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Figure shows the displacement $(x)$ and time $(t)$ graph of the particle moving on the $x$ -axis.(a) The particle is at rest.(b) The particle is continuously going ...

Question

Figure shows the displacement $(x)$ and time $(t)$ graph of the particle moving on the $x$ -axis.(a) The particle is at rest.(b) The particle is continuously going along $x$ -direction.(c) The velocity of the particle increases up to time, $t_{0}$ and then becomes constant.(d) The particle moves at a constant velocity up to a time, $t_{0}$ and then stops.

Figure shows the displacement $(x)$ and time $(t)$ graph of the particle moving on the $x$ -axis.(a) The particle is at rest. (b) The particle is continuously going along $x$ -direction. (c) The velocity of the particle increases up to time, $t_{0}$ and then becomes constant. (d) The particle moves at a constant velocity up to a time, $t_{0}$ and then stops.



Answers

Figure shows the displacement $(x)$ and time $(t)$ graph of the particle moving on the $x$ -axis.(a) The particle is at rest. (b) The particle is continuously going along $x$ -direction. (c) The velocity of the particle increases up to time, $t_{0}$ and then becomes constant. (d) The particle moves at a constant velocity up to a time, $t_{0}$ and then stops.

So you know that the acceleration and turn to and Cho Santee you know that it starts from rest, so and 00 we now go to your zero The position of time. Zero three. Yeah. So first rule I have to 18. That's the rule. Cosette. Teo signed Too close inconstancy. This is No. This was a colossal time to we know that the velocity A time cereal which happens to me cereal A sequel signs, you know, close. See, which was just see? So that tells us two It's just fine. Two. Now, your team, which is a position of time, Tio is the rule. Yeah, Team two, which is the integral sign? No, which is negative. Kosaka two close some constant of integration was calling d What? Another three is equal to zero physical too Negative co signed here. Clos de Dozens. No, this is negative. One plus two that tells us that Big Siegel for yeah, in the tunes you have to Yeah, lto plus four. Oh, that makes sense.

We're given this graph here, which represents the velocity of some function, and we'LL be able to do it. It is first fine when the particles move forward and backward when it speeds up and when it slows down, so moves forward. What this means is that our lost is bigger than your room moves backward means when our velocity less than zero. So let's go ahead and just right he's out first. Go part way. Well, we're positive along this access here along here, So this can be from zero to one, actually. So the key is big in zero on zero to one so or burn is zero to one union with And then we're also positive from I to so this part over here on girl. So this would be five to seven backward. We want to find where this function is less than zero. So here, so actually do less than zero green. So all along here, the function is Weston zero from one to five, and it's not zero. So that will be where this function is lost. So then backward. There's going to be from one to five now, speeding up and slowing down So it's important that we're looking at the word speed and so or this we want to know when the velocity and acceleration are the same size and for slowing down. We want to know where they are opposite signs. So speed up on the block, so speed it up. So again, this means no, another way. I can say same side is beauty over eighty to be bigger than zero. So this is just saying that they're saying sign and so let's see. So I'm going to do what the acceleration is going to be in global reading over here. So a long this interval here we have a negative soap for velocity, so that means a of tea is going to be less than zero sense. Recall that the prime is the prime is a derivative of velocity is acceleration on this here and instead of buying, your mom's gonna write A is going to be zero on this interval here it's increasing. So is going to be bigger than zero. And the past this a is less than zero again. And then on this final interval here, a is equal to zero. So what we want to do is fine. Where velocity is both positive and exhilaration is positive as well as we're acceleration is negative and our philosophy is negative. So may be off on this Idol actually write or acceleration. So safety is bigger than zero where so this would be on so first three to six, So three to six and a T is lust zero on So the zero to two they're over there too. And then also six to seven Union six, two seven. So speeding up is going to be where they have the same sign. So what we want to do is take the intersection of these two graphs here. So the only intersection that zero one, five, seven and three six have in common are going to be by two six. So what? By two six on its back. And likewise, if we want to look at when I go for these are negative, I can look at the union of these two. I mean the intersection of these two. So let's see the intersection of one, five and zero to Well, this is going to be from one. You and the There's no other intersection between these two so we could go ahead and just re like this's one, too. A million six now for slowing down? Yep, Slowing down, down. We want to look at the opposite. We want to look at when beauty is this your over eighteen Or in other words, instead of taking the intersection. And they're both negative when they go. Father, we want the opposites. So first you want to look at the intersection. Oh, we're going forward. And when we're going or when we have a negative acceleration, so doing that will be so zero one intersect zero two suits would be zero one union and in five seven intersects with six six to seven, which is going to be by two six. And then we want to interest play this point backwards and when our acceleration is positive. And so the intersection of one five and three six is going to be three, two, five, and so we can rewrite this zero one union. You need five union by two six. And so it's important that we don't include five here, since these are open intervals and not closed, and Holtz let me just go ahead and box on so this is forward. This's for backwards. This is when we're speeding up, and this here is what we are slowing down. So that's everything for a part. Hey and part B, we actually already answered with this using our graph here. So this over here is B or already has became, and we want to determine when the particle is living at its greatest feed now for seeking. So no. When this is moving out's created speeds or remembers, bead is equal to the absolute value of the lost. And depending on how you look at this, there's two contenders that might be the largest value. So for the positive, it's at time is equal to zero. And for negative. It's on this interval right here. So these is the most negative and up top here is our most positive velocity, and depending on which one of these actually has the bigger distance from the X axis Moby, our velocity. So I actually tried to measure this, and at least when I was measuring it. The distance from here to here does not seem as big as the distance from zero to our negative value, so I'm not too sure what this is the but the largest speed occurs hers on the interval two two three. But since there's really no values here telling, that's what these velocities are supposed to be, depending on how you look at that, you may think the top one is the larger speed. And then, lastly, we want to know When does the particle stand still for more than an instant. So what the is that asking? So stand still means velocity is equal to zero and Mohr than an instant just means or a time bigger than just pretty much at one point. So let's see where our velocity is equal zero. So at one is equal to zero. But it's only zero hat one, because before and after it, we are no longer a zero. And then over here it five or the lost easy from zero. But then, after that, it's no longer a zero but bin from seven to line. This function is still zero, and so that our velocity will also be so the park particles ten still for more than an instant at seven to nine seconds. So this is for D and I productive box or answer over here for a seat

