Position as a function of position?

[asap9993]

Homework Statement

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?

Homework Equations

a = dv/dt

v = dx/dt

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?

 

[rl.bhat]

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]

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]

The integration adx = ax and integration of vdv = v^2/2. Now proceed.

 

[SammyS]

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]

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]

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]

OK. That is correct.