Compound relative uncertainty

Question

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?

Answer 1

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 $V = \dfrac{L}{t}$

Your measurements are:

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

We evaluate

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

and apply RSS

$\sigma^2_V = \left(\dfrac{\partial V}{\partial L}\right)^2 \sigma_L^2 + \left(\dfrac{\partial V}{\partial t}\right)^2 \sigma_t^2 = \dfrac{1}{\overline{t}^2} \sigma_L^2 + \dfrac{\overline{L}^2}{\overline{t}^4} \sigma_t^2$

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 $ $\qquad \rightarrow \qquad$ $\dfrac{\sigma_V}{\overline{V}} = 0.0539 = 5.39 \%$