Acceleration from x vs. t graph

[Ben50275]

Homework Statement

Describe how to use the position vs. time graph to determine the numerical value of the particle's acceleration.

Homework Equations

None, these are Lab questions after we released a cart on a track on an incline.

The Attempt at a Solution

I know that you can derive the position vs. time function into the velocity vs. time function, then derive that to the acceleration vs time function and that would be acceleration, but is that a reasonable answer to this question when it doesn't mention the position vs. time function?

Thanks for all the help

 

[Bystander]

I know that you can derive the position vs. time function into the velocity vs. time function, then derive that to the acceleration vs time function and that would be acceleration, but is that a reasonable answer to this question when it doesn't mention the position vs. time function?

What are the specific computational steps you'll take?

 

[CWatters]

The position vs time graph is just a graphical representation of the position vs time function.

See bystanders post. Perhaps use your graph to plot another for the velocity vs time?

 

[GwtBc]

if the acceleration is constant, you can simply pick a point on your graph and then use $s = v_{i}t+\frac{1}{2} at^2$ where $v_{i} = 0$ to find the acceleration.

 

[Bystander]

doesn't mention the position vs. time function?

@GwtBc There's an even more basic approach.

 

[David Lewis]

(I)s that a reasonable answer to this question when it doesn't mention the position vs. time function?

If I understand correctly, the problem asks for a generalized solution no matter what x=f(t) happens to be. You're given a line on a piece of paper that is a picture of position as a function of time, so you might want to just figure out the situation where x versus t is not a straight line because that solution will work in all cases.

 

[Ray Vickson]

Homework Statement

Describe how to use the position vs. time graph to determine the numerical value of the particle's acceleration.

Homework Equations

None, these are Lab questions after we released a cart on a track on an incline.

The Attempt at a Solution

I know that you can derive the position vs. time function into the velocity vs. time function, then derive that to the acceleration vs time function and that would be acceleration, but is that a reasonable answer to this question when it doesn't mention the position vs. time function?

Thanks for all the help

If you have a graph of $x = f(t)$ (position $x$ vs. time $t$) at a sequence of distinct times $t_1, t_2, t_3, \ldots$ you can try various "differencing" techniques to estimate the velocity $v = g(t)$, then do it again on the $v$ vs. $t$ graph to estimate the acceleration $a = h(t)$.

However, this is likely not a very good way, because you do not REALLY have values of $f(t)$ at different measured values of $t$; you have an error-corrupted table of $f(t)$ values, with some inevitable experimental errors and/or limited-precision measurements. So, standard differencing techniques can be very, very misleading and can yield results that are highly erroneous. A better approach might be to perform a best-fit equation $x = F(t)$ to your measured values $(t_1,x_1), (t_2,x_2), \ldots$ and then take the second-derivative of $F$ as your estimate of acceleration $a$. To do that you would need to make some types of hypotheses as to the form of $f(t)$; for example, for constant acceleration the form would be $f(t) = \gamma + \beta t + \frac{1}{2} \alpha t^2$, where $\alpha$ is the unknown constant acceleration. Using some form of "curve fitting" you could obtain estimates $c, b, a$ of the parameters to get $f_{\text{fit}}(t) = c + b t + \frac{1}{2} a t^2$. The value of $a$ would be your estimate of acceleration $\alpha$.

 

[David Lewis]

Is the problem asking for a graphical solution or an analytical solution? If position vs. time is not a simple mathematical function, it will be faster and easier to use a ruler (or CAD software) to measure the geometric properties in which you're interested. The solution will be approximate, but if more precise than experimental data, close enough.