1. ## Applications of Calculus

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 $15ms^{-1}$, find its velocity after 3s.

What I've done:

$a = v\frac{dv}{dx} = 3x -1$
$\frac{1}{2} v^2 = \frac{3}{2}x^2 + x + c$

when t=0, x=0, v = 15
$v^2 = 3x^2 + 2x + 225$

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 $\frac{dx}{dt} = \sqrt{3x^2 + 2x + 225} \Rightarrow \frac{dt}{dx} = \frac{1}{\sqrt{3x^2 + 2x + 225}}$ but it doesn't work well.

2. Originally Posted by Gusbob
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 $15ms^{-1}$, find its velocity after 3s.

What I've done:

$a = v\frac{dv}{dx} = 3x -1$
$\frac{1}{2} v^2 = \frac{3}{2}x^2 + x + c$

when t=0, x=0, v = 15
$v^2 = 3x^2 + 2x + 225$

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 $\frac{dx}{dt} = \sqrt{3x^2 + 2x + 225} \Rightarrow \frac{dt}{dx} = \frac{1}{\sqrt{3x^2 + 2x + 225}}$ but it doesn't work well.
I think you are overcomplicating things.

Consider

$v(t)=\int{a(t)dt}+v_0$

3. Thank you. I've tried that before, but didn't derive the v² equation first. But now that I have it, it works

4. Originally Posted by Mathstud28
I think you are overcomplicating things.

Consider

$v(t)=\int{a(t)dt}+v_0$
Originally Posted by Gusbob
Thank you. I've tried that before, but didn't derive the v² equation first. But now that I have it, it works
Just to make sure I wasnt ambiguous $v_0$ is the constant of integration with the integration of a(t), so dont add an extra c, the c is $v_0$

5. Originally Posted by Gusbob
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 $15ms^{-1}$, find its velocity after 3s.

What I've done:

$a = v\frac{dv}{dx} = 3x -1$
$\frac{1}{2} v^2 = \frac{3}{2}x^2 + x + c$

when t=0, x=0, v = 15
$v^2 = 3x^2 + 2x + 225$

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 $\frac{dx}{dt} = \sqrt{3x^2 + 2x + 225} \Rightarrow \frac{dt}{dx} = \frac{1}{\sqrt{3x^2 + 2x + 225}}$ but it doesn't work well.
$a=\frac{dv}{dt}=3x-1$

so differentiate this again:

$\frac{d^2v}{dt^2}=3v$

which is a constant coefficient homogeneous ODE, which you should be able to solve, to get:

$v(t)=Ae^{\sqrt{3}t}+Be^{-\sqrt{3}t}$

and so:

$
x(t)=\frac{A}{\sqrt{3}}e^{\sqrt{3}t}-\frac{B}{\sqrt{3}}e^{-\sqrt{3}t}+C
$

Now putting both of these back into the equation $\frac{dv}{dt}=3x-1$ shows that $C=0$.

Now apply the initial condition to find $A$ and $B$ and then its simple.

RonL

6. Originally Posted by Mathstud28
I think you are overcomplicating things.

Consider

$v(t)=\int{a(t)dt}+v_0$
What you mean is:

$v(t)=\int_0^t {a(\tau)~d\tau}+v_0$

and how do you proceed from this point?

RonL

7. Originally Posted by CaptainBlack
$a=\frac{dv}{dt}=3x-1$

so differentiate this again:

$\frac{d^2v}{dt^2}=3v$

which is a constant coefficient homogeneous ODE, which you should be able to solve, to get:

$v(t)=Ae^{\sqrt{3}t}+Be^{-\sqrt{3}t}$

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)

8. Originally Posted by Gusbob
Thank you for that, but I have no idea what "constant coefficient homogeneous ODE" means.
I explain this in my Differential Equations Tutorial. Read the section on Undetermined coefficients (Post #6), and see if this helps out.

9. Originally Posted by Gusbob
[snip]

And for Mathstud28's method, I integrated a = 3x + 1 in respect to t, treating 3x as a constant. [snip]
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 $a = \frac{d}{dx} \left( \frac{v^2}{2} \right)$ or $v \frac{dv}{dx}$ 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.

10. Originally Posted by Gusbob
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)
x is not a constant so no its not valid.

RonL