For 1c on the ws it is asking the instantaneous velocity for t=2s. How do we know what points to choose in order to find the change in position/ change in time
i tried looking up displacement and i found a few different ways to do it if you have a certain shape like a rectangle (lxw), but how would you calculate the displacement for 2e?
I'm still confused on how to find an instantaneous velocity? Also, how do you find displacement in a velocity vs. time graph if displacement has to do with position?
draw a tangent, pick points, calculate for instantaneous velocity
displacement is the change in position. the change in position over the change in time, so it can be related to velocity. so for 2e, use the equation 1/2at^2+Vo
How do you find a numerical value of acceleration? Is it just the slope of the velocity vs time graph? Would you start an acceleration vs time graph with the slope as the y value?
Mr. Crane, while doing the study guide I have questions that I definitely need to talk to you about, I am so regretful that I was part of the parcc test class. I am very confused on what instantaneous velocity is compared to just velocity. Can someone please explain this to me?
Jake, instantaneous velocity is if you choose one specific point on the graph and tell what the velocity is. Average velocity is what the velocity usually is throughout the graph.
For #2d I keep getting a negative answer, but I don't know if I either did it wrong or if I don't get how it applies. I used the v=at+starting velocity and plugged in my numbers, and I ended up getting two negative results. Is it supposed to be negative?
And is there an actual equation we could use for #4? I saw Sydney's response, but I didn't know if we went over an equation in class that would apply.
I'm also confused on #7i. How do you find out when the graphs will have the same slope? I noticed that they meet at 3 seconds, but I didn't know if that just meant they crossed paths, but not necessarily had the same slopes.
Daniel Nachtigall Mod 7 Can anyone explain how to convert a position time graph into velocity time and acceleration time? I don't understand how you know it at all...I'm clueless
The slope of a position v time graph would be y/x = d/t = velocity. To convert from position v time to velocity v time, just find the velocity at each second from the position graph and put a point on the velocity graph, then connect your dots. The same goes for velocity to acceleration; velocity over time = m/s/s = acceleration. If it helps you to think of it this way: Picture three graphs; position on the top, velocity in the middle, and acceleration on the bottom. Now picture four types of lines you might find on a graph; an evenly curved line, a straight, inclined line, a flat line not at zero, and a flat line on zero. Whichever line you see on the position graph, you will see the next one over on the velocity graph, and the line after that on the acceleration graph. For example: a curved line on the position graph would be a straight, inclined line on the velocity graph and a flat, nonzero line on the acceleration graph. (All of them would show a steadily changing speed.) I hope this helps and I didn't just confuse you even more. -Kevin Meglathery
For 1c on the ws it is asking the instantaneous velocity for t=2s. How do we know what points to choose in order to find the change in position/ change in time
ReplyDeleteYes, you have to draw the tangent at 2s. There is only one tangent@ 2s. A tangent also only uses 1 point.
DeleteThis comment has been removed by the author.
ReplyDeleteCan someone explain 2. B? I have no clue what I'm doing wrong
ReplyDeleteDan Nachtigall Mod 7
i tried looking up displacement and i found a few different ways to do it if you have a certain shape like a rectangle (lxw), but how would you calculate the displacement for 2e?
DeleteSlope=a for a velocity time graph.
DeleteAll you have to do is describe the slope of the velocity vs. time graph, because the slope equals the acceleration in this situation.
DeleteI'm not really sure what questions 2b-e are asking. What formula are we supposed to use? Is it the P=1/2at+Vt+P formula? What numbers do we use?
ReplyDeleteFind the acceleration using the v vs t graph 2b
Delete2e Look at your answer from the DAM #19 WS
So the a(t)+Vo
Deletei also dont know how to do number 4
ReplyDeleteI just subtracted .25 from 5.0 ten times
DeleteThat's the physics way!!!
DeleteI'm still confused on how to find an instantaneous velocity? Also, how do you find displacement in a velocity vs. time graph if displacement has to do with position?
