Acceleration as a function of position

[DrLich]

Why can't I simply integrate a(x)=m*x with respect to x to determine the speed of a particle as a function of position v(x)=1/2*m*x^2+A?

 

[Dale]

Acceleration is $a=dv/dt$. It is a derivative with respect to time. What would it mean to integrate that wrt position?

 

[DrLich]

Acceleration is $a=dv/dt$. It is a derivative with respect to time. What would it mean to integrate that wrt position?

I stumbled upon the following exercise:

The acceleration of a particle is given as a function of position:
a(x) = (2 s^-2)x

I needed to calculate the velocity at a given position x. I initially thought I could solve this by integrating the acceleration function with respect to x to find the velocity as a function of position, and then put in the x-value to get the answer. Sorry if this is a silly question.

 

[Ibix]

I initially thought I could solve this by integrating the acceleration function with respect to x to find the velocity as a function of position, and then put in the x-value to get the answer.

Well, $\int a(x)dx$ has dimensions of $\mathrm{L^2T^{-2}}$, so that clearly isn't a velocity.

I assume $s$ is a constant. Have you come across simple harmonic motion in your studies yet? If so, do you recognise the form of your equation? If not, I'd suggest a quick google of the phrase and see if that helps you any.

 

[erobz]

$$ a(x) = \frac{dv}{dt} $$

Use the Chain Rule on the RHS to change the independent variable from time $t$ to position $x$ before you separate variables for the integration.

 

[erobz]

I assume $s$ is a constant.

I think $s$ is the dimension.

$$a(x) = 2 x [ \text{s}^{-2}] $$

 

[fresh_42]

I stumbled upon the following exercise:

The acceleration of a particle is given as a function of position:
a(x) = (2 s^-2)x

I needed to calculate the velocity at a given position x. I initially thought I could solve this by integrating the acceleration function with respect to x to find the velocity as a function of position, and then put in the x-value to get the answer. Sorry if this is a silly question.

I suspect that $s^{-2}$ means Hertz squared.

If you have a position $x_0$, then it is a position at a certain time $t_0$, namely $x_0=x(t_0).$ What we have is $a(x(t))=\overline{a}(t)=(2s^{-2})x(t)=\dfrac{dv(t)}{dt}$ since $x(t)$ is a function of time and $a(x)=a(x(t))=\overline{a}(t).$
This means
\begin{align*}
\int_{t_0}^t \overline{a}(t)\,dt &=\int_{t_0}^t \dfrac{dv(t)}{dt}\,dt= \int_{t_0}^t dv(t) =v(t)-v(t_0)=(2s^{-2})\int_{t_0}^{t} x(t)\,dt
\end{align*}
and
$$
v_0=v(t_0)= v(t)-(2s^{-2})\int_{t_0}^{t} x(t)\,dt = \left.\dfrac{d}{dt}\right|_t x(t) -(2s^{-2})\int_{t_0}^{t} x(t)\,dt
$$
I have no idea what the integral of position over time would stand for, but that is your velocity at time $t_0$ and position $x_0.$ But you need a second information $t$ for the integral and the velocity at that time.

You do not need to call $t$ time. You can as well call it a parameter that parameterizes the path of your positions.

 

[kuruman]

I needed to calculate the velocity at a given position x. I initially thought I could solve this by integrating the acceleration function with respect to x to find the velocity as a function of position, and then put in the x-value to get the answer.

$$\begin{align} & a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=2~(\text{s}^{-2})x \nonumber \\
&\implies v~dv= 2~(\text{s}^{-2})x~dx. \nonumber \end{align}$$Need I continue?

 

[kuruman]

I suspect that $s^{−2}$ means Hertz squared.

Me too.

 

[jbriggs444]

I initially thought I could solve this by integrating the acceleration function with respect to x to find the velocity

The [path] integral of instantaneous force times incremental displacement is known as "work".

Just give the accelerating object a unit mass. Then you can bring a whole pre-built set of intuitions to bear on the problem.

 

[kuruman]

Just give the accelerating object a unit mass. Then you can bring a whole pre-built set of intuitions to bear on the problem.

Like noting that the net force acting on the mass is conservative: $F_{\text{net}}=m~(2\text {s}^{-2})~x.$

 

[PeroK]

I stumbled upon the following exercise:

The acceleration of a particle is given as a function of position:
a(x) = (2 s^-2)x

Note that acceleration is the second derivative of displacement. So, we have:
$$\frac{d^2x}{dt^2} = \alpha^2 x(t)$$That might look familiar!

 

[kuruman]

That might look familiar!

It is also useful for finding $x(t)$ and hence $v(t)$. However, here we need to find $v(x)$ and for that I think the work-energy theorem is the least circuitous method.

 

[Ammaniya]

You can't, because it’s missing details. To find speed from acceleration, you need more information. Integration needs initial conditions to give a complete answer. Just knowing acceleration isn't enough.

 

[robphy]

a(x)=m*x

has an exponential solution since a=\frac{d^2 x}{dt^2} and you have a 2nd-order differential equation for x when m is a constant.

By the way, if you know v as a function of x,
then you can write v as \frac{dx}{dt}.
Your original equation has been re-expressed as a 1st-order ordinary differential equation for x.

 

[Charles Link]

This one could use the specification of the initial conditions. The solution of the second order homogeneous differential equation given in post 12 has a form that we should all know=two linearly independent exponential type solutions with two arbitrary multiplicative constants that get determined by the initial conditions, but it would be nice to hear from the OP.