How Does Integration Affect Particle Acceleration Calculations?

[tandoorichicken]

How do I go about this? I finished it but I think its off a bit. I think it has to do with the extra constant that comes out of integration that I didn't take into account.

Problem: The position of a particle at times t=10 sec and t=5 sec are known to be, respectively,
\vec{r}(10) = 10\vec{i}+5\vec{j}-10\vec{k}
\vec{r} (5) = 3\vec{i}+2\vec{j}+5\vec{k}
What is the acceleration of the particle at time t=5 sec if the velocity vector has the form
\vec{v}=C_1\vec{i}+C_2 t^2\vec{j}+C_3\ln{t}\vec{k}
where C_1, C_2, and C_3 are constants and t is time in seconds?

What I did:
integrated v(t) to get r(t), \vec{r}(t)=C_1 t\vec{i}+\frac{1}{3}C_2 t^3\vec{j}+C_3(t\ln{t}-t)\vec{k} then integrated once more to get a(t)= 2C_2 t\vec{j} +\frac{C_3}{t}\vec{k}
I then substituted the initial equation for r(5) into the equation for r(t) to find the constants C_1, C_2, C_3, and then using these constants, found the acceleration for t=5 to be .48\vec{j}+.33\vec{k}
The values I got for the constants were .6 for C_1, 6/125 for C_2, and 1/(ln(1)-1) for C_3.

 

[lightgrav]

You NEED to include a constant term for each coordinate.

x(10 sec) = 10 , not 6 as your formula suggests.

You have 2 equations for each coordinate, you can solve for 2 unknowns.

 

[sporkstorms]

I think it has to do with the extra constant that comes out of integration that I didn't take into account.

I think that's probably the problem. If you take the values you obtained for r(5) and plug them in for r(10) [or vice versa], I think you'll find it's incorrect.

As you're solving it, for each if i j and k you should have 2 equations and 2 unknowns.

Adding the integration constant to your expression, we have:
\vec{r}(t)=\left(C_1 t + k_1\right)\hat{i}+\left(\frac{1}{3}C_2 t^3 + k_2\right)\hat{j}+\left(C_3(t\ln{t}-t) + k_3\right)\hat{k}

Now solve for each of the i,j,k components using the initial values provided. Example for the x-component:
t=10: C_1*10 + k_1 = 10
t=5: C_1*5 + k_1 = 3

We have 2 equations and 2 unknowns. Solve for C_1 and k_1, plug it back into the expression you got by integrating, and we'll end up with an x-component of the form at+b