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.