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Tuesday, November 27, 2012
Period 9 GRAPHICAL ANALYSIS SUMMARY
Ok. We have had the discussion. Generated graphs and equations. Interpreted slopes and intercepts.
Now you need to summarize and look for the key concepts.
Use this review.
Now you need to summarize and look for the key concepts.
Use this review.
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Mr. C could you explain number 11? I understand the slope is positive so if the answer E because both graphs have a positive slope as they decrease towards zero?
ReplyDeleteHey Mr Crane! I also agree with Jon could you please go over #11, I don't completely understand why the answer is Graph E. In Graph H, the slope is decreasing but is positive, but I don't get why E represents the next graph
ReplyDeletei thought the answer to number 4 would obviously be D and E but i got that wrong. can anyone explain
ReplyDeleteThe graphs shown are of velocity, not position, so you are not looking for the graph to just become more positive. The answers are D and F. For D, the graph starts at 0. This means that the slope of the position graph starts at 0, which means that it is horizontal. Graph D then rises to a positive number, which means a positive slope for the position graph. The position graph would end up looking like the right half of an upward parabola. Because it is concave up, it is positive. This is because the object would be traveling forwards, which you can tell because the position graph rises. For graph F, it starts at a positive number and goes to 0. This means that the position graph would start with a positive slope and flatten out. It would still be moving forward though, and so it the object would be traveling positively
Deletepositive is when they move in the same direction
ReplyDeleteJade, Number 4 asks for the velocity graph when an object is moving in a positive direction. Graph D is correct, but so is Graph F as it starts positively. Graph E starts as negative.
ReplyDeleteRather, Graph F has a positive slope, not that it starts positively.
DeleteWow, I looked wrong, both start positively :D.
DeleteMr Crane, can we go over 9, 10, 11? I don't understand how you decide if the acceleration is positive or negative? I guess I jus dont understand the positive and negative parts of this assignment..
ReplyDelete9: You're looking for a positive acceleration, so the key is to remember the concave up/concave down thing (http://dl.dropbox.com/u/56817828/physics%20concept.png). Then you just match the line you're given to the line you have in hand:
Delete- H matches the first half of the concave up line, which means it goes from (-) -> (0), which is a positive change. This means the acceleration is positive.
- I matches the second half of the concave up line, which means it goes from (0) -> (+), which a positive change, so the acceleration is also positive.
- J and K match the 1st and 2nd halves of the concave down line, which goes from (+) -> (0) and then (0) -> (-), both of which are negative changes. Therefore, the acceleration for both lines would be negative.
Answer: Graph H and Graph I
10: The line (J) has a negative acceleration [(+) -> (0)], so you're looking for a velocity-time graph that not only shows negative acceleration, but specifically the negative acceleration shown in Graph J (in English: a graph that starts positive and goes to zero).
- D starts at zero and goes up
- E starts negative and goes to zero
- F starts positive and goes to zero
- G starts at zero and goes down
Answer: Graph F
11: This question is basically #10 with a different graph. H has a positive acceleration [(-) -> (0)], so you're looking for a velocity-time graph that shows that particular positive acceleration (in English: a graph that starts negative and goes to zero).
- D starts at zero and goes up
- E starts negative and goes to zero
- F starts positive and goes to zero
- G starts at zero and goes down
Answer: Graph E
Kind of a weird question but we talked about velocity and acceleration as "increasing negatively", is there a difference between this and just simply decreasing and if so what is it?
ReplyDeleteThat question wasnt related to the review but I also dont really understand #11
DeleteSure.
DeleteIncreasing negatively is definitely a better statement contextually.
decreasing gives the idea that the velocity is just getting smaller, this is not always true.
increasing means the value is getting larger and
(+) (-) only indicate in what direction the increase is occuring
Hey Mr. Crane i was having trouble with number 9 i didn't understand how it was positively accelerating when the graphs seem to go in different directions
ReplyDeleteFor number 9 the correct answer is graph H and I. These make sense because for the acceleration to be positive the slope has to be positive. The slope of graph H goes from negative (at its starting point) and approaches 0 (at its end point) so it gets "more positive" even though it started as a negative. For I, the graph starts at a slope of 0 and goes to a slope of 1 so again, more positive.
