Homework Help: Simple Error Analysis Problem

1. Apr 7, 2005

flower76

I'm having a problem with something I know should be simple, but my answer is off so I'm doing something wrong.

I need to find the amount of error for an acceleration that was found using the formula a=2d/t^2. Where d represents distance travelled. There is no uncertainty in the distance measurement, only the time.

2. Apr 7, 2005

dextercioby

$$\Delta a=\left|\Delta \left(\frac{2d}{t^{2}}\right)\right| =4dt^{-3} \Delta t$$

Daniel.

3. Apr 7, 2005

xanthym

Just in case you're also interested in the SIGN of the error "Δa" in "a" for a given error "Δt" in "t":

$$1: \ \ \ \ \Delta a \ = \ \Delta \left(\frac{2d}{t^{2}}\right) \ = \ \left ( \frac{\color{red} \mathbf{-} \color{black} 4d}{t^{3}} \right ) \Delta t$$

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4. Apr 7, 2005

dextercioby

There are no such things as negative errors.Errors always add...

I'm not interested in that minus...

Daniel.

5. Apr 7, 2005

Data

Yes, reported errors are standard deviations (or they should be), and hence are always positive (the definition of standard deviation of $X$ is $\sqrt{\mbox{Var} X}$).

6. Apr 7, 2005

xanthym

The term "error" alone can be ambiguous. "Standard Deviation" and "Variance" are much more specific, and they are always positive and always "add":

$$1: \ \ \ \ \ \ \ \color{blue}\mbox{Var(a)}\color{black} \ = \ \overline { \left ( \Delta a \right )^{2}} \ = \ \overline{ \left ( \Delta \left(\frac{2d}{t^{2}}\right) \right )^{2} }\ = \ \left ( \frac{-4d}{t^{3}} \right )^{2} \overline{ \left ( \Delta t \right )^{2} } \ \ + \ \ \left ( \frac{2}{t^{2}} \right )^{2} \overline{ \left ( \Delta d \right )^{2} }$$

$$: \hspace{9cm} \left ( For \ \ \overline{\Delta a} = \overline{\Delta t} = \overline{\Delta d} = \overline{\Delta t \Delta d} = 0 \right )$$

$$2: \ \ \ \ \color{red}(\mbox{Standard Deviation})\color{black} \ = \ +\sqrt{\color{blue} \mbox{Var(a)}}$$

The question here is what the OP had in mind. (We don't know what the OP originally meant by the term in Msg #1.) You may not be interested in the (-) sign, but the OP might have been ... thus the clarification in Msg #3.

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