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 August 16th, 2009, 02:53 PM #1 Newbie   Joined: Jul 2009 Posts: 6 Thanks: 0 Constants? We had a problem last week with a group submission. acceleration: a(t)=cos(t) At time t=0, the position is x=4 in the question, and we were asked to find the velocity, position, and the "values for t when the particle was at rest". We got that, except the constants. None of us could figure out how to find the constants. She hasn't gotten back to us with comments or grades, and we don't expect them anytime soon since she went on vacation. Out of curiosity could you tell me/us how to do this?
 August 16th, 2009, 03:22 PM #2 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: Constants? This problem isn't doable without more information. You are given a second-order differential equation $\frac{d^2x}{dt^2}=\cos t$ with one boundary condition $x(0)=4.$ The solution of the differential equation can be found by integrating twice: $\frac{dx}{dt}=\sin t+C;$ substituting $t=0$ in here gives $C=v_0,$ the initial velocity. Integrate again: $x(t)=-\cos t+v_0t+C#39;;$ again, substitute in $t=0$ to get $4=-1+C#39;,$ or $x(t)=5-\cos t+v_0t.$ Without $v_0,$ or at least a value of $\frac{dx}{dt}$ for some value of $t,$ you cannot determine fixed points - indeed, there will not be any if $|v_0|>1.$
August 18th, 2009, 04:59 AM   #3
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Re: Constants?

Thanks for the help. I thought maybe I should type in the question exactly as she gave it in case that helps any. I'm not an expert at this enough to know if I left something important out.

Quote:
 A particle, initially at rest, moves along the x-axis in a way that it's acceleration at time t>0 is: a(t)=cos(t) At the time t=0, its position is x=4 (necessary information to solve for the integration constants). A) Find the expression for the particle's velocity. B) Find the expression for the particle's position. C) Find the values of t for which the particle is at rest.

I know from the directions that there are constants we need to find. We just didn't know how to go about it. I'm getting seriously annoyed at her for running off so late in the class.

 August 18th, 2009, 06:46 AM #4 Global Moderator   Joined: Dec 2006 Posts: 17,221 Thanks: 1294 You now state the particle was initially at rest, so $a(t)\,=\,\frac{d^2x}{dt^2}\,=\,\cos(t)$ with initial conditions $x(0)=4\text{ and }x#39;(0)\,=\,0.$ Hence mattpi's solution becomes $x(t)=5\,-\,\cos(t),$ $v(t)\,=\,x#39;(t)=\sin(t).$ The particle is therefore at rest whenever t is a non-negative integer multiple of pi.

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