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

- blackstreet23

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- schrodinger

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- 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

@abhisar
@Shalante
@freckles

- anonymous

Are you aware of the relationships between displacement, velocity, and acceleration given by calculus?

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## More answers

- 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

yes. Displacement is the integral of velocity right?

- anonymous

Correct, what does the integral of a function mean?

- blackstreet23

area under the curve

- anonymous

Why does it mean that? Pretend I don't know and explain to me where that idea comes from please :D

- anonymous

Think first, how do you calculate the area under a curve?

- 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

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

no, I was answering a friend's text. Sorry

- 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

Oh no problems :D :D

- anonymous

Correct... how do you integrate though? It isnt a magical process (strictly speaking).... do recall how an integral is defined?

- blackstreet23

as the summation of ....

- blackstreet23

taking the limit

- blackstreet23

sigma

- blackstreet23

delta xk

- blackstreet23

i dont remember totally lol

- 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

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

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

What does it stand for in terms of the problem at hand?

- blackstreet23

f(x)

- anonymous

yes

- blackstreet23

the change of velocity over time?

- anonymous

correct

- blackstreet23

or an exact point of velocity in a certain time?

- blackstreet23

so acceleration?

- anonymous

Oh wait no sorry I just saw velocity

- blackstreet23

but i thought that the slope of the graph meant acceleration which is f'(x). I am confused

- anonymous

It stands for velocity i.e. the rate of change of position over time

- anonymous

Well formally speaking f(x) can stand for any function and as you noted the displacement is the integral of velocity.

- 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

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

sooooooo. f(t) in the graph of velocity vs time. Is still velocity?

- anonymous

Yes

- anonymous

if f(t) gives is velocity and t is time then, the following tells me how velocity changes over time:

- blackstreet23

but that is the definition of acceleration

- blackstreet23

how velocity changes over time

- anonymous

|dw:1442106917949:dw|

- blackstreet23

I guess it would be instantaneous velocity right?

- anonymous

Sorry it is really hard to draw using this thing

- blackstreet23

I mean velocity at a certain point

- blackstreet23

but so far with your explaination. I guess the answer is 1 lol

- 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

kind of makes sense. I mean which what i know of integrals

- blackstreet23

i was refering to answer #1

- anonymous

Ok then so it is clear the meaning of f here is the velocity curve correct?

- 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

Thank you for the medal btw

- 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

right?

- anonymous

Yes but why is area under the curve (aka the integral of the velocity function) equal to the net displacement?

- anonymous

Wait 2 medals? lol who else is in here?!?!

- anonymous

I had to refresh because my backspace key ceased to function for some reason.

- blackstreet23

umm. Could you repeat the question? sorry lol

- blackstreet23

i do not understand what you are asking me :/

- 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

because the velocity is a vector, so if the velocity is negative its integral is negative right?

- anonymous

\[\int\limits_{a}^{b} f(t) dt \approx \sum_{n=0}^{N} f(t^{*}_{n}) \Delta t_{n}\]

- blackstreet23

I am actually lost with the formula lol

- 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

Ok no this is good we have identified your hang up :D feel good this means progress can ensue

- blackstreet23

but the answer is #1 right?

- blackstreet23

i'll submit

- anonymous

Ok so the first term: \[ f(t^{*}_{n}) \] represents the velocity evaluated at the time\[ t^{*} \] on the nth subinterval

- 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

ohhh i see

- 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

yuhu! yeah it was correct!

- blackstreet23

but yeah. I think I got it

- anonymous

|dw:1442108030512:dw|

- anonymous

Ok sorry the drawing is a mess it is so hard to draw in this thing

- 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

hey btw could you help me with another problem?

- anonymous

Therefore the area of the rectangle is given by width * height which is:
\[\ f(t^{*}_{n}) \Delta t = velocity * time = distance \]

- anonymous

Does this make sense?

- 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

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

This graph will be a right triangle, which you can caluclate the area exactley by applying the formula 1/2 base times height

- anonymous

Checking the units wil reveal that this is indeed a unit of distance

- anonymous

Does this make sense to you?

- anonymous

Sure go ahead

- 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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