Question

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."

Answer 1

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?

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."

Answer 2

@PlasmaFuzer

@Shalante

@freckles

Answer 3

Ok so I take it you are able to select multiple answers that are correct yes?

Answer 4

Oh darn nm you are supposed to select the INCORRECT statements.... ok then no problem

Answer 5

So let me ask you a question... If velocity changes with respect to time does this mean there is an acceleration?

Answer 6

Assuming of course the mass of the object isn't changing that is :D

Answer 7

yes

Answer 8

ok then what are the units of velocity and what are the units of acceleration.... and how are the two related?

Answer 9

units of velocity are m/s and of acceleration m/s^2

Answer 10

Correct... how are the two related to one another?

Answer 11

they are related because the derivative of one of them gives us the other one

Answer 12

right?

Answer 13

Exactley 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?

Answer 14

And before you start sweating this is actually a lot easier to imagine and explain than the integral before :D

Answer 15

umm. the derivative of a curve point?

Answer 16

Yes.... 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 :
\[ a(T)=\frac{ d }{ dt } v(T) \]

Answer 17

Geometrically, what is its meaning... Just like the integral is the area under the curve... what does this derivative at the point T mean?

Answer 18

If you dont know you can just say btw

Answer 19

No judgements here :D we all have to learn it at some point I don't hold it against people for not knowing

Answer 20

Or getting it wrong or making a mistake for that matter

Answer 21

the derivative is the slope of the tangent

Answer 22

Bingo 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

Answer 23

fits*

Answer 24

if the acceleration is decreasing, the jerk is negative? is that so?

Answer 25

I 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?

Answer 26

ohh 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?

Answer 27

Exactley :D :D

Answer 28

if the velocity is decreasing is accelaration has a different sign

Answer 29

Different?

Answer 30

like if the velocity is negative and acceleration positive, it starts backwards, slow down and then goes to the front and vice-versa

Answer 31

ohh so the same principle applies to acceleration?

Answer 32

Exactley correct, but note the use of the word DECREASING.... if the velocity is negative and is DECREASING it is getting more negative no?

Answer 33

And in this case the acceleration is also negative so it happens to be the SAME sign as that of velocity

Answer 34

i guess

Answer 35

Your situation is when velocity is negative and begins to INCREASE.... then it has a positive acceleration

Answer 36

but if the acceleration is positive and the jerk negative, the acceleration decreases?

Answer 37

Exactley :D:D

Answer 38

but also if the acceleration is NEGATIVE and DECREASING the jerk is also negative
Does this make sense?

Answer 39

and if the acceleration is negative and the jerk is negative as well it becomes more negative, just as the same as (velocity and acceleration)

Answer 40

change in acceleration over time is what is important not the absolute sign

Answer 41

yes it does!

Answer 42

i mean yes it does make sense!

Answer 43

Oh yes sorry didnt see that post... yes that is correct

Answer 44

I get it now :D

Answer 45

Ok so is option 1 true or false?

Answer 46

true

Answer 47

1-T

Answer 48

bingo.... next

Answer 49

because we are of course looking for what is INCORRECT

Answer 50

2 is the false one right?

Answer 51

Correct yes that is the answer

Answer 52

jerk is the derivative of the graph (acceleration vs time) and not the integral of it (area under the curve)

Answer 53

What would that represent?

Answer 54

that would be velocity

Answer 55

Exactley on both posts :D :D

Answer 56

So I take it you got it but just to be thorough do the units make sense of jerk?

Answer 57

You are a physics genius lol. I am going to look for you in any question i ask haha

Answer 58

Hahahaha why thank you, but unfortunately I am in school myself so I only come here when I have some spare time to help out :(

Answer 59

ohh no why the units are m/s^3?

Answer 60

But 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

Answer 61

Ahh well let me ask you a question... why are the units of acceleration m/s^2 ?

Answer 62

\[\frac{ m }{ s } * \frac{ 1 }{ s }\]

Answer 63

velocity / time

Answer 64

so it would be (acceleration / time)

Answer 65

so you really mean:
\[\frac{ \frac{m}{s}}{s} \]
... Yes exactley

Answer 66

Yes exactley....

Answer 67

\[\frac{ \frac{m}{s^2}}{s}\]

Answer 68

ohh ok. Thanks a lot! Btw can i ask what is your major and if you are getting your Associates, Bachelors or Masters? lol

Answer 69

Im actually in a Ph.D program for plasma physics

Answer 70

sorry just got curious

Answer 71

Its alright :D

Answer 72

ohhh ok. I am talking to a Doctor! that is so cool!!!

Answer 73

hahaha not yet no but maybe in a few years

Answer 74

But what is plasma physics about?

Answer 75

Well I have always been interested in fusion energy... So I decided to pursue it and study man made plasmas specifically tokamak plasmas

Answer 76

Its 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.

Answer 77

This should be a good place to start: http://fusionforenergy.europa.eu/understandingfusion/

Answer 78

And this one also: https://en.wikipedia.org/wiki/Tokamak if you want to get a little more

Answer 79

But 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

Answer 80

haha yeah i know ...

Answer 81

and btw what motivated you to go for your masters?

Answer 82

sorry P.h.D

Answer 83

Hmmm 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.

Answer 84

A 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.

Answer 85

But 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).

Answer 86

ohh i loved your answer is pretty enlightening!

Answer 87

btw 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

Answer 88

Sure no problem, but give me a min I have been going back and forth from the kitchen I'm making something to eat.

Answer 89

ohh no problem

Answer 90

are u here?