meters per second. And so that is our answer to this part of the question. Let’s also write down this piece of information up here. And let’s look at the next part of the question.

What is the object’s acceleration?

Now in order to calculate the acceleration of the object, we can think about it in one of two ways. Firstly, we can recall that an object’s acceleration is defined as the change in velocity of the object divided by the time taken for that change in velocity to occur. And so in this particular scenario, we can say that the acceleration of our object is equal to the final velocity which is 50 meters per second minus the initial velocity which is 20 meters per second divided by the time over which this acceleration occurs, which we know is 15 seconds from earlier. And by the way, the change in velocity is equal to the final velocity minus the initial velocity. That’s why we’ve subtracted the initial velocity from the final velocity in order to give us ∆𝑣 or change in velocity.

But anyway, so evaluating this on the right-hand side, we firstly see that the numerator is going to have units of meters per second because both of these quantities have units of meters per second and we’re dividing this by the unit of seconds. So our final unit for the acceleration is going to be meters per second divided by seconds or, in other words, meters per second squared, which is perfect because that is the unit of acceleration. So evaluating this, which becomes 50 minus 20 divided by 15 and the unit of course is meters per second squared, we find that the acceleration of the object is two meters per second squared. In other words, every second the velocity of the object increases by two meters per second. So it starts out at 20 meters per second. Then one second later, the velocity is 22 meters per second. Another second later which is 24 meters per second, and so on and so forth. Until 15 seconds later, its velocity is 50 meters per second.

Now another way to find the acceleration of this object is to find the slope of the straight line. Because the line is representing the motion of the object and the slope of a velocity time graph will give us the acceleration of the object. In fact, mathematically, it ends up being the same. Because in order to find the slope, we find the change in the vertical axis coordinates and we divide this by the change in the horizontal axis coordinates. And the change in vertical axis coordinates end up being 50 minus 20 meters per second and the change in horizontal axis coordinates ends up being 15 minus zero seconds. So, mathematically, we do exactly the same calculation and get exactly the same answer. And so we can say that the object’s acceleration is two meters per second squared. Writing that information up here, we can move on to the final part of the question.

What is the object’s displacement?

Okay, so now, we’re being us to find how far the object moves in a straight line as it accelerates from 20 meters per second to 50 meters per second. Now there are two ways to do this. The first way is to remember that the area underneath a velocity-time graph is going to give us the displacement of the object in question. And so if we want to find the object’s displacement, then we need to find this area here, in other words, between zero seconds and 15 seconds and the line itself, the blue line, and the horizontal axis.

Now the simplest way to do this is to split up the area into two chunks. The first chunk is the rectangle that we’ve labeled here, and the second chunk is this triangle. So let’s label each chunk. Let’s say that the rectangle is labeled as area one and the triangle is labeled as area two. Then we can say that the displacement of the object which we will call 𝑠 is equal to the total area under the line. And so, we can say that this total area is equal to area one plus area two. So let’s start by finding area one, the area of the rectangle.

We can recall that the area of a rectangle is equal to its width multiplied by its height. Now the width of the rectangle is equal to 15 minus zero seconds, which ends up being 15 seconds. And the height of the rectangle is equal to 20 minus zero meters per second, which ends up being 20 meters per second. Now in both of these cases, we’ve remembered to include the units because remember the horizontal axis is the time axis and the vertical axis is the velocity axis. But anyway, so we can say that area one is equal to the width, which is 15 seconds, multiplied by the height, which is 20 meters per second. And when we evaluate this, we find that the unit of seconds here cancels with the unit of seconds in the denominator. And what we’re left with is the unit of meters.

So area one, the area of the rectangle on the graph, is representing 15 multiplied by 20 meters. And this ends up being 300 meters. Now, these 300 meters or, in other words, the area of the rectangle — area one — represents the total distance covered by the object if it had not accelerated and if it instead had stayed at a constant 20 meters per second all the way through its journey. But we know that the object is accelerating. It’s speeding up. And so, it’s going to cover a larger distance than usual. And that extra distance covered due to the object’s acceleration is given by area two, the area of the triangle. So let’s go about calculating area number two now.

To do this, we can recall that the area of a triangle, which we will call 𝐴 subscript triangle, is equal to half multiplied by the base of the triangle multiplied by its height. Now the base of the triangle is simply the same length as the base of the rectangle. It’s equal to 15 seconds. And the height is this distance here or, in other words, this length here. And that length ends up being 50 minus 20 meters per second. And so, in other words, the height of the triangle is 30 meters per second. And so, now, we can say that the area of the triangle — area number two — is equal to half multiplied by the base which we’ve seen is 15 seconds multiplied by the height which we’ve seen is 30 meters per second. And when we evaluate this, we see once again that the unit of seconds cancels with the unit of seconds in the denominator here. So we’re left overall with the unit of meters. And all of this evaluates to 225 meters.

And so at this point, we’ve calculated area one and area two. So now all that’s left for us to do to calculate the total displacement of the object as it accelerates is to add the two quantities, area one plus area two. But before we do this, it’s worth noting that we’ve said that we’ve calculated the area under this graph. And yet, each of these areas have units of meters. Well, the reason for this is that we’ve actually calculated the physical area underneath this curve, but these areas are representing a displacement. And in order to find these physical areas, we’ve multiplied units of meters per second by seconds. In other words then, these quantities do not represent areas in real life. They represent a displacement. The only area that we’re actually thinking about is if we were to draw this graph out on a piece of paper, then the actual physical area on that piece of paper underneath the graph is what we’ve calculated.

But anyway, so, coming back to our calculation, we can say that the displacement of the object is equal to area number one which is 300 meters plus area number two, which is 225 meters. And all of this ends up being 525 meters. So at this point, we found the object’s displacement.

However, there is another way that we could have done it. We could have used the following kinematic equation: 𝑠 is equal to 𝑢𝑡 plus half 𝑎𝑡 squared, where as we’ve already seen 𝑠 is the displacement of the object. 𝑢 is the initial velocity, which we know already is 20 meters per second. 𝑡 is the time over which the object accelerates, which once again we know it’s 15 seconds. And 𝑎 is the acceleration of the object itself, which we’ve calculated to be two meters per second squared. And using this kinematic equation will give us the exact same answer. Try it for yourself. Pause the video and use these quantities in this kinematic equation to find the displacement of the object. For now, though, we found the answer to the final part of our question. The object’s displacement is 525 meters.