B Acceleration as a function of position

AI Thread Summary
Integrating acceleration a(x) with respect to position x to find velocity v(x) is incorrect because acceleration is defined as a derivative with respect to time. The integration of a(x)dx results in dimensions that do not correspond to velocity. To correctly find velocity as a function of position, one must use the chain rule to relate time and position, requiring additional information such as initial conditions. The discussion emphasizes that without knowing the time variable or initial conditions, integration alone won't yield a complete solution for velocity. Understanding these concepts is crucial for solving problems involving motion and acceleration.
DrLich
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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?
 
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Acceleration is ##a=dv/dt##. It is a derivative with respect to time. What would it mean to integrate that wrt position?
 
Dale said:
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.
 
DrLich said:
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.
 
$$ a(x) = \frac{dv}{dt} $$

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.
 
Ibix said:
I assume ##s## is a constant.
I think ##s## is the dimension.

$$a(x) = 2 x [ \text{s}^{-2}] $$
 
DrLich said:
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.
 
Last edited:
DrLich said:
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?
 
fresh_42 said:
I suspect that ##s^{−2}## means Hertz squared.
Me too.
 
  • #10
DrLich said:
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.
 
  • #11
jbriggs444 said:
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.##
 
  • #12
DrLich said:
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!
 
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  • #13
PeroK said:
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.
 
  • #14
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.
 
  • #15
DrLich said:
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.
 
  • #16
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.
 
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