Question:
The acceleration of a particle is given by a = 3x+1 (where a is acceleration, x is displacement from origin). If the particle is initially at the origin and moving at a velocity of , find its velocity after 3s.
What I've done:
when t=0, x=0, v = 15
I'm not sure where to go on from here. I know if I find t in terms of x, I can substitute it into that equation. I've tried but it doesn't work well.
Thank you for that, but I have no idea what "constant coefficient homogeneous ODE" is.
And for Mathstud28's method, I integrated a = 3x + 1 in respect to t, treating 3x as a constant. I then substituted t = 3 to get the a relation between v and x at that time. Then solved simultaneously with the equation I derived. Is that a valid method? or is it incredible coincidence that I got the right answer to 1 dp. (16.1)
I explain this in my Differential Equations Tutorial. Read the section on Undetermined coefficients (Post #6), and see if this helps out.
Sorry, but that's completely wrong. x is the displacement of the object. I assume that the object is actually moving and so x is changing. Therefore .......
x is NOT a constant and CANNOT be treated as one.
This is not the only error. Correct me if I'm wrong but doesn't the original question say a = 3x + 1? Why then do you say a = 3x - 1 in your very first line of working.
No offence, but this whole thread is a comedy of errors as a consequence of post #2 (and to a lesser extent the confusion of what a is actually meant to be).
CaptainB's approach is correct and efficient. But you don't have background to use this approach (not your fault). So:
The bottom line is that you must first use one of the expressions or to get v = v(x). You did this but then lost your way.
You then have to do the following:
dx/dt = v(x) => dt/dx = 1/v(x).
Integrate to get x = x(t).
dx/dt gives you v as a function of time.
Substitute t = 3 and you're done.