Friday, October 11, 2013
Week 7 Blog, Due Sunday, October 13th, by 11:00 pm
Last week we started Unit 3: Uniform acceleration. We are still graphing position vs. time(p-t) and velocity vs. time (v-t). I want you to compare and contrast the p-t and v-t graphs from the constant velocity unit (unit 2) with p-t and v-t graphs from the uniform acceleration unit (unit 3). Discuss the differences and explain why you think the graphs look differently (specifically, what is going on with the object's motion to cause the change in the graphs.) Since I am posting this blog late, I am going to give you a few extra days to get it done. Your comments and replies need to be posted by Wednesday night. If you have questions, please let me know. Remember, if you have trouble posting, you can post your comments and replies on edmodo.com under the link to the blog. You can also email your response to me through the district website or gmail, respond on twitter, or text your responses. In order to receive full credit, you must post an ORIGINAL comment PLUS a REPLY to another person's report. Keep it friendly and clean and most of all, no texting language. Happy Blogging!
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ReplyDeleteIn Unit 2, both the position vs. time and velocity vs. time graphs displayed linear equations, as the velocity of each buggy was constant, only differentiating in directions and relative speed in respect to the other buggy. However in Unit 3, only the velocity vs. time graphs displayed linear equations, while the position vs. time graphs varied greatly, as the push carts accelerated, decelerated, and reversed directions, causing significant changes in position, while leaving velocity linear at all times.
ReplyDeleteThe only particular thing that my graphs had a difference in was the position vs. time graph that we did for Unit three due to the fact that we had to push the cars the opposite direction at an inclined plane.Therefore the car had a decreasing negative slope on the p-t graph that was recorded. Everything else for both units were the same since majority had a constant increasing velocity with a positive slope.
ReplyDeleteThe same happened in our group when pushing the car in an opposite direction, the p-t graph was the only one with an actual change.
DeleteEither having a constant or a changing slope effects how the graph of the v-t or p-t looks.
DeleteThere are also many formulas that you can use to figure out the different ways to to relate acceleration to variables. Examples of those formulas were on the last packet we did.
DeleteThere wasn't really a difference between the p-t graphs & v-t graphs in unigrapht 2 because they were both linear. On the other hand in unit 3 only the v-t graph remained linear and the p-t graph changed depending on how you pushed the object.
ReplyDeleteYes, when graphing the position of an object relative to time, position is entirely dependent on its starting point, velocity, and acceleration.
DeleteYup, the P-T graph in unit 3 demonstrated a non-linear relationship between x and y. The reason for this can be traced back to the object's constant acceleration.
DeleteThe graph grows exponentially that is why the graph isn't linear.
DeleteYeah because the initial push made it accelerate and as it went up the ramp it decelerated.
DeleteYeah a lot of people were thrown off by the fact that we were adding acceleration into the mix.
DeleteThere was a difference between unit 2 p-t & the unit3 v-t because in unit 3 we actually pushed the car from both side of the ramp, inlined or not.
ReplyDeleteAlso that the unit 2 was linear and unit 3 was exponential.
DeleteTrue, Joceline. Because we gave that little blue car a push, the car either accelerated or decelerated, which is what unit 3 was all about.
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ReplyDeleteThe notable difference from unit 2 (constant velocity) and unit 3 (constant acceleration) was that the graph of P-T had a linear relationship in unit 2 and non-linear in unit 3. This can be explained by considering the velocity that the object traveled in reference to these graphs. In Unit 2 the velocity was constant; therefore the graph of P-T would have been linear since the velocity had remained the same throughout the total time. However, in unit 3 the velocity is no longer constant but instead steadily increasing or decreasing because the object has a constant acceleration, thus leading to a non-linear relationship of x and y in the P-T graph of unit 3.
ReplyDeleteThanks a lot for explaining this to me in class today ! :D
DeleteThe difference from the Position graph in Unit 2 to the one in Unit 3 is that in Unit 2's is usually a linear graph while Unit 3's is usually growing exponentially. And the Velocity graph for Unit 2 was usually horizontal in comparison to Unit 3 which tended to be linear graphs.
ReplyDeleteThe differences and similarities between Unit 2 & 3. A similarity was that in both Unit 2 & 3 the position vs time graphs were both linear graphs. A difference was that the velocity vs time graphs were not both linear. In Unit 2 the velocity vs time graph is linear but in Unit 3 it is not. Uniform Acceleration (Unit 3) is when the speed of an object changes at a constant rate. In Uniform acceleration the direction changes but the acceleration remains constant. due to this change, the graph for velocity vs time would be an exponential graph.
ReplyDeletethank you i was looking for the word that means increasing quickly which was exponentially
DeleteI thought that only the position graphs in unit 2 were linear not in unit 3.
DeleteUnit 2 it's that of the position is going linear and the velocity. How constantly it goes. However in unit 3 the acceleration is on incraments and how it goes faster by second with the little car and goes faster than before. Therefore shows a curve and becomes an exponential graph rather than constantly linear.
ReplyDeleteunit 2 the Velocity was constant. unit 3 was constant acceleration. With this acceleration the object would get faster and faster or slower and slower depending on where it started. the graph would not be linear because the car would have traveled more distance (position) as time increase and will travel more and distance at each second it moves.
