 A rabbit is trying to cross the road, its velocity v as a function of time t is given in this graph, where right-wards is the positive velocity direction. At what time does the rabbit have the same position as t equal to 0? Answer with two significant digits. As always, why don't you pause the video first and try this one on your own. Okay, hopefully you have given this a shot. Now let's try and break down this question. Let's see what the question is trying to say. We have one rabbit which is trying to cross the road and velocity v is given as a function of time t and right-wards is the positive velocity direction. So this is the positive velocity direction. At what time does the rabbit have the same position as t equal to 0? What does this mean? We know that at t equal to 0, the rabbit is at a position of 0. It's at the origin. It means it has no displacement at time t equals to 0. Delta x is 0 at t equals to 0. The question is asking, another time instant could be 1 second, 1.5 second, 2, 2.5, 3, 3.5. At what time does the rabbit again have a displacement of 0? So when does delta x again become equal to 0? What I try to do with these questions is try to draw a rough setup. Actually draw a rabbit. It doesn't matter if it's a rabbit. You can draw a circle for an object. Draw a road and see what the question is trying to say. So here we have a rabbit and the rabbit is trying to cross this road. We can say that this time instant is t equals to 0. The rabbit is just beginning to cross the road. So this rabbit starts moving. It starts crossing the road, crossing the road, crossing the road. Maybe stops and then decides to come back. Decides to come back to where it was in the beginning. This means the displacement would be 0 because it moved to the right. There was some positive displacement. Then it moved to the left. There was some negative displacement. The difference between those two displacements, the positive displacement, let's show that with a green arrow, this much positive displacement and the negative displacement, they should be equal in magnitude. This difference must be equal to 0. And we know that in a VT graph, the area under the curve that gives us the displacement of the object in motion. Here it will give us the displacement of the rabbit. Part of the area we see is above the x-axis. That means there is some positive area, which means there is some positive displacement. And part of the graph is below the x-axis, which means there is some negative area. That means there is some negative displacement. So let us write this. Positive displacement minus the negative displacement that should be equal to 0. That is only when delta x will be equal to 0. And we need to figure out at what time this happens. So for the displacement to be equal to 0, positive area and negative area, those two should be equal. The difference between the two should be equal to 0. Let's try and figure out what the areas are. And also let's try and make a rabbit move in that way. Now at t equals to 0, this is t equal to 0. There is some positive displacement, there is some positive area. Let's shade that. Here it is. We can figure out the area because this is a triangle. Triangles area is half into base into height. So this would be half into 2, that is the base, 2 into 10, that is the height. The area is 10 meters. That means the displacement is 10 meters. So for the first two seconds when the rabbit has a positive displacement, the rabbit will be moving, it will be moving like this. It traveled 10 meters. This is when time is equal to 2 seconds and the displacement is equal to 10 meters. Now the rabbit changed its mind and started to go back because we see some negative area, some negative displacement. Let's try and shade that. We see one triangle, some rectangle, it's a combination of two shapes. We can divide it, maybe we can focus first on this triangle and think about its area. That would be half into 1, the base into height that is 10. So the area would be 5 meters. Okay, this means when the rabbit started to go back, it traveled a distance of 5 meters, maybe this much length, and it took one second to do that. So far at this instant, time would be equal to 3 seconds, this would be t equal to 3. And this arrow would be slightly shorter. Let's do that, let's change that. This would be till right here. Okay, till now the displacement at delta x is not equal to 0 because we see the rabbit is not at its original position. It must travel some more distance. And we know how much it has to travel now because it has traveled positive 10, negative 5. It means right now it is at a position of positive displacement of 5 meters. Right now the positive displacement is 5 meters. This is plus 5. So it has to travel back 5 meters also. And there is, we can see there's a lot of scope for that. Rabbit in fact moved, seems like it moved a lot in the negative side. But we only want to know when it traveled some extra 5 meters to the left, some extra negative displacement of 5 so that the net displacement again becomes equal to 0. And if we try to figure out that from the graph, we can see that we only have rectangle on our hands. The area is length into breadth. One side is 10, so if the other side is 0.5, then the area would be 5. That means if you think about this area right here, the length, the one side is 10 and this side is just 0.5. So the area here would be 0.5 into 10, that is 5 meters. And if you think about this time instant, this would be 3.5 seconds. Now this means the rabbit has moved some more to the left. It has traveled this much to the left, some extra negative displacement. Now the net displacement, we can see the length of the arrows also is the same to the right and to the left. Now the net displacement is equal to 0. And this happens at time t. This happens at time t equal to 3.5 seconds. Alright, you can try more questions from this exercise in this lesson. And if you are watching on YouTube, do check out the exercise link which is added in the description.