# Position of a particle given an equation for acceleration

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1. Sep 4, 2011

### CaYn

1. The problem statement, all variables and given/known data

The acceleration of a particle is given by a(t)= - 2.1 m/s2 + (3.1 m/s3)t

Find the initial velocity v0 such that the particle will have the same x-coordinate at time t=3.96 as it had at t=0.

What x-coordinate will the particle have at time t=3.96

2. Relevant equations
I'm assuming integration plays a factor in this, but I'm just not sure how to make the particle have the same position at t=3.96 and t=0

3. The attempt at a solution

Integral of 2.1+3.1t= 2.1t+3.1/2t^2+c
integrating again should give us:
1.05t^2+(3.1/6)t^3 +ct +k
I dont know where to go from there
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 5, 2011

### SteamKing

Staff Emeritus
You are confused because you are throwing out equations without regard for what they represent. In the first equation you wrote, the constant of integration c should be equal to v0, the initial velocity. Similarly, in your second equation, k will be equal to s0, the initial displacement.

If you rewrite your equations in the form v(t) = something and s(t) = something else, the re-read the problem statement and see if you can solve for v0.

3. Sep 5, 2011

### Rayquesto

you find out the position of the particle at t=0 which you must do by integrating your equation twice in which v0t will be part of the x(t) equation. Since you know at t=0 that there must be some initial x, then well do you get the idea?

4. Sep 5, 2011

### Rayquesto

from what you wrote out, youre c would be v0 and k is your initial position which you can assume is zero at t=0.

5. Sep 5, 2011

### Rayquesto

so if at t=0 k=0 and x(t)=0 then you could say that x(t)=0 when t=3.69 seconds plus those factors in and you should get v0

6. Sep 5, 2011

### DiracRules

You've almost there. However, I think you should use different letters for the constant to see it through (sometimes, with different letters it's easier to see the solution).

I'm just doing your passages in a more rigorous way so that you can see properly the physical interpretation of the integration (usually in physics we tend to do indefinite integrations, but it has hardly a physical meaning: the proper integration is the definite one) :
$a(t)=\alpha+\beta t$, where alpha=-2.1 and beta=3.1 so,
$a(t)=\frac{dv(t)}{dt}\Rightarrow\int_{v_0}^{v(t)} dv=\int_{t_0}^{t} a(t)dt\Rightarrow v(t)-v_0=-\alpha t+\frac{\beta}{2}t^2\Rightarrow v(t)=v_0-\alpha t+\frac{\beta}{2}t^2$
$v(t)=\frac{dx(t)}{dt}\Rightarrow x(t)=x_0+v_0 t-\frac{\alpha}{2} t^2+\frac{\beta}{6}t^3$ just like you did.

Now, you have only to pose that the particle has the same position in t=0 and t=3.96, that is $$x(0)=x(3.96)$$, as requested from the problem :D