blackstreet23
  • blackstreet23
Shakina and Juliette discuss the relationship between the velocity vs. time graph and the displacement of a particle. Which of the following statements by Shakina is true? 1- "The displacement of a particle is the area between the curve of the velocity vs. time graph and the horizontal axis for a given time interval, with areas above the horizontal axis counted as positive and areas below the horizontal axis counted as negative." 2- "The displacement of a particle is the slope of the velocity vs. time graph." 3- "The displacement of a particle is the area in the first quadrant between t
Physics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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blackstreet23
  • blackstreet23
The question above is incomplete. Shakina and Juliette discuss the relationship between the velocity vs. time graph and the displacement of a particle. Which of the following statements by Shakina is true? 1- "The displacement of a particle is the area between the curve of the velocity vs. time graph and the horizontal axis for a given time interval, with areas above the horizontal axis counted as positive and areas below the horizontal axis counted as negative." 2- "The displacement of a particle is the slope of the velocity vs. time graph." 3- "The displacement of a particle is the area in the first quadrant between the curve of the velocity vs. time graph and the horizontal axis for a given time interval." 4- "The displacement of a particle is the area in the third quadrant between the curve of the velocity vs. time graph and the horizontal axis for a given time interval."
blackstreet23
  • blackstreet23
@abhisar @Shalante @freckles
anonymous
  • anonymous
Are you aware of the relationships between displacement, velocity, and acceleration given by calculus?

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anonymous
  • anonymous
I'm just curious because if you are not, then there isn't much more I can do but just tell you the answer which I would prefer not to do if I don't have to. Stating answers without explanations may save you on this problem, but in the long run lack of understanding will cause you to get things wrong on tests where it matters.
blackstreet23
  • blackstreet23
yes. Displacement is the integral of velocity right?
anonymous
  • anonymous
Correct, what does the integral of a function mean?
blackstreet23
  • blackstreet23
area under the curve
anonymous
  • anonymous
Why does it mean that? Pretend I don't know and explain to me where that idea comes from please :D
anonymous
  • anonymous
Think first, how do you calculate the area under a curve?
anonymous
  • anonymous
Please just state what you know, dont go referencing books or google.... consider me an interactive google in a sense on this topic :D
anonymous
  • anonymous
I don't see you typing which means you are looking up an answer, please I am curious to know where you stand currently
blackstreet23
  • blackstreet23
no, I was answering a friend's text. Sorry
blackstreet23
  • blackstreet23
So the area under the curve can be found by obtaining the integral of a function when is above the x -axis and continuos
anonymous
  • anonymous
Oh no problems :D :D
anonymous
  • anonymous
Correct... how do you integrate though? It isnt a magical process (strictly speaking).... do recall how an integral is defined?
blackstreet23
  • blackstreet23
as the summation of ....
blackstreet23
  • blackstreet23
taking the limit
blackstreet23
  • blackstreet23
sigma
blackstreet23
  • blackstreet23
delta xk
blackstreet23
  • blackstreet23
i dont remember totally lol
anonymous
  • anonymous
You have the pieces lets put it all together: \[\int\limits_{a}^{b} f(x) dx \approx \sum_{n=0}^{ N } f(x^{*}_{n}) \Delta x_{n}\] (one way it can be defined at least) Now the limit process you allude to would result in the width (delta x) tending to zero which implies the number of sub intervals (N) goes to infinity. Intuitively, this results from the fact that as each sub division decreases in size it takes more subdivisions to make up the entire interval.
anonymous
  • anonymous
But the formal limiting process isnt strictly needed here for the explaination because the main concept here lies in that approximate definition. Please tell me what f(x) stands for here.
anonymous
  • anonymous
And for simplicity the * on the x we will just take to mean the function evaluated at the midpoint of the interval delta x
anonymous
  • anonymous
What does it stand for in terms of the problem at hand?
blackstreet23
  • blackstreet23
f(x)
anonymous
  • anonymous
yes
blackstreet23
  • blackstreet23
the change of velocity over time?
anonymous
  • anonymous
correct
blackstreet23
  • blackstreet23
or an exact point of velocity in a certain time?
blackstreet23
  • blackstreet23
so acceleration?
anonymous
  • anonymous
Oh wait no sorry I just saw velocity
blackstreet23
  • blackstreet23
but i thought that the slope of the graph meant acceleration which is f'(x). I am confused
anonymous
  • anonymous
It stands for velocity i.e. the rate of change of position over time
anonymous
  • anonymous
Well formally speaking f(x) can stand for any function and as you noted the displacement is the integral of velocity.
anonymous
  • anonymous
So for this problem f(x) stands for velocity... and also I probably should be using f(t) to avoid confusion... I apologize... Please substitute t for all the x's in the formula above
anonymous
  • anonymous
Also, your second velocity comment was correct.... since f(t) stands for the velocity then.... \[f(t^{*}_{n})\] stands for the velocity evaluated at a certain point of time
blackstreet23
  • blackstreet23
sooooooo. f(t) in the graph of velocity vs time. Is still velocity?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
if f(t) gives is velocity and t is time then, the following tells me how velocity changes over time:
blackstreet23
  • blackstreet23
but that is the definition of acceleration
blackstreet23
  • blackstreet23
how velocity changes over time
anonymous
  • anonymous
|dw:1442106917949:dw|
blackstreet23
  • blackstreet23
I guess it would be instantaneous velocity right?
