A community for students.
Here's the question you clicked on:
 0 viewing
blackstreet23
 one year ago
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
 one year ago
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

This Question is Closed

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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Are you aware of the relationships between displacement, velocity, and acceleration given by calculus?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'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
 one year ago
Best ResponseYou've already chosen the best response.0yes. Displacement is the integral of velocity right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Correct, what does the integral of a function mean?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0area under the curve

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why does it mean that? Pretend I don't know and explain to me where that idea comes from please :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Think first, how do you calculate the area under a curve?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Please just state what you know, dont go referencing books or google.... consider me an interactive google in a sense on this topic :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't see you typing which means you are looking up an answer, please I am curious to know where you stand currently

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0no, I was answering a friend's text. Sorry

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0So the area under the curve can be found by obtaining the integral of a function when is above the x axis and continuos

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh no problems :D :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Correct... how do you integrate though? It isnt a magical process (strictly speaking).... do recall how an integral is defined?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0as the summation of ....

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0taking the limit

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0i dont remember totally lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You 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
 one year ago
Best ResponseYou've already chosen the best response.0But 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
 one year ago
Best ResponseYou've already chosen the best response.0And for simplicity the * on the x we will just take to mean the function evaluated at the midpoint of the interval delta x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What does it stand for in terms of the problem at hand?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0the change of velocity over time?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0or an exact point of velocity in a certain time?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0so acceleration?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh wait no sorry I just saw velocity

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0but i thought that the slope of the graph meant acceleration which is f'(x). I am confused

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It stands for velocity i.e. the rate of change of position over time

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well formally speaking f(x) can stand for any function and as you noted the displacement is the integral of velocity.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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
 one year ago
Best ResponseYou've already chosen the best response.0Also, 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
 one year ago
Best ResponseYou've already chosen the best response.0sooooooo. f(t) in the graph of velocity vs time. Is still velocity?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if f(t) gives is velocity and t is time then, the following tells me how velocity changes over time:

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0but that is the definition of acceleration

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0how velocity changes over time

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442106917949:dw

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0I guess it would be instantaneous velocity right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry it is really hard to draw using this thing

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0I mean velocity at a certain point

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0but so far with your explaination. I guess the answer is 1 lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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
 one year ago
Best ResponseYou've already chosen the best response.0kind of makes sense. I mean which what i know of integrals

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0i was refering to answer #1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok then so it is clear the meaning of f here is the velocity curve correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes 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
 one year ago
Best ResponseYou've already chosen the best response.0Thank you for the medal btw

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0umm. I mean the displacement is the integral of velocity. And when an integral in above the xaxis is positive and when is below is negative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes but why is area under the curve (aka the integral of the velocity function) equal to the net displacement?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wait 2 medals? lol who else is in here?!?!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I had to refresh because my backspace key ceased to function for some reason.

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0umm. Could you repeat the question? sorry lol

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0i do not understand what you are asking me :/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Its ok... You are correct that the integral of velocity is equal to the area under the velocity curve, where it is + above the xaxis and  below the xaxis, 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
 one year ago
Best ResponseYou've already chosen the best response.0because the velocity is a vector, so if the velocity is negative its integral is negative right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{a}^{b} f(t) dt \approx \sum_{n=0}^{N} f(t^{*}_{n}) \Delta t_{n}\]

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0I am actually lost with the formula lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes velocity is a vector, but if we were just in onedimension (i.e. if we only had an xvelocity) 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
 one year ago
Best ResponseYou've already chosen the best response.0Ok no this is good we have identified your hang up :D feel good this means progress can ensue

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0but the answer is #1 right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok so the first term: \[ f(t^{*}_{n}) \] represents the velocity evaluated at the time\[ t^{*} \] on the nth subinterval

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok 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
 one year ago
Best ResponseYou've already chosen the best response.0yuhu! yeah it was correct!

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0but yeah. I think I got it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442108030512:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok sorry the drawing is a mess it is so hard to draw in this thing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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
 one year ago
Best ResponseYou've already chosen the best response.0hey btw could you help me with another problem?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Therefore the area of the rectangle is given by width * height which is: \[\ f(t^{*}_{n}) \Delta t = velocity * time = distance \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Does this make sense?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Applying 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
 one year ago
Best ResponseYou've already chosen the best response.0The 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
 one year ago
Best ResponseYou've already chosen the best response.0This graph will be a right triangle, which you can caluclate the area exactley by applying the formula 1/2 base times height

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Checking the units wil reveal that this is indeed a unit of distance

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Does this make sense to you?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Actually 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
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.