DeleteYou need to draw the graphs.
ReplyDeletelook at and define the slopes...
you cant just look at the graphs...
you are looking at the direction of the graph which is not an indicator of the acceleration.
How do you determine the acceleration of a position vs. time graph? The easiest way is to draw the V vs.t graph
mr crane
ReplyDeletewhat are the main equations we are going to need to know for tomorrow's test?
i'm pretty sure it's this: basically know how slope applies to each graph so on a position graph m=V= p2-p1/t2-t1 and just use algebra to be able to solve for each variable ; understand that position or distance = velocity x time ; and then our last equation which is like the combination of all equations for each graph is
Deletep(footnote2) = .5a(t squared) + V(footnote1)T + P(footnote1) . . .sorry that was hard to type, hopefullly that makes some sense
How do you know if a position vs time graph starts in the fourth quadrant, or somewhere other than the first quadrant? Is it only for moving backwards or is there more to it?
ReplyDeletei thought position versus time graphs always started in the first because you either go closer or farther from an object. So unless you get closer to an object AND proceed to travel past it then i dont think it's ever in the fourth quadrant. It might not even be in the fourth in that case because distance cant be negative so it should just go back up even if you travel past it
ReplyDeleteMaybe I'm wrong, I honestly don't know, but I was looking at the example on the worksheet about backing up a car. For some reason I thought that started at 0 and went down, I think because it is going backwards maybe. Does that make sense or am I messing myself up because how else you you draw it?
Deleteactually it probably does go to the fourth i just havent seen it
ReplyDeletecan we use any notes/wksts on the test tomorrow?
ReplyDeleteprobably cant use notes im guessing
DeleteThis comment has been removed by the author.
ReplyDeleteDoes a object uniformly gaining speed in a negative direction have a position vs time graph of a concave down parabola, right?
ReplyDeleteCorrect, the second half of the concave down parabola. the slope starts at 0 and gets more negative, which means the speed (velocity) starts at 0 and gets faster, in a negative direction
DeleteCan someone explain #12?
ReplyDeleteanyone care to explain the whole concave up and down concept... confused
ReplyDeleteWhen something is concave up, it is an upward parabola. If you look at the start of that parabola, it has a negative slope. The slope continues being negative until it is at the bottom of the parabola, where it becomes 0. After it reaches 0, the slope becomes positive. So overall it goes from negative to positive, which makes it increasing. This goes for anything which is concave up. And the opposite is true for anything concave down (the slope decreases).
DeleteAlso confused for number 8 how H 's acceleration isnt increasing positively
ReplyDeleteNumber 8 doesn't ask about acceleration? For number 9, H's acceleration is increasing positively. the slope it going from negative to 0
DeleteI am confused on the equation which we were supposed to prove where the velocity final comes from because I have it copied down as Velocity 1 not delta velocity or anything like that. Did I copy it down wrong or is there something I am missing
ReplyDeleteIf a square root function gave the graph of a position function, would the velocity graph just be a positive linear line descending in the negative direction? Thus the acceleration graph would just be a horizontal line in Quadrant 1?
ReplyDeleteThe acceleration part of ur comment is correct, but i am not sure about the velocity part of it.
DeleteThe acceleration part of ur statement is correct, but i am not sure about the velocity?
DeleteAre we supposed to do all twenty problems of this homework u gave us mr crane
ReplyDeleteI'm confused about the homework. Did you mean pages or problems 122-133 on the study guide or textbook?
ReplyDeleteHey Mr. C this is not about the homework, but about if there were any assignments that we had to do during the break. I have not noticed any at all, and I don't want to assume that there isn't.
ReplyDelete