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For a lab that I have to do, the teacher has given me a data table of the displacement of a cart on a ramp after successive intervals of time. I have used this data to derive the velocities and accelerations of the cart.

This is all fine, the problem is that I am now being asked to find the angle of the ramp. I don't know how to figure this out because the acceleration is not constant. After every half a second the acceleration appears to switch between -1.68 m/s2 and -1.72 m/s2. There is no indication that any other forces are involved aside from gravity. How can this be possible if the acceleration of gravity is constant?

My initial thought was that it might be a circular or spiral ramp of some sort, but the displacement never decreases as I would expect after the cart would reach a displacement equal to the diameter of the circle. What else could it be and how could I find the angle from it?

Here is the data that I am given:

Time (s) | Displacement (m)
0.00     | -10.00
0.50     | -5.21
1.00     | -0.85
1.50     | 3.09
2.00     | 6.60
2.50     | 9.69
3.00     | 12.35
3.50     | 14.59
4.00     | 16.40

And here is the data that I have calculated:

Time (s) | Velocity (m/s)
0.50     | 9.6
1.00     | 8.73
1.50     | 7.88
2.00     | 7.02
2.50     | 6.18
3.00     | 5.32
3.50     | 4.48
4.00     | 3.62

Time (s) | Acceleration (m/s^2)
0.75     | -1.7
1.25     | -1.68
1.75     | -1.72
2.25     | -1.68
2.75     | -1.72
3.25     | -1.68
3.75     | -1.72
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  • $\begingroup$ Error is bound in some practical calculations, I guess. You can take geometric mean of all accelerations and equate to $g\sin\theta$. Geometric mean is better as $AM\geqslant GM\geqslant HM$ $\endgroup$ – evil999man Mar 9 '14 at 7:27
  • $\begingroup$ But I guess your professor wants you to take $AM$ as that will come out to be $-1.70$ as there are equal number of $.68$ and $.72$...:D Btw write $-1.7$ as $-1.70$ $\endgroup$ – evil999man Mar 9 '14 at 7:32
  • $\begingroup$ @Awesome Your approach is probably what the teacher expects from us. As for the -1.7 vs -1.70, it should be -1.7 shouldn't it? There are only two significant digits in time 0.00 and time 0.50 values so to me that means that AM should also be to two significant digits. $\endgroup$ – scribblemaniac Mar 9 '14 at 15:27
  • $\begingroup$ Note that the input data are inconsistent with the significant digits, some of the numbers have four (eg 14.59) and some have three (eg 9.69). $\endgroup$ – kevinsa5 Mar 9 '14 at 17:33
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    $\begingroup$ @kevinsa5: The displacements do not have to follow sig figs since they're measurements (so they'd follow precision), the velocities & accelerations ought to follow sig fig rules (and they don't seem to follow it) $\endgroup$ – Kyle Kanos Mar 9 '14 at 18:40
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You are right in that gravity did not change during data collection. You are a victim of uncertainty, which is a very important part of experimental physics. I'm sorry in advance for the "wall of text", and I hope that this clears up some confusion.

The problem is that $1.50$ may not be exactly $1.500000000...$. Because the numbers are provided rounded, they are not exact and you have lost information. Imagine that my watch can only report time to the nearest second and my measuring tape only reports to the nearest meter. If I measure a car's movement and it moves 2.3 meters in 0.8 seconds (true measurements, 2.875 m/s) I am required to round all my data to the nearest round number (2 meters in 1 second, 2.0 m/s). So, if I calculate a number based on my rounded data, it won't perfectly reflect reality because my data do not perfectly reflect reality. Even though your numbers are more precise than my example (accurate down to hundredths of a second and centimeters), there is still some amount of uncertainty in the numbers.

Feel free to skip to the "What it all means" section. The stuff after this is pretty dry and obtuse.

Quantitative Explanation

Note that I'm going to call displacement $x$ and time $t$, just for convenience.

Let's have a look at a subset of the data as an example:

Time (s) | Displacement (m)
1.50     | 3.09
2.00     | 6.60

You know the displacement to two decimal places-- that means that, for $t = 1.50$, $x$ could be anywhere from $3.085$ to $3.094999... \approx 3.095$, and it would still be okay to call it $3.09$, as long as you're rounding to that number of significant digits. Similarly, the time might not be exactly $1.5$! So, if you calculate anything based on those numbers, there's a certain amount of uncertainty in the result. Since there are two variables ($t$ and $x$), they can both vary at once. With the equation $v = (x_f - x_i)/ \Delta t = $, $v$ is biggest when $\Delta x$ is maximized and $\Delta t$ is minimized (dividing by a smaller number yields a larger number), and $v$ is smallest when $\Delta x$ is smallest and $\Delta t$ is biggest. We don't know the true, unrounded values of the two variables, so any possible combination of them is just as good as any other. Here's the range of possible velocities based on the above displacements:

$t = 1.5 \rightarrow 2.0$:

$$ v_{max} = \frac{6.605 - 3.085}{1.995 - 1.505} \approx 7.18 \, \mathrm{m/s} $$

$$ v_{mid} = \frac{6.600 - 3.090}{2.000 - 1.500} = 7.02 \, \mathrm{m/s} $$

$$ v_{min} = \frac{6.595 - 3.095}{2.005 - 1.495} \approx 6.86 \, \mathrm{m/s} $$

Quite a range! Let's do it again for the next time so that we can get a range of accelerations:

