# 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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