ReplyDeletedraw a tangent, pick points, calculate for instantaneous velocity
Deletedisplacement is the change in position. the change in position over the change in time, so it can be related to velocity. so for 2e, use the equation 1/2at^2+Vo
But how do you calculate for it? is there a formula?
Deletethe slope formula
DeleteOh okay. For number 7, we should probably just make our own scale on the x and y axis?
Deletethe slope formula
ReplyDeleteWould the average velocity for number 7 just be the slope of the line, since the velocity equals the slope?
ReplyDeleteHow do you find a numerical value of acceleration? Is it just the slope of the velocity vs time graph? Would you start an acceleration vs time graph with the slope as the y value?
ReplyDeleteAlso, for 2 d, would you use 1/2at^2+Vot+Po to find the answer?
ReplyDeleteJake Babb, Mod: 7
ReplyDeleteMr. Crane, while doing the study guide I have questions that I definitely need to talk to you about, I am so regretful that I was part of the parcc test class. I am very confused on what instantaneous velocity is compared to just velocity. Can someone please explain this to me?
See the previous post on 1-14
Deletehttp://physics30.edcentre.ca/kindyn/lessoni_4_4.html
Jake, instantaneous velocity is if you choose one specific point on the graph and tell what the velocity is. Average velocity is what the velocity usually is throughout the graph.
ReplyDeleteFor #2d I keep getting a negative answer, but I don't know if I either did it wrong or if I don't get how it applies. I used the v=at+starting velocity and plugged in my numbers, and I ended up getting two negative results. Is it supposed to be negative?
ReplyDeleteAnd is there an actual equation we could use for #4? I saw Sydney's response, but I didn't know if we went over an equation in class that would apply.
I'm also confused on #7i. How do you find out when the graphs will have the same slope? I noticed that they meet at 3 seconds, but I didn't know if that just meant they crossed paths, but not necessarily had the same slopes.
Does P4 Tuesday mean that Mod 4's test is on Tuesday?
ReplyDeleteYes
ReplyDeleteHow would you go about explaining number seven letter i?
ReplyDeleteDaniel Nachtigall Mod 7 Can anyone explain how to convert a position time graph into velocity time and acceleration time? I don't understand how you know it at all...I'm clueless
ReplyDeleteThe slope of a position v time graph would be y/x = d/t = velocity. To convert from position v time to velocity v time, just find the velocity at each second from the position graph and put a point on the velocity graph, then connect your dots. The same goes for velocity to acceleration; velocity over time = m/s/s = acceleration.
DeleteIf it helps you to think of it this way:
Picture three graphs; position on the top, velocity in the middle, and acceleration on the bottom.
Now picture four types of lines you might find on a graph; an evenly curved line, a straight, inclined line, a flat line not at zero, and a flat line on zero.
Whichever line you see on the position graph, you will see the next one over on the velocity graph, and the line after that on the acceleration graph.
For example: a curved line on the position graph would be a straight, inclined line on the velocity graph and a flat, nonzero line on the acceleration graph. (All of them would show a steadily changing speed.)
I hope this helps and I didn't just confuse you even more.
-Kevin Meglathery
Mr. Crane where is the review packet answers link
ReplyDeleteI can't find the link either
DeleteI also can't find the answers
Delete~Emily Mack
DeleteYou have to click the word answers it's in bold and in uppercase
ReplyDeleteI clicked on that, and I don't think those are the answers to the review packet we got.
DeleteMorgan Schutz
Yeah for some reason it's not even letting me view it at all
DeleteWhen I click ANSWERS it says the file cannot be opened
ReplyDeleteI clicked answers and the packet you have attached isn't the review sheet
ReplyDeletemr. crane, the packet you posted is not the test review packet
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteI can't find the link to the answers for the review packet
ReplyDeleteMr. Crane, I am confused as to why we use the v=vt+.5at^2 for number 2e
ReplyDelete