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Let’s say we have a scenario of a ball being released from the top of the building. This can be modeled simply with the kinematics equation $S=ut +\frac{1}{2}at^2$, which reduced to $S=\frac{1}{2}at^2$. We are given $\Delta t, t, \Delta S, S$, are we are to find $a, \Delta a$.

Firstly, I have no problems calculating the absolute portion of the uncertainty.

Here is my problem: Differentiating $S=\frac{1}{2}at^2$ gives me $\frac{\Delta S}{S}=\frac{\Delta a}{a}+2\frac{\Delta t}{t}$. However, substituting these values gives me a wrong value of $\Delta a$.

The correct approach should have been to rearrange the equation to $a=\frac{2S}{t^2}$, and then solve $\frac{\Delta a}{a}=\frac{\Delta S}{S}+2\frac{\Delta t}{t}$. As can be seen, there appears a contradiction.

Further substitution of $S=82m,\Delta S=1m,t=4.1s, \Delta t=0.2s$ to solve for $a, \Delta a$ using the second equation and then putting this value back into the first gives me a contradiction.

I would like to know which one is correct and which should be used because both seem correct to me.

I have discovered that the addition/subtraction of uncertainties is as follows. Let’s say $(A\pm\Delta A)+(B\pm\Delta B)=(C\pm\Delta C)$.

Then $C_{max}=(A+\Delta A)+(B+\Delta B), C_{min}=(A-\Delta A)+(B-\Delta B)$. Referring back to the definition of uncertainty, $C+\Delta C$ is the average of the minimum and maximum of $C$, thus giving us $C=A+B$ and $\Delta C=\Delta A+\Delta B$.

Using this principle, I am however confused by what I get. $C_{max}=(A+\Delta A)(B+\Delta B), C_{min}=(A-\Delta A)(B-\Delta B)$. Expanding, I got $C=AB +\Delta A\Delta B$, which was contradictory to what I have learnt. I got $\Delta C=A\Delta B + B\Delta A$, which was correct though... This raises a new problem, as I am now unsure as to why the rule applies to multiplication.

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    $\begingroup$ One time I answered a question which seems related or might be a duplicate, I'm not sure. $\endgroup$
    – David Z
    Oct 6 '18 at 3:28
  • $\begingroup$ @AaronStevens I have looked through them. can I ask a question? Let’s say we have $(6\pm1)(5\pm1)$. Using what I’ve learnt, the result is $(30\pm11)$. However, this range does not cover $7\times6$! There appears a contradiction! $\endgroup$
    – QuIcKmAtHs
    Oct 6 '18 at 4:11
  • $\begingroup$ Have you looked into adding in quadrature? $\endgroup$ Oct 6 '18 at 4:16
  • $\begingroup$ What is that? I’m not sure how that is even relevant here. $\endgroup$
    – QuIcKmAtHs
    Oct 6 '18 at 4:25
  • $\begingroup$ Something to keep in mind is that when we write something as $A\pm \Delta A$ we are making a estimate on the actual value according to some confidence interval (typically based on the s standard deviation of our measurements). We are not saying that all measurements will lie within the $\Delta A$ range. So in your example there is actually no contradiction. $\endgroup$ Oct 6 '18 at 4:26
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So I am pretty sure the difference between your two equations depends on what you actually measured and what you are calculating from those measurements.

Think of it this way: One of your two equations (before your addition to the question), tells you how to calculate an uncertainty from the uncertainties of your measurements. In other words, these equations are not relationships between any uncertainties you want. You use the measured quantities on the right to find the uncertainty of the value on the left. You never actually measured the value on the left. (Although if you did a different experiment to directly measure the value on the left, then you would want it to be consistent with said calculations).

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My speculation is that uncertainty refers to the range at which the possible value might lie in. Therefore, when we calculate uncertainties, we always add the absolute/fractional uncertainty.

I have discovered that the addition/subtraction of uncertainties is as follows. Let’s say $(A\pm\Delta A)+(B\pm\Delta B)=(C\pm\Delta C)$.

Then $C_{max}=(A+\Delta A)+(B+\Delta B), C_{min}=(A-\Delta A)+(B-\Delta B)$. Referring back to the definition of uncertainty, $C+\Delta C$ is the average of the minimum and maximum of $C$, thus giving us $C=A+B$ and $\Delta C=\Delta A+\Delta B$.

Using this principle, I am however confused by what I get. $C_{max}=(A+\Delta A)(B+\Delta B), C_{min}=(A-\Delta A)(B-\Delta B)$. Expanding, I got $C=AB +\Delta A\Delta B$, which was contradictory to what I have learnt. I got $\Delta C=A\Delta B + B\Delta A$, which was correct though...

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Aaron Stevens' answer above has already answered the question but I wanted to add something extra that might help others who see this question.

To calculate uncertainty for a multiplication or division, we add the fractional uncertainties in quadrature.

So, considering $S=\frac{1}{2}at^2$, we have \begin{align} \frac{\Delta S}{S} = \sqrt{ \left(\frac{\Delta a}{a}\right)^2 + 2\left(\frac{\Delta t}{t}\right)^2} \end{align} whereas when considering $a=2\frac{S}{t^2}$ we have \begin{align} \frac{\Delta a}{a} = \sqrt{ \left(\frac{\Delta S}{S}\right)^2 + 2\left(\frac{\Delta t}{t}\right)^2} \end{align} In both expressions, the multiplicative factor of $2$ comes from the power of $t$. The multiplicative factor in the expressions for $S$ or $a$ has no bearing on the calculation of fractional uncertainty.

Why if we put $\frac{\Delta S}{S}$ equal in both formulae do we find a contradiction? As Aaron Stevens above explains in his answer above, this is to do with which variables were measured. The $\Delta a$ and $\Delta t$ describe uncorrelated, random uncertainty. That is, it's normally distributed. Therefore, the error will resultant quantity $\frac{\Delta S}{S}$ will have a different value than if you measured $\Delta S$ and $\Delta t$ and tried to determine $\frac{\Delta a}{a}$.

Aside: You mention that you differentiated and saw a contradiction.

Starting with \begin{align} S = \frac{1}{2}at^2 \end{align} and considering the differential $dS$ gives \begin{align} dS = \frac{1}{2}t^2da + atdt \end{align} which, by dividing through by $S$, can be expressed as \begin{align} \frac{ds}{S} = \frac{da}{a} + 2\frac{dt}{t} \end{align} Now, rearranging $S$ to make $a$ the subject so that \begin{align} a = 2\frac{S}{t^2} \end{align} and performing a similar calculation gives \begin{align} da = 2\frac{dS}{t^2} -4S\frac{dt}{t^3} \end{align} which can be expressed as \begin{align} \frac{da}{a} = \frac{dS}{S} - 2\frac{dt}{t} \end{align} This differs in your assertion by a minus sign in the second term. So, just looking at differentials there seems to be no contradiction. However, uncertainties are not calculated by differentiating.

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