Question Video: Interpreting a Velocity-Time Graph Physics • 9th Grade

The graph shows the change in velocity of an object with time. The object reverses its direction of motion twice during the 60 seconds for which it moves. The initial direction of motion is toward the north. After 60 s of motion, is the object moving north or south?

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Video Transcript

The graph shows the change in velocity of an object with time. The object reverses its direction of motion twice during the 60 seconds for which it moves. The initial direction of motion is toward the north. After 60 seconds of motion, is the object moving north or south?

In this question, we’re being asked to consider a graph that plots velocity on the vertical or 𝑦-axis against time on the horizontal or 𝑥-axis. If we look at the scale on the vertical velocity axis, we can see that there are both positive values and negative values of velocity. To understand this, we can recall that velocity is a vector quantity. And this means that it has both a magnitude and a direction. In this question, we’re told that the initial direction of the motion is toward the north.

We can see from the velocity–time graph that initially, so that’s at a time value of zero, the velocity has a positive value. Since we’ve been told that at this point in time the object is moving toward the north, then that means that in this case we’re treating north as the positive direction of motion. On the graph that we’re given then, whenever the velocity has a positive value, so that’s whenever the blue line is above this horizontal axis, the object is moving toward the north.

Now, if north is the positive direction, then this makes the opposite direction, which is south, the negative direction. Then on the velocity–time graph, whenever the velocity has a negative value, so that’s whenever the blue line is below the horizontal axis, then the object is moving toward the south. We can notice that in the question we’re told that during the 60 seconds for which the object moves, it reverses its direction of motion twice. On the graph, that’s these two positions where the line crosses the horizontal axis, which corresponds to a velocity value of zero.

If we consider the first of these two points, we can see that initially, before this instant in time, the object has a positive value of velocity, which means that it’s moving toward the north. Then, as time goes on and we get closer and closer to this point, the magnitude of the object’s velocity decreases. So it’s still moving toward the north, but it’s getting slower and slower. Eventually, at this point where the line crosses the axis, the object instantaneously has a velocity of zero. At this point then, the object has stopped moving north, and it immediately starts moving back in the opposite direction to the south.

As the object’s velocity becomes more and more negative, it’s increasing the magnitude of its velocity in the negative southward direction. It’s then the same idea again at this second point in time at which the direction of motion gets reversed. We can see that before this time the object’s velocity is negative, which means that it’s moving toward the south. The magnitude of that negative velocity decreases as the object gets slower and slower. And then at this point, it reverses its direction and starts moving back toward the north, getting faster and faster as it does so.

Okay, now that we’ve taken some time to understand what it is that the graph is showing us. Let’s turn our attention to this first bit of the question.

We’re being asked which direction, so north or south, the object is moving in after 60 seconds of motion. If we look on the velocity–time graph at a time value of 60 seconds, we can see that at this point the blue line is above the horizontal axis. That means that the velocity of the object at a time of 60 seconds has a positive value. Now we’ve already established that north is the positive direction, which means that a positive velocity means the object is moving toward the north. We can say then that after 60 seconds of motion, the object is moving north.

Let’s now clear some space on the board and look at the second part of the question.

After 60 seconds of motion, what is the distance of the object from its starting position?

Okay, so the second bit of the question is asking us how far away the object is from its starting position after it’s been moving for 60 seconds. The first thing that we need to be clear on here is that this is not the same thing as the total distance traveled by the object during these 60 seconds. The reason for this is that the object reverses its direction during this time. The object is initially traveling to the north, and then it turns around and starts moving back to the south for a bit before changing direction once more and going back toward the north.

The total distance traveled by the object during this time is equal to the distance that it travels to the north plus the distance that it then travels to the south plus the distance that it travels to the north for the second time. However, the quantity that we want to find is the distance of the object away from its starting position that it ends up after 60 seconds of motion.

So if these three arrows represent the displacement of the object during the three sections of its motion, then the object starts out at this position here, it moves north, then south, and then north again, ending up at this position here. If we draw an arrow from the start position to the end position, then this arrow represents the overall displacement of the object during these 60 seconds. The distance that the object ends up away from its starting position is then the length of this arrow. In other words, the quantity that we’re trying to find here is the magnitude of the displacement of the object after 60 seconds.

In order to work this out, we’re going to want to make use of this kinematic equation. This equation says that if an object takes an amount of time 𝑡 to accelerate from an initial velocity 𝑢 to a final velocity 𝑣, then the displacement 𝑠 of the object after that time 𝑡 is equal to a half times 𝑢 plus 𝑣 times 𝑡. Now the kinematic equations come with a couple of strings attached, specifically the object has to be moving in a straight line and it has to have a constant value of acceleration.

In this question, the object is only moving north and south. And since both these directions lie along the same line, then we can safely say that the object is moving in a straight line. However, the other condition is a little more tricky. We can recall that if an object’s motion is shown on a velocity–time graph like this, then the slope of that graph gives us the acceleration of the object. The slope of this velocity–time graph changes first at this point and then again at this point here. So then this graph doesn’t have a single constant slope for all time values, which means that the acceleration is not constant throughout the 60 seconds.