ReplyDeleteI think the most obvious difference between the unit 2 and unit 3 graphs was that the Velocity graph in Unit 2 always remained horizontally flat with a slope of zero; it remained constant. In Unit 3 velocity graph was now linear with a slope either less than or greater than zero. There were no longer any horizontal lines on the v-t graph.
ReplyDeleteYour right Immer it's obvious with the velocity graphs in unit 2 would be horizontal meaning the object has a constant velocity.
DeleteIn unit 2 & Unit 3 we are still dealing with linearization with both graphs. However in unit 3 the velocity vs. time graphs can determine the slope of the velocity(acceleration)
ReplyDeleteTrue! So basically linear.
DeleteAna M.
Well, you can always look at a line and determine its slope. In unit 2, the velocity graphs were horizontal meaning that the slope was zero, therefore, the acceleration was zero.
ReplyDeleteOkay, well the way you explained to me was that in Unit 2 the position graph was curved and the velocity graph was horizontal. Whereas, in Unit 3 the uniformed acceleration was horizontal and the velocity graph was diagonal. So to wrap it all up in Unit 2 the graphs are more linear and the graphs in unit 3 are more curved.
ReplyDeleteSo in Unit 2, the position graph is curved because if the velocity is moving at a constant rate the position graph is also moving at a constant velocity making the graph curved. And in Unit 3, since the velocity is increasing the acceleration is positive.
Honestly I didn't understand the graphs during unit 2. Until we got to unit 3 with acceleration and we got the Kinematics Graphs I now understand the graphs now. With position and velocity the graphs the postion graph would be horizontal or diagonal and the the velocity would just be horizontal. Once we got to unit 3 with acceleration the position graph would be at a curve, the velocity would be a a diagonal and the acceleration graph would be at a horizontal line. Also we have to determine is the graph is speeding up or slowing down postive or negative direction.
ReplyDeleteTo determine if the object is speeding up or slowing down we can look at how the slope changes along a curve line from the position graph. For example, if it goes from small steep to greater steep, then it is speeding up and if vice versa it is slowing down.
DeleteI was and honestly still am a bit confused as well with understanding the graphs. I did understand the way that the curve could be utilized in order to determine changes in the slope.
DeleteThe difference between unit 2 and unit 3 took me a bit of a struggle to understand it. On unit 2 it was uniformed velocity and in unit 3 it dealt with uniformed acceleration.
ReplyDeleteI agree. I struggled to understand the difference but now I realize that the same relationship p(t) and v(t) had in unit 2 is the same relationship that v(t) and a(t) have in unit 3.
DeleteThe p(t)-v(t) graphs from the constant velocity unit showed no acceleration since the position graphs were always linear thus representing constant velocity, therefore the velocity graphs had only flat horizontal lines. On the other hand, the p(t)-v(t) graphs from the uniform acceleration unit clearly show an increase or decrease in acceleration, because now the position graphs show curve lines with positive or negative slope, thus causing the velocity graphs to represent linear lines instead of flat horizontal ones.
ReplyDeleteThe difference between the unit 2 and unit 3 graphs were the relationships in the lines shown. In Unit 2, the line at rest signified that the velocity was left on constant. On the other hand, Unit 3 has a constant acceleration which allows us to infer that the velocity is either increasing or decreasing. We were also able to see differences between the relationships within each graph in terms of linearization.
ReplyDeleteThe difference between unit 2 and unit 3 was that unit 2 only dealt with position and velocity time graphs. Most of the time, the position time graphs were linear (diagonal lines). That then told us that the velocity was constant (horizontal lines). However in unit 3, we dealt with position, velocity, and acceleration time graphs. Now, the position graphs were curved which then showed a linear velocity time graph(diagonal lines) and a constant acceleration (horizontal lines).
ReplyDeleteThe velocity graph in unit 2 had a slope of zero, which meant it was constant. However, in unit 3, the slope/velocity was greater or less than zero.. in other words, linear.
ReplyDeleteAna M.
The major difference between these graphs was that there was no acceleration present during unit 2 (hence, constant velocity). That is why the v-t graphs were horizontal lines. Also, the p-t graphs from unit 3 were less linearized than unit 2, due to the fact that the object is now accelerating or decelerating. During unit 2, when we would do labs, we always did them with buggies because they moved at a constant velocity. When we did labs during unit 3, we used the little blue car on a ramp because the ramp caused it to accelerate or decelerate (depending on how the car was set up).
ReplyDeleteIn unit 2 it was mainly content due to the zero slope and in unit 3 it's linear which it can be both positive and negative. It was bigger than zero or smaller than zero.
ReplyDeleteThe way they are different is that I kmow, by taking calculus last year, that velocity is the derivative of position so for most of the graphs I just take the derivative and imagine the graph. I just need more practice a bit but I should get it during class.
ReplyDeleteI'm taking calculus this year. I don't like it that much but I'm grateful that it is helping me in Physics.
DeleteIt was a little bit easier for me to understand the p-t graphs and the v-t graphs because I took calculus last year. The difference is that one is the derivative of the other.
ReplyDelete