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Position as a function of position?

by asap9993
Tags: function, position
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asap9993
#1
Feb17-11, 11:34 AM
P: 18
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?
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rl.bhat
#2
Feb17-11, 11:55 AM
HW Helper
P: 4,435
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.
asap9993
#3
Feb17-11, 12:26 PM
P: 18
To rl.bhat,

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

rl.bhat
#4
Feb17-11, 09:09 PM
HW Helper
P: 4,435
Position as a function of position?

The integration adx = ax and integration of vdv = v^2/2. Now proceed.
SammyS
#5
Feb17-11, 10:27 PM
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Quote Quote by asap9993 View Post
To rl.bhat,

Shouldn't it be (x^2 - 1) instead of (x-1)? And I did the integration. It gives a natural log equation that doesn't give me the right answer
To asap9993

Yes, x2‒1

Also, It's v2 like you had, not v2/2.

Altering rl.bhat's solution with those changes gives:

√[1.8(s‒2)]dt = dx/√[(x2 - 1)]
Integrate to get t ≈ 1.287 s
rl.bhat
#6
Feb17-11, 11:19 PM
HW Helper
P: 4,435
[QUOTE=SammyS;3143763]To asap9993

Yes, x2‒1

Also, It's v2 like you had, not v2/2.

QUOTE]
Integration of xn = x(n+1)/(n+1)
SammyS
#7
Feb17-11, 11:35 PM
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P: 7,801
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 → kx2/2 + C/2 = v2/2

→ (1.8s‒1)x2 + C = v2 , which is what asap9993 had.
rl.bhat
#8
Feb17-11, 11:55 PM
HW Helper
P: 4,435
OK. That is correct.


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