However, we can notice that there is a constant slope between zero seconds and 20 seconds, and then there’s a different second constant slope between 20 seconds and 35 seconds, and a third constant slope again different between 35 seconds and 60 seconds. That means that we can split the graph up into three sections, which we’ve labeled as one, two, and three. Then within each of these sections, the acceleration is constant, and that means that we can use this equation to find the displacement for each section separately.

If we label the individual displacements during the sections one, two, and three as 𝑠 one, 𝑠 two, and 𝑠 three, respectively, then the total displacement after 60 seconds, which we’ve labeled as 𝑠 subscript 𝑇 is equal to 𝑠 one plus 𝑠 two plus 𝑠 three. Now a good thing about using this kinematic equation is that it means we don’t need to get caught up worrying about whether the object is moving north or south at any particular time.

On the right-hand side of the equation, the velocities 𝑢 and 𝑣 are vector quantities. This means that they have an associated direction. And as we’ve already seen, velocities to the north will be positive and velocities to the south will be negative. The effect of this is that the whole business of whether the object is moving north or south is automatically taken care of for us in the signs of these two quantities. We can notice that all three of these quantities on the right-hand side of the equation are quantities that we can read off from the velocity–time graph for each of the three sections of motion.

Let’s begin by looking at the first section of the graph. That’s the section between a time of zero seconds and a time of 20 seconds. In this case, we’ll label all the quantities in this equation with a subscript one. The amount of time 𝑡 one that this first section of motion lasts for is equal to the time at the end of the section, which is 20 seconds, minus the time at the start of it, which is zero seconds. 20 seconds minus zero seconds is simply 20 seconds. So this is our value for the quantity 𝑡 one.

We can see from the graph that the object starts this first section with a velocity of positive 10 meters per second. So that’s our value for 𝑢 one, the initial velocity of this section. Then the final velocity at the end of this section is the velocity at this point on the graph. We can see that this is equal to 30 meters per second. Again, this value is positive, and we have then that 𝑣 one is equal to 30 meters per second.

When we substitute these values into this equation, we find that 𝑠 one is equal to a half multiplied by 10 meters per second plus 30 meters per second multiplied by 20 seconds. Adding together the two velocities of 10 meters per second and 30 meters per second, we get a result of 40 meters per second. Since the velocity is in units of meters per second and the time is in units of seconds, then the seconds and per second will cancel out. And that just leaves us with units of meters. We have then that 𝑠 one is equal to a half times 40 times 20 in units of meters. Evaluating this expression then gives a result for 𝑠 one of 400 meters.

Let’s clear some more space now so that we can do the same thing to find the values of 𝑠 two and then 𝑠 three.

Since we are now considering the second section of the graph, we’ve labeled all the quantities in this equation with a subscript two. This second section of the graph is this bit here with a downward slope between 20 seconds and 35 seconds. We have then that the length of time 𝑡 two is equal to 35 seconds, that’s the time at the end of the section, minus 20 seconds, the time at the start of it. This works out as a value of 15 seconds.

The value for 𝑢 two is the initial velocity at the start of this section. So that’s the velocity at this point here on the graph. This velocity must be the same as the final velocity at the end of the first section, which, we already found, has a value of 30 meters per second. The final velocity 𝑣 two for this section is the velocity at this point here on the graph. We can see that this has a value of negative 10 meters per second. We have then that 𝑣 two is equal to negative 10 meters per second.

Substituting these values into this equation, we end up with this expression here. Looking at the velocities, we can see that we’ve got 30 meters per second plus negative 10 meters per second. We can write that more simply as 30 meters per second minus 10 meters per second. Doing the subtraction then gives a result of 20 meters per second. If we then cancel the units of seconds and per second, we have that 𝑠 two is equal to a half times 20 times 15 meters. This comes out as 150 meters.

Now that we found our value for the quantity 𝑠 two, let’s clear some space again so that we can work out 𝑠 three. We’re now looking at this third section of the graph between a time of 35 seconds and a time of 60 seconds. This means that the length of time 𝑡 three must be equal to 60 seconds minus 35 seconds. And that then works out as a time of 25 seconds.

The initial velocity for this third section of the graph, that’s the value of 𝑢 three, is the velocity at this point here. And that’s the same as the final velocity for the second section, which we found had a value of negative 10 meters per second. Then the final velocity 𝑣 three is the velocity at this point. So that’s the point that corresponds to a time of 60 seconds. We can see that the velocity at this point has a value of positive 25 meters per second.

Substituting these values into the kinematic equation, we get this expression for 𝑠 three. Negative 10 meters per second plus 25 meters per second works out as 15 meters per second. And then multiplying this by a half and by 25 seconds, we find that 𝑠 three is equal to 187.5 meters. We are now ready to use this equation to find the value of 𝑠 subscript 𝑇. That’s the distance of the object from its starting position after 60 seconds. Let’s clear some space one last time so that we can do this.

If we take this equation and substitute in that 𝑠 one is equal to 400 meters, 𝑠 two is 150 meters, and 𝑠 three is 187.5 meters, we end up with this expression for the quantity 𝑠 subscript 𝑇. Adding together these three terms, we get a result of 737.5 meters. We have found then that after 60 seconds of motion, the distance of the object from its starting position is equal to 737.5 meters.