So the instantaneous velocity at any time is the slope of the ex position versus time graph at that specific time and so we can compute the slope. So for the instantaneous velocity, AT T equaling 1.0 seconds, this would be equaling 10.0 meters minus zero meters, divided by 2.0 seconds minus zero seconds. This is giving us 5.0 meters per second for that instantaneous velocity. And then we have tea equaling 3.0 seconds. But here the instantaneous velocity would be 5.0 minus 10.0 units meters divided by 4.0 minus 2.0 units seconds. And so this is giving us negative 2.50 meters per second for part C. We have the, uh the instantaneous velocity at T equal in four and 1/2 seconds and that would be equaling 5.0 meters minus 5.0 meters, divided by 5.0 seconds minus 4.0 seconds. And this is giving us zero meters per second and finally four part d. We have the instantaneous velocity at t equaling 7.50 seconds. So this would be equaling 20 minus negative 5.0 meters and this would be divided by 8.0 minus 7.0 seconds. And so this is giving us 5.0 meters per second. These would be our four answers. That is the end of the solution. Thank you for watching.

Given the graph of the position function for this problem, we're gonna look to see where the slopes are positive, negative or zero for the position function, and that's going to help us determine part a here. So what are the slopes? Positive? It looks like from zero up to time is 1.5, and then it slopes back down negative and then again from 6 to 7. So that's where therefore the velocity is positive. The particle is moving to the right or up. Let's do the next one. So velocity is negative. That would be between 1.5 and two, since the slope is negative there and then again from 5 to 6. Where is the velocity? Equal to zero? Well, that is where the slopes are zero and that occurs at two points at the 20.1 point five. So at time is 1.5 at time six and then along the whole interval from two 25 So the particle was stopped from two seconds to five seconds, and then it temporarily stopped because it turned around at 1.5 and six. Whenever an object turns around, its velocity is zero just for an instant. So using that we could then go ahead and jump to part C because C. In order to find that out, we're gonna look at the com cavity of this graph that will immediately tell us about the acceleration or the second derivative. So it's con cave, down from 0 to 1.5. So that means that it's acceleration is negative, since the slopes are decreasing. And so we'll go ahead and write that. So from zero toe, 1.5 negative acceleration does the slopes are decreasing, um, slopes air increasing the whole time from 5 to 7 because they're getting bigger, bigger, bigger or another way to say that is it's calm. Cave up. So notice this guy here, come cave down, the next guy con cave up. If it's concave up, acceleration is positive. And so that is from five 27 And then, of course, in between there, we could say the slope is neither con cave up nor con cave down from 2 to 5. So the acceleration is zero. Which makes sense, because if it stopped like we just said that it should have no acceleration on it. Um, so then we also want to graph the velocity function. Um, to do this again, we want to look at the slopes here, so we know that the velocity starts off with a really big value because the slopes air really big. But then it goes towards see euro. And then it finally hit zero, and then it actually dips into, um, negative values. The velocity does. But then all of a sudden it jumps because it's like we just all of a sudden went from, you know, negative speed. So then we just immediately stopped. And so that's why the velocity going to be a flat line Wrong. 2 to 5 seconds, I'll go ahead and that in the times here, you 25 seconds and then from there, what happens? The slopes then jump way negative. And then they dio positive and sorry. They go to zero and then, um, they increase Thio really positive values again. And so the velocity doesn't make much sense looking at it. But, um, we know that it was positive up until 1.5. So let's go ahead and put this 0.1 point five and then from 2 to 5 seconds the velocity with zero, which is why it's in the middle there. So you're really looking to see if it's negative that's moving backwards, velocities negative and above the X axis. It's positive here, and then it starts off a big negative because the slopes negative. So that's why we started down there and then went all the way up at time. Is seven to we switch back toe moving forward again with positive velocities? Um, using this graph, we can often find out these last two, which is when it's speeding up, slowing down anytime that it's moving towards velocity zero. It's actually slowing down, so it's slowing down the whole time from zero up until 1.5 0 to 1.5, because another way to say that is, um, it's velocity is positive, but its acceleration is negative. If B and a or opposites of each other, then it's actually slowing down. And then again, that happens from 5 to 6, because from 5 to 6, the velocities negative. But the acceleration is positive. And so let's put that from 5 to 6. It's also slowing down, and I think that's it. And then the other times here it's moving away from being at rest Azaz, well assed here, moving away from being at rest. So this is where it's speeding up. So from 1.5 to to beating up and then from six, that was that on there, 0.6 from 6 to 7. It is also speeding up because the velocity is positive here and the acceleration is positive. So if DNA have the same sign, then it's speeding up because they're working with each other, So I hope that was helpful there.


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