(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A particle with a mass of 3kg is at rest at x = 3m, and then a force = (12 N/m)x is applied to it. What is the acceleration of the particle when it reaches x = 5m? Determine the v(t) for the particle.

2. Relevant equations

[tex]

\sum F = ma

[/tex]

3. The attempt at a solution

The first part was simple. I just solved for acceleration and got the same answer as the book, 20 m/s^2. However there was no answer for the second part so I want to know if I solved for v(t) properly.

[tex]

a = (4 \frac {1}{s^2}) x

[/tex]

[tex]

a = \frac {d^2x} {dt^2}

[/tex]

[tex]

\frac {d^2x} {dt^2} - (4 \frac {1}{s^2}) x = 0

[/tex]

Which is just a 2nd order differential equation, whose general solution is:

[tex]

x = C_1 e^(\lambda_1 t) + C_2 e^(\lambda_2 t)

[/tex]

Where lambda is the solution to the auxiliary equation,

[tex]

\lambda^2 - 4\lambda = 0

[/tex]

Therefore

[tex]

x = C_1 e^(2 t) + C_2 e^(-2 t)

[/tex]

and v(t) is

[tex]

v = 2 C_1 e^(2 t) - 2 C_2 e^(-2 t)

[/tex]

The problem stipulates that at x=3m, v=0, so when if I set t=0 at x=3m, then I can solve for C1 and C2

[tex]

3 = C_1 + C_2

[/tex]

[tex]

0 = 2 C_1 - 2 C_2

[/tex]

[tex]

C_1 = \frac {3}{2} and C_2 = \frac {-3}{2}

[/tex]

So v(t) becomes

[tex]

v = (3 \frac {m}{s}) ( e^(2 t) + e^(-2 t) )

[/tex]

Then I wanted to make it look nicer so I rewrote it as:

[tex]

v = (6 \frac {m}{s}) \cosh {(2 \frac {1}{s} t)}

[/tex]

Is this the proper solution to the question?

If not could someone please explain to me how to get the correct one. Thank you for the help.

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# Homework Help: Velocity function of a variable force

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