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blackstreet23
 one year ago
In this exploration, acceleration was assumed to be constant. Consider the case where the acceleration is changing with time. This is known as "jerk." Jerk is often used in engineering, especially when constructing roller coasters. Humans can be safely subjected not only to a maximum acceleration, but also to a maximum jerk. Which of Shakina's statements regarding jerk is not correct?
"If the acceleration is decreasing over time, the jerk is negative."
"Jerk is the area under the acceleration vs. time graph."
"The units of jerk are m/s3."
blackstreet23
 one year ago
In this exploration, acceleration was assumed to be constant. Consider the case where the acceleration is changing with time. This is known as "jerk." Jerk is often used in engineering, especially when constructing roller coasters. Humans can be safely subjected not only to a maximum acceleration, but also to a maximum jerk. Which of Shakina's statements regarding jerk is not correct? "If the acceleration is decreasing over time, the jerk is negative." "Jerk is the area under the acceleration vs. time graph." "The units of jerk are m/s3."

This Question is Closed

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0In this exploration, acceleration was assumed to be constant. Consider the case where the acceleration is changing with time. This is known as "jerk." Jerk is often used in engineering, especially when constructing roller coasters. Humans can be safely subjected not only to a maximum acceleration, but also to a maximum jerk. Which of Shakina's statements regarding jerk is not correct? 1 "If the acceleration is decreasing over time, the jerk is negative." 2 "Jerk is the area under the acceleration vs. time graph." 3 "The units of jerk are m/s3." 4 "Jerk is the slope of the acceleration vs. time graph."

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0@PlasmaFuzer @Shalante @freckles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok so I take it you are able to select multiple answers that are correct yes?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh darn nm you are supposed to select the INCORRECT statements.... ok then no problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So let me ask you a question... If velocity changes with respect to time does this mean there is an acceleration?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Assuming of course the mass of the object isn't changing that is :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok then what are the units of velocity and what are the units of acceleration.... and how are the two related?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0units of velocity are m/s and of acceleration m/s^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Correct... how are the two related to one another?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0they are related because the derivative of one of them gives us the other one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Exactley the derivative of the velocity function gives us the acceleration function.... tell me what does it mean to find the derivative of a curve at a point?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And before you start sweating this is actually a lot easier to imagine and explain than the integral before :D

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0umm. the derivative of a curve point?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes.... so we said that: \[a=\frac{ dv }{ dt }\] But to be more precise it is really: \[ a(t)=\frac{ d }{ dt } v(t) \] For an object who has a constant mass.... Therefore at the point t=T what is the meaning of : \[ a(T)=\frac{ d }{ dt } v(T) \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Geometrically, what is its meaning... Just like the integral is the area under the curve... what does this derivative at the point T mean?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you dont know you can just say btw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No judgements here :D we all have to learn it at some point I don't hold it against people for not knowing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Or getting it wrong or making a mistake for that matter

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0the derivative is the slope of the tangent

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Bingo nailed it.... So you should have all the pieces to get this one... Please go through the choices for me and explain to me why one first or why one doesnt please? :D

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0if the acceleration is decreasing, the jerk is negative? is that so?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I dont know is it? Please try and reason it out.... Start with this... We already agreed on the relationship between velocity and acceleration... If the velocity is decreasing what can you say about acceleration?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0ohh I think i got it. Since the derivative means the change of something over time. That means that the jerk has to be the change of acceleration over time. Right?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0if the velocity is decreasing is accelaration has a different sign

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0like if the velocity is negative and acceleration positive, it starts backwards, slow down and then goes to the front and viceversa

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0ohh so the same principle applies to acceleration?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Exactley correct, but note the use of the word DECREASING.... if the velocity is negative and is DECREASING it is getting more negative no?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And in this case the acceleration is also negative so it happens to be the SAME sign as that of velocity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Your situation is when velocity is negative and begins to INCREASE.... then it has a positive acceleration

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0but if the acceleration is positive and the jerk negative, the acceleration decreases?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but also if the acceleration is NEGATIVE and DECREASING the jerk is also negative Does this make sense?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0and if the acceleration is negative and the jerk is negative as well it becomes more negative, just as the same as (velocity and acceleration)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0change in acceleration over time is what is important not the absolute sign

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0i mean yes it does make sense!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh yes sorry didnt see that post... yes that is correct

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok so is option 1 true or false?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because we are of course looking for what is INCORRECT

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.02 is the false one right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Correct yes that is the answer

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0jerk is the derivative of the graph (acceleration vs time) and not the integral of it (area under the curve)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What would that represent?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0that would be velocity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Exactley on both posts :D :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I take it you got it but just to be thorough do the units make sense of jerk?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0You are a physics genius lol. I am going to look for you in any question i ask haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hahahaha why thank you, but unfortunately I am in school myself so I only come here when I have some spare time to help out :(

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0ohh no why the units are m/s^3?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But I do check my emails so if I have a moment and I see your post I will try to help, but sometimes I wont have the same amount of free time so keep that in mind

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ahh well let me ask you a question... why are the units of acceleration m/s^2 ?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ m }{ s } * \frac{ 1 }{ s }\]

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0so it would be (acceleration / time)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so you really mean: \[\frac{ \frac{m}{s}}{s} \] ... Yes exactley

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ \frac{m}{s^2}}{s}\]

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0ohh ok. Thanks a lot! Btw can i ask what is your major and if you are getting your Associates, Bachelors or Masters? lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im actually in a Ph.D program for plasma physics

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0sorry just got curious

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0ohhh ok. I am talking to a Doctor! that is so cool!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hahaha not yet no but maybe in a few years

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0But what is plasma physics about?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well I have always been interested in fusion energy... So I decided to pursue it and study man made plasmas specifically tokamak plasmas

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Its hard to give a simple explanation, but if you think it is interesting I invite you to check out wikipedia.... It is an invaluable resource to get a once over viewpoint of a topic, though I should not you always need to go deeper if you are asked to actually write a paper or do research since it is only really a summary of things.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This should be a good place to start: http://fusionforenergy.europa.eu/understandingfusion/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And this one also: https://en.wikipedia.org/wiki/Tokamak if you want to get a little more

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But I would suggest saving it till you have a little free time.... make sure to finish your hw first so you dont have it hanging over your head... that is a terrible feeling

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0haha yeah i know ...

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0and btw what motivated you to go for your masters?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmmm well that's a good question and it doesn't have a simple answer; but I guess it ultimately came down to the fact that I decided to pursue physics so I could help out in this field. If I were to stop at a bachelors level then I felt I couldn't really contribute the way I wanted to (or at least doing the things I liked to do). So yea, suffice it to say I always found this stuff interesting and it always drove me to want to know more.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A real turning point was when I stopped looking at hw problems and study problems as just a means to an end (a good grade) and started looking at them as valuable opportunities to practice my reasoning skills on a specific topic... If I couldn't do it then obviously I didn't understand it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But at the end of the day I'm with Feynman on this one.... I just got interested in this stuff and decided to want to pursue it. I really feel anyone can do it, it just comes down to finding an interest and passion for it that can carry you through the dull bits in order to be able to enjoy the more satisfying bits (which also happen to be the more difficult bits).

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0ohh i loved your answer is pretty enlightening!

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0btw do you have to go. Because it would be helpful if you could explain to me how to do the previous problem. The one I told you to skip. Without the simulator haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sure no problem, but give me a min I have been going back and forth from the kitchen I'm making something to eat.
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