The position-time graph for a ball on a track is shown below.(a) What is the ball's velocity at 4s?(b) At what time(s) is the ball approximately travelling at -10m/s?(c) From t = 3 to 7s, what is the sign of the acceleration?

(a) +5 m/s; (b) t= 7 s; (c) negative

The position-time graph for a ball on a track is shown below. (a) What is the ball's velocity at 4s? (b) At what time(s) is the ball approximately travelling at -10m/s? (c) From t = 3 to 7s, what is the sign of the acceleration?

Hey, guys. Let's get some more practice with some curvy looking position time diagrams. We've got a position graph for a ball that's on a track, and the first part asks us what's the ball's velocity at t equals 4. Now we're asked for the velocity, which means we're looking for the slope, but it's at a specific moment in time, not between two points, Which means we know we're gonna be looking for the instantaneous velocity. So we're gonna be looking at the tangent lines of the graph. So let's just go ahead and figure out what the tangent line at t equals 4 is going to look like. So we're gonna look for the value in the graph, and now we have to figure out what the tangent line looks like. Remember the tangent line is if you were to sort of tracing, you know, the lines over here and keeping them straight. It's basically the tangent line, one way you can think of it, is if I got to this point over here, and instead of following the curvature, I would've just keep going in a straight line, then you would actually just sort of go off in this direction like this. And so we can figure out what this line looks like by just extending it a little bit further. So even though this is the line at a specific point, now we actually have to calculate the slope of this line. Now how do we do that? Well, remember, we can just use the two points that are at the ends over here and then calculate the rise of the run of this slope. So that means that the velocity, when t is equal to 4, we're still gonna use delta x over delta t, the rise of the run, even though this is gonna be the velocity at a specific point. So what is the rise of the run? Well, between these two points, this is the run. This is 2 seconds. And between these two points, the change in the position is actually 10. So that means my delta x over delta t is just 10 over 2, and I get plus 5 meters per second. The plus means that it's positive velocity, which makes sense because it's an upward slope. Alright. So that's pretty much it. Cool. So now moving on to the second part. Now at what time or times is the ball approximately traveling at negative 10 meters per second? So now the answer or the question is sort of flipped on us. Instead of using a time to find a velocity, now we're actually using the velocity to find the time. So it's gonna be the same concept, just a little bit backwards. So basically what we have to do is figure out at what times is the tangent line equal to negative 10. And so first we have to figure out what negative 10 even looks like on the graph. For instance, we know that 5 meters per second looks like this because we just calculated the velocity over here to be 5 meters per second. So what does negative 10 meters per second look like? Well, if I were to just grab any point in the graph like this I'm just gonna use this as an example. Then negative 10 meters per second, the slope of that would be as if I'm going 10 downwards, and then one over. So that means that the slope of the line would be looking like this over here. So basically I have to figure out where on the graph does the slope look like that line. So now I'm just gonna trace along the graph over here. I've already got my values here, here. I know at 5 it's just going to be a flat line, and then at 6 it starts to go a little bit negative, but this sort of this line here does not look as steep enough as this line yet. But what about at 7? Well, at 7, what happens is, if I were to go off in a straight line like this, then now what happens is, I'm gonna get a value, or I'm gonna get a slope that looks very, very, very similar to the value that I get over here. So remember, this is equal to negative 10 meters per second over here, and that actually happens right at this point. And so if you look at down here, this time, the time is gonna be 7 seconds where the velocity is equal to approximately negative 10. So that's our answer, t equals negative 7. Now finally for this last part here, what is, the sign of the acceleration from t equals 3 to 7 seconds? So basically, between these two points over here, what is the sign of the acceleration? Well remember, when it comes to acceleration on a position time diagram, that's related to the curvature. And we have two rules. Remember, if you are curving down like this, then it's going to be negative acceleration. And if you're curving up like this, it's going to be positive acceleration. So let me just go ahead and write that down. So negative acceleration like this, and then positive acceleration is when you are having a smiley face, right? And so, which one of these makes sense for t equals 3 to 7? It's obviously going to be negative acceleration here. So this is our answer choice. One way you can make sense of this, by the way, is we know that upward values or upward slopes are gonna be positive, flats are gonna be negative, and then downwards are gonna be, or sorry. Flats are gonna be, horizontal or 0. And then downwards are gonna be negative. So basically, the slopes are becoming more negative as you go along here. That's, that means that your acceleration is negative. Anyway, so that's it for this one. Let me know if you guys have any questions.

Physics with Algebra 때문에 힘들어하고 계신가요?

The position-time graph for a ball on a track is shown below.(a) What is the ball's velocity at 4s?(b) At what time(s) is the ball approximately travelling at -10m/s?(c) From t = 3 to 7s, what is the sign of the acceleration?

다음 시청

The position-time graph for a ball on a track is shown below.
(a) What is the ball's velocity at 4s?
(b) At what time(s) is the ball approximately travelling at -10m/s?
(c) From t = 3 to 7s, what is the sign of the acceleration?