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Physics 14 Online
OpenStudy (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

OpenStudy (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."

OpenStudy (blackstreet23):

@abhisar @Shalante @freckles

OpenStudy (anonymous):

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

OpenStudy (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.

OpenStudy (blackstreet23):

yes. Displacement is the integral of velocity right?

OpenStudy (anonymous):

Correct, what does the integral of a function mean?

OpenStudy (blackstreet23):

area under the curve

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (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

OpenStudy (blackstreet23):

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

OpenStudy (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

OpenStudy (anonymous):

Oh no problems :D :D

OpenStudy (anonymous):

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

OpenStudy (blackstreet23):

as the summation of ....

OpenStudy (blackstreet23):

taking the limit

OpenStudy (blackstreet23):

sigma

OpenStudy (blackstreet23):

delta xk

OpenStudy (blackstreet23):

i dont remember totally lol

OpenStudy (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.

OpenStudy (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.

OpenStudy (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

OpenStudy (anonymous):

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

OpenStudy (blackstreet23):

f(x)

OpenStudy (anonymous):

yes

OpenStudy (blackstreet23):

the change of velocity over time?

OpenStudy (anonymous):

correct

OpenStudy (blackstreet23):

or an exact point of velocity in a certain time?

OpenStudy (blackstreet23):

so acceleration?

OpenStudy (anonymous):

Oh wait no sorry I just saw velocity

OpenStudy (blackstreet23):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (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

OpenStudy (blackstreet23):

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

OpenStudy (anonymous):

Yes

OpenStudy (anonymous):

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

OpenStudy (blackstreet23):

but that is the definition of acceleration

OpenStudy (blackstreet23):

how velocity changes over time

OpenStudy (anonymous):

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