$t = 2.0 \rightarrow 2.5$:

$$ v_{max} = \frac{9.695 - 6.595}{2.495 - 2.005} \approx 6.36\, \mathrm{m/s} $$

$$ v_{mid} = \frac{9.690 - 6.600}{2.500 - 2.000} = 6.18 \, \mathrm{m/s} $$

$$ v_{min} = \frac{9.685 - 6.605}{2.505 - 1.995} \approx 6.04 \, \mathrm{m/s} $$

Again, quite a range. Now, to find the max,min accelerations possible for your data, you take the same approach. For $a_{max}$, divide the biggest possible $\Delta v$ by the smallest possible $\Delta t$, and the reverse for the min. For the above numbers, you get accelerations like so:

$$ a_{min} = \frac{6.36 - 6.86}{2.505 - 1.995} \approx -0.98 \, \mathrm{m/s}^2 $$

$$ a_{mid} = \frac{6.18 - 7.02}{2.5 - 2} = -1.68 \, \mathrm{m/s}^2 $$

$$ a_{max} = \frac{6.04 - 7.18}{2.495 - 2.005} \approx -2.33 \, \mathrm{m/s}^2 $$

where "min" and "max" are used (somewhat sloppily) to mean "greatest in magnitude", not the true meanings of "minimum" and "maximum". If I haven't messed up the numbers, the intermediate accelerations could be anywhere in the above range!

What it all means: Now you should be able to recognize that, because your accelerations are derived from numbers of limited certainty and they fall within the range given above, they cannot really be said to be different-- they are said to be "within uncertainty" of each other. It would be appropriate to take the average of all your values and call that the best guess you have of the acceleration. If you wanted to be a true scientist about it, you might have a go at calculating the uncertainty in acceleration, which is nontrivial, or you could go the lazy route (like a lot of scientists :)) and use something like the standard error of all your acceleration values. Either way, the end-all is that the acceleration didn't change during the data collection, it only looks like it did because you didn't measure position and time precisely enough :)

If you're interested in reducing the uncertainty, read on!

These numbers are not acceptable, the range is too big. The problem is that our uncertainties ($0.005 \mathrm{m}$ and $0.005 \mathrm{s}$) are very large compared to our numbers: we're using time steps of half a second, but we have an uncertainty of $2 \cdot 0.005 = 0.01$ in each variable (twice the uncertainty because you're taking the difference of two values). You can reduce the effect by evaluating the difference across more than one time step (ie find the acceleration from $t=0 \rightarrow 1$ instead of $0 \rightarrow 0.5$. This way, the change in time (and displacement) is bigger, but the uncertainty remains the same, so it impacts the result less!

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  • $\begingroup$ Thank you for the "wall of text", I enjoyed learning about the effect that uncertainty had on the results. I knew there would be some uncertainty in the calculations, but I never guessed that it would have made such a large impact on the numbers! I will try to follow your suggestions to attain numbers with larger timesteps. Would it be better if I calculated the maximum and minimum allowed values for acceleration based on the uncertainty and averaged those together to come up with my average acceleration or should I just take the mean of the calculated accelerations? $\endgroup$ – scribblemaniac Mar 9 '14 at 15:42
  • $\begingroup$ Just taking the mean of your calculated accelerations is fine. Another (slightly better) way would be to plot the velocities $v_{mid}$ vs time and take a least-squares linear regression line, as KvdLingen said. You will get a similar number with this method, but because it uses a slightly different method to find the average (least squares vs arithmetic), some people find it preferable. $\endgroup$ – kevinsa5 Mar 9 '14 at 17:21
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I agree with kevinsa5 that the variation in $a$ is due to rounding errors, but I'd like to suggest a better way to analyse the data.

Generally speaking, the best way to analyse data is to find a way to convert it to a straight line, then you can graph it and do a linear regression. In this case the way to procede is to note that if the acceleration is constant the distance is given by the SUVAT equation:

$$ s = ut + \tfrac{1}{2}a t^2 $$

where $u$ is the initial velocity and $a$ is the acceleration. The graph of $s(t)$ isn't a straight line, but we can convert it to a straight line by dividing through by $t$ to get:

$$ \frac{s}{t} = u + \tfrac{1}{2}a t $$

So a graph of $s/t$ against $t$ will be a straight line with gradient $\tfrac{1}{2}a$ and intercept $u$. There's a minor wrinkle that in your data when $t = 0$ $s = -10$, however we can add $10$ to all the distances to make $s(0) = 0$ as long as we remember to subtract if off again at the end.

If we do this to your data we get:

 t    s (+10)  s/t
0.0    0.00   
0.5    4.79   9.58
1.0    9.15   9.15
1.5   13.09   8.73
2.0   16.60   8.30
2.5   19.69   7.88
3.0   22.35   7.45
3.5   24.59   7.03
4.0   26.40   6.60

and a graph of $s/t$ against $t$ looks like:

Graph1

The blue squares are your data and the red line is a linear regression. The data obviously lies on the straight line, and the linear regression gives:

Gradient   -0.8508
Intercept   10.003

So the initial velocity is $u = 10.00$ and the acceleration is $a = -1.70$ (rounding everything to 2 decimal places).

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The best thing to do is draw a graph of velocity versus time. Knowing the acceleration is constant, you draw the best straight line. The inclination will give you the acceleration. You can even draw the maximum inclined and least inclined line to determine your uncertainty.

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