Solving Error Analysis for Acceleration: Troubleshooting with Simple Formula

[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.

Could someone please help.

 

[dextercioby]

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

Daniel.

 

[xanthym]

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

Daniel.

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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[dextercioby]

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

I'm not interested in that minus...

Daniel.

 

[Data]

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

 

[xanthym]

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

I'm not interested in that minus...

Daniel.

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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