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cgbork
 one year ago
A particle's position is r(vector) =(ct^2−2dt^3)i^+(2ct^2−dt^3)j^,
where c and d are positive constants. Find expressions for times t > 0 when the particle is moving in the xdirection.
Express your answer in terms of the variables c and d.
can anybody walk me through how to even approach this problem? I am totally lost.
cgbork
 one year ago
A particle's position is r(vector) =(ct^2−2dt^3)i^+(2ct^2−dt^3)j^, where c and d are positive constants. Find expressions for times t > 0 when the particle is moving in the xdirection. Express your answer in terms of the variables c and d. can anybody walk me through how to even approach this problem? I am totally lost.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Have you taken calculus by any chance? And If so is this a calculus based class?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can walk you through it but I need to think about how to tailor my approach based on your level.

cgbork
 one year ago
Best ResponseYou've already chosen the best response.0Yes. This is calc based. I believe the derivative needs to be taken to solve this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Correct the instantaneous rate of change of position is velocity, and that can be calculated by taking the derivative. This seems function seems to be fairly tame, do you recall the derivative rule for polynomials?

cgbork
 one year ago
Best ResponseYou've already chosen the best response.0not entirely. I am on a site for my class where it deducts 5% every time I guess wrong. I have put in d^2r/dt^2 and it said it didn't depend on dt or r

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry I didnt see that you had returned hold on...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok is this the function you gave:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ r(t)= (c*t^{2}  (2d)*t^{3})i+((2c)*t^{2}  d* t^{3})j \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Note how I made sure not to run the constants together with the independent variable... * denotes typical multiplication but I added it explicitly to make sure this is what you had in the problem statement

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Are you still there?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So if a particle is moving in the xdirection what does that mean?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It means the xcomponent of its velocity is nonzero... Please check to make sure you copied the question correctly because I suspect something is missing here.

cgbork
 one year ago
Best ResponseYou've already chosen the best response.0I believe it is correct

cgbork
 one year ago
Best ResponseYou've already chosen the best response.0My hint was this.. Hint 1. You will have to take a derivative with respect to time to get the velocity of the particle.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Correct, but what I am getting at is only requiring the xcomponent of the velocity v_x to be nonzero results in only one point where t>0 and v_x is zero and that is t=c/(3d).... I was expecting an interval or something more than a single point

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So in other words any time t>0 except fot that point t=c/(3d) the particle is moving in the x direction

cgbork
 one year ago
Best ResponseYou've already chosen the best response.0I already tried t=c/3d and it said I was missing something

cgbork
 one year ago
Best ResponseYou've already chosen the best response.0Specifically, Your answer either contains an incorrect numerical multiplier or is missing one.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No at that point there is NO motion in the xdirection. At that point v_x is zero... at all other times besides that point there is some motion in the xdirection

cgbork
 one year ago
Best ResponseYou've already chosen the best response.0Oh crap. That is because I found it when t=0 and it was asking for when it is greater than 0. So how would I find it when it isn't 0, but instead t>0?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No that is the answer for when t > 0. When t=0 the answer is just v_x = 0 plain and simple but when t >0 you get c3dt=0 or t = c/(3d)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Like I said I feel like this question is either ill posed or incorrect... If that isn't the answer.... all t =/= c/(3d)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Does it mean in the POSITIVE x direction maybe?

cgbork
 one year ago
Best ResponseYou've already chosen the best response.0It was asking for t>0 but told me that c/3d was wrong. This computer program for this question is on crack.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes because it IS wrong.... The answer is all t NOT equal to that value

cgbork
 one year ago
Best ResponseYou've already chosen the best response.0So I need to make an equation that excludes that point

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Assuming any motion in the xdirection (either way) is permitted.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sure try 0<t<c/(3d) and c/(3d)<t<infinity

cgbork
 one year ago
Best ResponseYou've already chosen the best response.0Part a is asking for the x direction and part b is asking for the y direction. So I'm hoping once I understand how to get part a that part b will be a breeze. Are you sure it is asking for all of that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No Im not sure like I said I feel like this question is missing some information or poorly written or something. I mean have you tried the above answer?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Having only a single point seems kind of silly to me... I feel like it should be an interval or something.... Asking only about POSITIVE xdirection or something would make more sense

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But that is just a guess

cgbork
 one year ago
Best ResponseYou've already chosen the best response.0All of 0<t<c/(3d) and c/(3d)<t<infinity? No, I haven't put that in yet. The whole MasteringPhysics online program is definitely confusing. What would I have to do if it was in positive x direction? I'm going to assume it is because it didn't say anything differently. Thank you for your help btw!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if it were in the positive xdirection then the as long as 2ct > 6dt^2 or as long as 2c >6dt or c/(3d) > t

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So for all times 0 < t < c/(3d)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0After that the velocity becomes negative (moves in the xdirection) and stays that ways forever

cgbork
 one year ago
Best ResponseYou've already chosen the best response.0I don't know if I can put it in like that

cgbork
 one year ago
Best ResponseYou've already chosen the best response.0I figured it out. Yay for physics roommate! Thank you for trying!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So what did the answer end up being?
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