# Compound relative uncertainty

Problem

From a word problem:

A car drives 20km with a relative uncertainty of 5%. The time elapsed is measured with an uncertainty of 2%.

What is the relative uncertainty of the average velocity?

Thoughts

If the time elapsed had a given value of, say, 600 seconds, then I feel like calculating

$$v = \frac st = \frac{20,000m \ \pm \ 1000m}{600 \ \text{sec} \ \pm \ \frac{2}{100}600 \ \text{sec}}$$

would have yielded a $$[\min(v), \ \max(v)]$$ interval that I could have used to calculate the compound uncertainty.

But with the time elapsed being unknown, I'm not sure how to solve this.

Does one even need to consider the values in the question, besides just the percentages?

• You can eliminate both values (distance and time) from your calculation, using the fact that $2\% = 0.02 \ll 1$. Sep 15, 2022 at 18:02
• @warlock it's better to use a rigorous method, instead. Please, see the answer below Sep 15, 2022 at 18:34

## Approach

Use RSS method, assuming the measurement of length and time are independent. For details about the RSS method, see https://physics.stackexchange.com/q/727302.

## Solution

You know that $$L = V t$$, so that you can write $$V = \dfrac{L}{t}$$

• space: $$\overline{L} \pm \sigma_L$$
• time: $$\overline{t} \pm \sigma_t$$

We evaluate

• $$\dfrac{\partial V}{\partial L} = \dfrac{1}{t}$$
• $$\dfrac{\partial V}{\partial t} = -\dfrac{L}{t^2}$$

$$\sigma^2_V = \left(\dfrac{\partial V}{\partial L}\right)^2 \sigma_L^2 + \left(\dfrac{\partial V}{\partial t}\right)^2 \sigma_t^2 = \dfrac{1}{\overline{t}^2} \sigma_L^2 + \dfrac{\overline{L}^2}{\overline{t}^4} \sigma_t^2$$
And the relative uncertainty, dividing by $$\overline{V}^2$$ and remembering that $$\overline{V} \overline{t} = \overline{L}$$, reads
$$\dfrac{\sigma^2_V}{\overline{V}^2} = \dfrac{\sigma_L^2}{\overline{L}^2} + \dfrac{\sigma^2_t}{\overline{t}^2} = 0.05^2 + 0.02^2$$ $$\qquad \rightarrow \qquad$$ $$\dfrac{\sigma_V}{\overline{V}} = 0.0539 = 5.39 \%$$