Analyzing straight-line motion graphically

A particle moves along a number line not shown for t is greater than or equal to 0. Its position function, s of t, is shown in blue. So this is its position as a function of time. Its velocity function, v of t, is in red. That's velocity. And its acceleration function, a if t, is in green. All are graphed with respect to time t in seconds. With the graphs as an aid, answer the questions below. So that's what's going on here. So it's position as a function of time. Actually, let me just draw their number line that they did not depict just so we can really think about this. So let's say that's our number line. Let's say this right over here is 0. That's 1. That is 2. This is negative 1. So we're defining going to the right as the positive direction. So what's happening here? So at time equals 0 right over here, s of 0 is 0. And then as time increases, our position increases all the way until time equals 1. At time equals 1, our position is 2. So at time equals 1, our position is 2. And then our position, s of t, starts decreasing. So one way to think about it is, and you see we move up, we move to the right really fast. We get to 2. We stop at 2, and then we start moving to the left. So at time equals 0, the first second looks like this. We go zoom, oh, slow down and stop. And then we start moving the other way. And then we start drifting. Notice our position is decreasing. So our position is decreasing, but it's decreasing at ever slower, slower, and slower and slower rates. It's not clear if we'll ever get back to the origin. So that's what's going on here. And we see that no matter which graph we look at. Our position function is definitely telling that story. Our velocity function, which is the derivative of the position function, is telling that story. Out the gate, we have a high positive velocity, but we decelerate quickly. And at 1 second, our velocity is 0, and then we start having a negative velocity, which means we're moving to the left. So fast rightward velocity, but we decelerate quickly, stop at time equals 1 second, and then we start drifting to the left. And the acceleration also shows that same narrative. But anyway, let's actually answer the questions. The initial velocity of the particle is blank units per second. I encourage you to pause this video and answer that. Well, we just said the velocity, let's see, at time equals 0, we're at 8 units per second. So we'll just put 8 right over there. The particles moving to the right when t is in the interval, and since they're doing this as a member of, they really want this kind of in the set notation t is a member of the interval. Well, when are we moving to the right? We already went over that. We're moving it to the right-- there's a couple of ways to think about it. When our velocity is greater than-- so we're moving to the right when v of t is greater than 0. When v of t is less than 0, we're moving to the left. When v of t is equal to 0, we're stationary. So when is v of t greater than 0? Well, it's between t being 0, velocity is definitely positive, all the way to t is 1, but not including t is 1. So I'll put a parentheses there. So this is equivalent to saying-- so t is a member of that interval is equivalent to saying that 0 is less than or equal to t is less than 1. Once again, the first second, at time 0, we're going fast, slow down, and then stop for an infinitesimal moment, and then we start drifting back. That happens at time equals 1. We start drifting back. The total distance traveled by the particle for t in the interval between 0 and 3 is blank units. So once again, I encourage you to pause the video and try to answer that, the total distance. So this is interesting. Don't get distance confused with displacement. If I were to move three to the right and then I were to move back one to the left, the total distance I've traveled is four. The distance I traveled is four, while the displacement would be a positive. We could maybe put a minus 1 there. We moved one to the left. So three to the right, one to the left. Well, our displacement would be a net of positive 2. So they're asking, what's the total distance traveled? So between time 0 and time 1, we have moved two to the right. And then between time 1 and time 3, we move back, or to the left, we move half. So to the left, we move half. So what's our total distance? It's going to be two to the right plus half to the left, which is going to be 2.5 units. And we're done.