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Mathematics 8 Online
OpenStudy (anonymous):

please help me with this The rectilinear motion of a particle if given by 〖s=v〗^2-9 where s is in feet and v in feet per second. When t=0, s=0 and v= 3 fps. Find the s-t, v-t and a-t relations.

OpenStudy (anonymous):

that is s= v^2 - 9

OpenStudy (mathmale):

Josh: As stated (s=v^2-9), this is a highly unusual problem. Have you any way of taking and sharing a photo of the original problem, so that I could see where you're starting from? Is this problem from a Calculus class?

OpenStudy (anonymous):

its physics actually

OpenStudy (mathmale):

That's OK. Same question: could you share an image of the original problem with me, so that I could see where you're coming from?

OpenStudy (anonymous):

its a .doc file send by my teacher

OpenStudy (mathmale):

Thanks. Great how we can share docs so easily. Josh: Have you any class notes, worked examples, examples in your textbook or the like, pertaining to "variable acceleration?" While familiar with physics and the calculus involved here. I'm having trouble visualizing what we need to do. Can you share anything else with me that would give us cues?

OpenStudy (anonymous):

ok wait for a sec i;ll take pics on my notes

OpenStudy (anonymous):

here it is

OpenStudy (anonymous):

sorry its quite blurry... my camera sucks

OpenStudy (mathmale):

Josh: I've had a connection problem. But now I can view your photo and will look it over.

OpenStudy (mathmale):

I need to know whether or not you've had experience with solving first order linear differential equations.

OpenStudy (anonymous):

ok mathmale

OpenStudy (anonymous):

yes i've had experience solving first order linear D.E

OpenStudy (mathmale):

Cool. That may be the key to solving this problem.

OpenStudy (anonymous):

ok i will try it

OpenStudy (anonymous):

is it possible??? to use D.E instead of diff. cal or integral?

OpenStudy (mathmale):

Josh: Not getting very far, so I'll share with you where I am now and let you decide how to proceed next. If s=v^2-9, and v=ds/dt (which is certainly true), then \[s=(\frac{ ds }{ dt })^{2}-9.\]. Since the derivative is squared, this is a NON-LINEAR differential equation. Agreed? I'm well equipped to solve first order linear d. e.'s, but not non-linear ones. Have you any experience with solving non-linear d. e.'s?

OpenStudy (mathmale):

I see in your notes that your teacher has mentioned the relationship \[v ^{2}=v _{0}^{2}+2as.\] I've certainly used that before, in cases where the acceleratin is constant, but am wondering whether it applies when the acceleration is NOT constant. I hope these questions and suggestions are of some help to you. I'd be interested in knowing your thoughts after you've had some time to review what I've written and to reflect upon it.

OpenStudy (anonymous):

the other forms i forgot some of it mathdale

OpenStudy (mathmale):

Sorry, Josh, I'm not sure I understand what you're trying to say. What "other forms"?

OpenStudy (anonymous):

other forms aside from linear d.e

OpenStudy (mathmale):

I have no recent experience in solving non-linear differential equations; have y ou any experience? No? OK. Do you have a classmate with whom you could share ideas regarding the solution of this problem?

OpenStudy (mathmale):

It seems to me that if we knew how to solve nonlinear differential equations, we'd quickly have a solution to \[s=\left( ds/dt \right)^{2}-9.\]

OpenStudy (mathmale):

This relationship does relate s and t, as was required. But it'd be so much nicer were we able to solve that nonlinear d. e. Josh, happy to work with you. Thought provoking problem. Sorry I'm not able to do any more beyond this point. If you have other problems to work on, submit some either through OpenStudy.com or through a private message to me. Good luck!

OpenStudy (anonymous):

....ok tnx mathmale!!! i really appreciate it

OpenStudy (mathmale):

Happy to do this! Again, I wish so much I were able to help you reach a definitive answer. Bye for now!

OpenStudy (anonymous):

ok tnx bro

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