geniuses needed!

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thread starter soccer2rulz Feb 06 2010 + PM | QUOTE | PERMALINK | REPORT

i've been doing this question since yesterday and I can't seem to solve it D= The acceleration of a particle at time t is given by a(t)=-8j (j being the symbol for the vertical part of the vector and "i" being the symbol of the horizontal part of the vector). For this particle r(2)= -4i+3j and r(5)= 11i+15j (r being the displacement). Determine the vector expression for the displacement of the particle at 't'.

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RexRaydon 02/06/10 + PM | QUOTE | PERMALINK | REPORT

Is this for a class? I don't remember doing anything like that in physics/whatever science that is.

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soccer2rulz 02/06/10 + PM | QUOTE | PERMALINK | REPORT

its high school mathematics

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john12345172 02/06/10 + PM | QUOTE | PERMALINK | REPORT

Calculus?

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renague 02/06/10 + PM | QUOTE | PERMALINK | REPORT

I got lost at "acceleration". -Braces for high schoo, failure-

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kaixian0011 02/06/10 + PM | QUOTE | PERMALINK | REPORT

vector = speed, Derived of acceleration. just derive the equation and yu get your vector. I didn't understand your equation.

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CaptObvious 02/06/10 + PM | QUOTE | PERMALINK | REPORT

similar to kai would help greatly if you said what subject

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FatelessNexus + PM | QUOTE | PERMALINK | REPORT

[quote] kaixian0011 : vector = speed, Derived of acceleration. just derive the equation and yu get your vector. I didn't understand your equation. [/quote] Velocity = speed. Vector just tells direction. She's asking for a vector equation of the displacement (how far it traveled) of the particle based on the equation. @below: my bad =P got lazy considering no one reads technical explanations anyways.

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cb000 02/06/10 + PM | QUOTE | PERMALINK | REPORT

@above: Velocity changes if direction changes. Speed is the magnitude of velocity.

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Gamesta64 02/06/10 + PM | QUOTE | PERMALINK | REPORT

oh god we're doing this right now too, but in physics! I did this last year in Calculus but there was only one direction, so we didn't have to take into account that it was a vector really...Um I'll try it and let you know if I can solve it.

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hero159 02/06/10 + PM | QUOTE | PERMALINK | REPORT

[quote] john12345172 : Calculus? [/quote] I think it has a hint of integration but mostly vectors. For the question, I can't help you, I've forgotten most of my math.

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6ReDevil6 + PM | QUOTE | PERMALINK | REPORT

r(t)=(5t -14)i + (constant*t -4t*t)j The problem is incorrect for the j component since no constant will allow those positions for those times with that constant accelaration, even assuming any initial conditions.

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wizlid 02/06/10 + PM | QUOTE | PERMALINK | REPORT

It's actually genii o.o Just had to share even though it doesn't matter.

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soccer2rulz 02/06/10 + PM | QUOTE | PERMALINK | REPORT

[quote] 6ReDevil6 : r(t)=(5t -14)i + (constant*t -4t*t)j The problem is incorrect for the j component since no constant will allow those positions for those times with that constant accelaration, even assuming any initial conditions. [/quote] the first part is correct but can you tell me how you got that? D= @above: Lol sorry i'm not geniouusss, that's why i can't spell it properly ;D

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cb000 + PM | QUOTE | PERMALINK | REPORT

The acceleration of a particle at time t is given by a(t)=-8j (j being the symbol for the vertical part of the vector and "i" being the symbol of the horizontal part of the vector). For this particle r(2)= -4i+3j and r(5)= 11i+15j (r being the displacement). Determine the vector expression for the displacement of the particle at 't'. Integrate -8j. -8tj+C The 'i' component is a constant, so you take your givens and calculate. 11i-(-4i)/(5-2) = 15i/3 = 5i But there could also be an initial velocity for the j vector, so we'll leave that as A. So v(t) = 5i+(-8t+A)j. Integrate once more to get the position. r(t) = 5ti+ (-4t^2+At)j+C Again, you have to solve for C, so you take your givens and calculate. r(2)=5(2)i+(-4*2^2+2A)j+C=-4i+3j r(2) = 10i+(-16+2A)j+C = -4i+3j C = -14i+(19-2A)j r(t) = (5t-14)i+(-4t^2+At+19-2A)j Now we need that final A. We do this with r(5). I'll only do the 'j' component since the 'i' component works for both r(2) and r(5). -4(5)^2+A*5+19-2A = 15. -100+5A+19-2A= 15. -81+3A = 15. 3A = 96. A = 32. So we get r(t) = (5t-14)i+(-4t^2+32t-45)j. Checking with r(2) we get, r(2) = (5*2-14)i+(-4*2^2+32*2-45)j = (10-14)i+(-16+64-45)j = -4i+3j. It works. Edit: Had to patch up a few mistakes. Sorry.

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