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The graphs in Figure 13 represent the positions $s$ of moving particles as functions of time $t .$ Match each graph with a description:(a) Speeding up(b) Speeding u...

Question

The graphs in Figure 13 represent the positions $s$ of moving particles as functions of time $t .$ Match each graph with a description:(a) Speeding up(b) Speeding up and then slowing down(c) Slowing down(d) Slowing down and then speeding up

The graphs in Figure 13 represent the positions $s$ of moving particles as functions of time $t .$ Match each graph with a description: (a) Speeding up (b) Speeding up and then slowing down (c) Slowing down (d) Slowing down and then speeding up



Answers

The graphs in Figure 13 represent the positions $s$ of moving particles as functions of time $t .$ Match each graph with a description:
(a) Speeding up
(b) Speeding up and then slowing down
(c) Slowing down
(d) Slowing down and then speeding up

Here we have the graph of a position function and we want to know when the particle is speeding up and when it's slowing down, it'll be speeding up when the velocity and acceleration have the same sign, either both positive or both negative. And it will be slowing down when they have different signs one positive and one negative. So what we're going to do is break our time, intervals down and analyze what's happening with the velocity and acceleration in each interval. So we have 0 to 1 1 to 2, 2 to 3 and 3 to 4. Something is changing for each interval. So now we're going to find the sign of the velocity and the sign of the acceleration. When the position graph is increasing, the velocity is positive. So we see it increasing from 0 to 1, and we see it increasing from 3 to 4. When the position equation is decreasing, the velocity is negative, so we see a decreasing from 1 to 3. Now, the acceleration is the second derivative of position, and the second derivative relates to the con cavity of the curve con que but versus conch it down. If the curve is conquered down. The second derivative will be negative. And if the curve is Khan gave up, the second derivative will be positive. So we see that this curve is conclave, down from 0 to 2. So even negative acceleration. And then it's Khan gave up from 2 to 4, so we have a positive acceleration. Now let's look and see when both of these signs are positive. Velocity and acceleration are both positive from 0 to 4, and they're both negative from 1 to 2. So that's an indication of speeding up. So we have speeding up from times 1 to 2 and from times 3 to 4. Now, look for when they have opposite signs from 0 to 1 and from 2 to 3. So that means it's slowing down 0 to 1, 2 to 3. Okay, let's do something similar for the second graph. So again, we're going to be analyzing the velocity and the acceleration to figure out whether they're positive or negative. So we have the interval from 0 to 1 from 1 to 2 from 2 to 3, and from 3 to 4, we have the velocity and we have the acceleration. Okay, So this graph is decreasing from 0 to 3, and so it has a negative velocity from 0 to 3. This graph is increasing from 3 to 4, so it has a positive velocity from 3 to 4. This graph is conclave, up from 0 to 1, so it has a positive acceleration. Then it's Con Cape, down from 1 to 2, so it has a negative acceleration. Then it's called Cave up from 2 to 4, so it has a positive acceleration again. Now let's look for where the signs are the same. The signs are the same from 1 to 2 and 3 to 4, so that would be when it's speeding up. The signs are different from 0 to 1 and 2 to 3. That would be when it's slowing down.

Sound you D'oh! When you dio is call me to grab shown So it's going to be a rough sketch. I'm the first thing will do in the what appears to be a cubic function the function starting at the origin. But the origin was back down the Xing a second would look up again fish. And we know you know that it goes up to eat it and have an accident before. Okay, um and we also know that I mean, all we know for sure. Okay, Next, let's do the problem. Uh, here's the Block 16 problem picking up here. It has often exit. Put the fork down below. Uh, that's problem. Where And finally we have a line. Something like, yes, looks pretty run down like that. Okay. And we need to identify which grab for then velocity with conservative position whenever the acceleration. Um, any idea that velocity? Uh, shut up, Dave. One of the functions is, uh are you calling it condition what they wanted A Their version of function 50. A position that means that f property of velocity and second derivative is acceleration, block and exploration. It went with the position so that means that bunks, which had been different you twice, they're the coloration and we differentiate panem mules. The degree get smaller. For example, that's a our original function of X cube breath driven. It would be the degree to pound on the good of the powerful second derivative would be degree one part of it. That means that the function graph here with the smallest degree Yeah, the highest derivatives that it's mosque. Every bunch of here is becoming your function, Which means this must be examination the function of Croatian. Um, it's a degree. One function, one degree up is one derivative less. So this problem out here must be the after that if people off and I mean that cubic function on his position. Okay, so next we need to identify when the particle speeds up and want to slow down. Well, speeding up means that glossies increasing. And that means that huh? The loss of increasing that meaning the particle is accelerating or a commission is positive. You always do that. Look for the federation function. If in line, it's positive. And if you see positive after this, what? Which, uh, not exactly shown in the graph a little bit after two the functions eating up going off is increasing when, um after to the right of this excellent step in and going down to the mud. And we can often look at the proble for the problem of in velocity. And if the function of beating up that means that philosophy is or the smoke is increase and, uh, start me not everything that the Holy Father is if they wanted and negative.

