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Position as a function of position? 
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#1
Feb1711, 11:34 AM

P: 19

1. The problem statement, all variables and given/known data
Here's the problem: Suppose the acceleration of a particle is a function of x, where a(x) = (1.8 s^2)x. a) If the velocity is zero when x = 1.0 m, what is the speed when x = 2.9 m? b) How long does it take the particle to travel from x = 1.0 m to x = 2.9 m? 2. Relevant equations a = dv/dt v = dx/dt 3. The attempt at a solution I managed to figure out how to do part a. a = dv/dt (dt/dx)a = (dv/dt)(dt/dx) Since v = dx/dt, the above equation becomes a(1/v) = dv/dx which becomes a dx = v dv. Integrating both sides gives you v^2 = (1.8s^2)(x^2) + C (s is seconds) Since v = 0 when x = 1.0 meters, C = (1.8s^2)(m^2) (m is meters) So the final function is v^2 = (1.8s^2)(x^2)  (1.8s^2)(m^2) Plugging in x = 2.9 m, gives 3.65 m/s which is the correct answer. I can't figure out part b. I tried using the same approach as in part a to find a "time function" : v = dx/dt so dt = (1/v) dx However. when I integrate this, it doesn't give me the correct answer which is 1.29 s. Can anyone please help me? 


#2
Feb1711, 11:55 AM

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P: 4,433

a dx = v dv.
Check the above integration. You have obtained v^2/2 = 1.8(s^2)(x  1) v = sqrt[2*1.8(s^2)(x  1)] dx/dt = sqrt[3.6(s^2)(x  1)] dt = dx/sqrt[3.6(s^2)(x  1)] Now you can find the integration. 


#3
Feb1711, 12:26 PM

P: 19

To rl.bhat,
Shouldn't it be (x^2  1) instead of (x1)? And I did the integration. It gives a natural log equation that doesn't give me the right answer 


#4
Feb1711, 09:09 PM

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P: 4,433

Position as a function of position?
The integration adx = ax and integration of vdv = v^2/2. Now proceed.



#5
Feb1711, 10:27 PM

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Yes, x^{2}‒1 Also, It's v^{2} like you had, not v^{2}/2. Altering rl.bhat's solution with those changes gives: √[1.8(s^{‒2})]dt = dx/√[(x^{2}  1)] Integrate to get t ≈ 1.287 s 


#6
Feb1711, 11:19 PM

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[QUOTE=SammyS;3143763]To asap9993
Yes, x^{2}‒1 Also, It's v^{2} like you had, not v^{2}/2. QUOTE] Integration of x^{n} = x^{(n+1)}/(n+1) 


#7
Feb1711, 11:35 PM

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In asap9993's equation, a dx = v dv, the quantity, a, is not a constant. a = kx, where k=1.8s^{‒1}
So, upon integrating, asap9993 dropped the 2 in the denominator on both sides. kx·dx = v·dv → kx^{2}/2 + C/2 = v^{2}/2 → (1.8s^{‒1})x^{2} + C = v^{2} , which is what asap9993 had. 


#8
Feb1711, 11:55 PM

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OK. That is correct.



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