anonymous
  • anonymous
Sorry it is really hard to draw using this thing
blackstreet23
  • blackstreet23
I mean velocity at a certain point
blackstreet23
  • blackstreet23
but so far with your explaination. I guess the answer is 1 lol
anonymous
  • anonymous
The definition of acceleration is the rate instantaneous rate of change of velocity with respect to time.... i.e. the derivative of velocity curve at a given point
blackstreet23
  • blackstreet23
kind of makes sense. I mean which what i know of integrals
blackstreet23
  • blackstreet23
i was refering to answer #1
anonymous
  • anonymous
Ok then so it is clear the meaning of f here is the velocity curve correct?
anonymous
  • anonymous
Yes but I want to connect the concepts for you while I have you here.... to give you something to think about next time you face something like this so please dont just be happy with the answer
anonymous
  • anonymous
Thank you for the medal btw
blackstreet23
  • blackstreet23
umm. I mean the displacement is the integral of velocity. And when an integral in above the x-axis is positive and when is below is negative
blackstreet23
  • blackstreet23
right?
anonymous
  • anonymous
Yes but why is area under the curve (aka the integral of the velocity function) equal to the net displacement?
anonymous
  • anonymous
Wait 2 medals? lol who else is in here?!?!
anonymous
  • anonymous
I had to refresh because my backspace key ceased to function for some reason.
blackstreet23
  • blackstreet23
umm. Could you repeat the question? sorry lol
blackstreet23
  • blackstreet23
i do not understand what you are asking me :/
anonymous
  • anonymous
Its ok... You are correct that the integral of velocity is equal to the area under the velocity curve, where it is + above the x-axis and - below the x-axis, and that this represents the net displacement. My question is, how do you know this is true? Why is this the case? Think physically, based on the definition I gave above.
blackstreet23
  • blackstreet23
because the velocity is a vector, so if the velocity is negative its integral is negative right?
anonymous
  • anonymous
\[\int\limits_{a}^{b} f(t) dt \approx \sum_{n=0}^{N} f(t^{*}_{n}) \Delta t_{n}\]
blackstreet23
  • blackstreet23
I am actually lost with the formula lol
anonymous
  • anonymous
Yes velocity is a vector, but if we were just in one-dimension (i.e. if we only had an x-velocity) the fact that it is a vector doesn't matter in the question I asked you. Therefore, higher dimensions it similarly doesn't affect the answer.
anonymous
  • anonymous
Ok no this is good we have identified your hang up :D feel good this means progress can ensue
blackstreet23
  • blackstreet23
but the answer is #1 right?
blackstreet23
  • blackstreet23
i'll submit
anonymous
  • anonymous
Ok so the first term: \[ f(t^{*}_{n}) \] represents the velocity evaluated at the time\[ t^{*} \] on the nth subinterval
anonymous
  • anonymous
Yes because #2 is a derivative, #3 would ignore the potential "negative" area in the 4th quadrant and answer #4 would ignore the "positive" area in the second quadrant
blackstreet23
  • blackstreet23
ohhh i see
anonymous
  • anonymous
Ok back to the explaination.... so that \[ t^* \] I have definied to simply mean the time at the halfway point on the sub interval.... ill draw it for N=2
blackstreet23
  • blackstreet23
yuhu! yeah it was correct!
blackstreet23
  • blackstreet23
but yeah. I think I got it
anonymous
  • anonymous
|dw:1442108030512:dw|
anonymous
  • anonymous
Ok sorry the drawing is a mess it is so hard to draw in this thing
anonymous
  • anonymous
So this drawing shows you i am APPROXIMATING the area under the curve by drawing N=2 rectangles that have a height of: \[ f(t^{*}_{n}) \] and a width of \[ \Delta t \] where I am assuming that all the widths are equal (aka all time intervals are the same amount)
blackstreet23
  • blackstreet23
hey btw could you help me with another problem?
anonymous
  • anonymous
Therefore the area of the rectangle is given by width * height which is: \[\ f(t^{*}_{n}) \Delta t = velocity * time = distance \]
anonymous
  • anonymous
Does this make sense?
anonymous
  • anonymous
Applying the formalism, taking the limit, etc. just refines this definition to instantaneous points on the curve and the infinite amount of resulting rectangles makes the approximation become an equality
anonymous
  • anonymous
The easiest way to become fully convinced is draw the graph of a constant velocity starting out at v=0 at t=0 and going to some value v=V at time t=T
anonymous
  • anonymous
This graph will be a right triangle, which you can caluclate the area exactley by applying the formula 1/2 base times height
anonymous
  • anonymous
Checking the units wil reveal that this is indeed a unit of distance
anonymous
  • anonymous
Does this make sense to you?
anonymous
  • anonymous
Sure go ahead
anonymous
  • anonymous
Actually if you dont mind could you start a new problem thread for your question (if it is unrelated) since this one is already really long? Also, so if my answer is good I can get another medal :D

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