So the party we want to drink the speed from figure 13. 30. Now, we can say that for the speed, the speed of the object will be equal to the magnitude of this slope of the I don't displacement or position rather vs time graph, uh, at particulate at that particular point. Now, looking at the graph, we can see that the lowest speed would be at sea an F because the magnitude of the slope would be zero in that case. So we have c equaling the speed at F and now thespian at B and E would be the next highest speed. So they would say that this would be less than be equal to E. And the highest speed would be at a and at D, so this would be less than a equaling d. So this would be for the speeds four part to be. The velocity is the same as the speed, except that we now account for direction. So here velocity, um, of course has a magnitude and direction. So now we will be taking count the magnitude of the slope of the Exorcist T graph. However, we need to take account if if if the slope is negative or if the slope is positive. So here the highest would be D because that's rather the lowest would be D because it has the highest magnitude of the slope. However, it is negative. So d would be our lowest philosophy, huh? However high magnitude speed next would be e having a lower magnitude, however still negative. This next would be c having essentially no magnitude essentially no direction in this case either. So see a equaling F because they're both Ah, horizontal slope and this would be less than be where here we have a positive slope. However, it is not, as the magnitude of the slope is not as great as the magnitude of the slope at point A, where a has the highest velocity, highest magnitude, and it is in the positive direction that for part C, now we have the acceleration and the acceleration. It's simply the rate, uh, change of the velocity. So here it's simply taking the essentially the second derivative of the Exorcist T graph, and we can say that here it is going to be proportional to the curvature of the Exorcist T graph so proportional to the curvature, uh, X versus T graph. And again, the higher the curvature, the greater the acceleration. So in this case, the highest, the lowest acceleration where we are. Actually I have a turning point from positive to negative would be at point C where C has the highest curvature. However, it has the lowest acceleration because again, acceleration is a vector we have to take into account direction. And at this point at this point, see, we haven't we're going from a negative to a positive. Next would be at B. Next would be a A and again, this is simply because there the curvature at a is is very low. However, it's still positive. That's why this would be equal to the curvature at D Point E would have a positive acceleration because now we're going from a negative velocity toe, a positive velocity, and we have the greatest acceleration at F because this is actually the turning point where the velocity changes from negative to positive. This would have the highest curvature on the graph, and it would have the same. They would have the highest acceleration. The magnitude, however, of the acceleration at F is equal to the magnitude of the acceleration at sea. However, they're on opposite sides simply because see has tthe e. C is negative and f it's positive that is the end of the solution. Thank you for watching.

All right. So this problem, we are given two graphs which depict the position of some sort of particle And what problems asking us to do is find the intervals in which this particle is speeding up and when it is slowing down. And so the key to doing this is finding the functions velocity and this acceleration, which we're going to do now. But first I feel like it will be good to talk about what speeding up and slowing down actually mean in the context of this problem. So all functions have a velocity and acceleration. When we talk about speeding up and slowing down, we need to remember that speed is different than velocity, where speed can't be negative, so speeds the absolute value of velocity. So when a particle has a positive velocity, so it's moving forward and has a positive acceleration, which means it's increasing, the velocity is increasing. That means we're speeding up because we are going in a positive direction and our acceleration is making us go more positive. It's increasing velocity. Similarly, if we have a negative velocity and acceleration is negative, that means we're going backward and our acceleration is decreasing the velocities was making it more and more negative. So even though the velocity is decreasing, our speed is increasing because we're going faster, backward and then for these other ones it's going to be slowing down because if they have a negative velocity and a positive acceleration, it means we're going backward. But the acceleration is making the velocity increase, so it's becoming a smaller negative number. We're moving backwards slower and same with the positive velocity and negative acceleration. We're moving forward, but the acceleration is negative. It's making us move forward slower, so wanting to keep in mind we're doing this problem is that if the signs for the acceleration and the velocity match, that means it's speeding up. But if the signs for the acceleration the velocity don't match, that means air particles going be slowing down. Let's hop over to part A this first problem and let's start marking the velocity and the acceleration of this function. So for the 1st 2nd here from zeros of one, we have a positive velocity because this graph is increasing and then we have a negative velocity for the next two seconds and we have a positive velocity after that because the part because probably up for the 1st 2nd down for the next two in and up for that last second. As for acceleration, there's little pneumonic or con cave up looks like a cup con cave down looks like a frown. So when it looks like it's smiling when was like a cup, the acceleration is positive. And what looks like a frown accelerations negative. So we can kind of see the frown for the 1st 2 seconds here along this path. Um, so that means for these 1st 2 seconds, we have a negative acceleration, and then we can see the smile for the last two seconds, which gives us a positive acceleration. So when it comes a speeding up and slowing down when you look at Wendy's signs match. So from 2nd 1 to second to their boast negative, which means we are speeding up from one 22 And that's how we read the interval from 1 to 2, with parentheses in common between them and in the union. To show there's another, uh, ranger that happens in their domain. Um, so the interval from 3 to 4, they're both positive, so it looks like this is when it's speeding up the domain or the interval from 1 to 2. You need the interval from 3 to 4, and it's slowing down everywhere else, so you can see from zero to a one. The velocity is positive of the acceleration is negative, so they don't match. And then from 2 to 3 as well. The velocities negative and the explorations positives they don't match. That means slowing down. This is our answer for the first problem. If we're looking into this 2nd 1 over here, let's see. We can do here. So the velocity for the 1st 3 seconds, it looks likely label this graph. The velocity for the 1st 3 seconds is negative because this is a decreasing function up until 2nd 3 when it starts going back up again. And then, as for acceleration, this is a little bit harder to see. So for this 1st 2nd right here, we can kind of see this left part of a smiley face. So that means it's conking up looking like a cup. The 2nd 2nd section, we can see the right part of a frowny face, so it looks like it's Khan came down and then for the last part is pretty easy to see everywhere where it's below that access only positive again. So it looks like we're speeding up from 1 to 2 again when her beaus negative and then from 3 to 4, when they're both positive and they're slowing down from 001 and from 2 to 3, when the velocity is negative and the acceleration is positive. So that's how you do problems when you're asking for speeding up and slowing down.


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