In uniform motion the distance s is measured
In uniform motion the distance $s$ is measured with an accuracy of $0.2\%$ and the time $t$ with an accuracy of $0.4\%$. With what accuracy can the velocity be computed?
Explanation of Method
Relative errors of a quotient add, they do not cancel.
Step 1: Take logarithms first
For a product or quotient, logarithmic differentiation separates the variables completely: \[v = \frac{s}{t} \;\Longrightarrow\; \ln v = \ln s-\ln t \;\Longrightarrow\; \frac{dv}{v} = \frac{ds}{s}-\frac{dt}{t}.\]
Step 2: Bound the worst case
The two measurement errors are independent, so they may have any signs; the largest possible discrepancy occurs when they push in opposite directions, and then the minus sign becomes a plus:
\[ \frac{\bigl|dv\bigr|}{v} \le \frac{\bigl|ds\bigr|}{s}+\frac{\bigl|dt\bigr|}{t}. \]
Step 3: Substitute
\[ \frac{\bigl|dv\bigr|}{v} \le 0.2\%+0.4\% = 0.6\%. \]
Step 4: Verification
\[\text{Take } s = 100,\ t = 10,\ v = 10. \text{ Worst case: } s = 100.2,\ t = 9.96\]
\[v = \frac{100.2}{9.96} = 10.0602, \quad\text{a rise of } 0.602\% \;\;\checkmark\]
\[\text{The small excess over } 0.6\% \text{ is the second-order term } \left(\frac{dt}{t}\right)^{2} = 0.0016\% \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\frac{\bigl|\Delta v\bigr|}{v} \le 0.6\%\,}\]
Remark
Remark. The general rule for $v = s^{\alpha}t^{\beta}$ is $\dfrac{\bigl|dv\bigr|}{v} \le |\alpha|\dfrac{\bigl|ds\bigr|}{s}+|\beta|\dfrac{\bigl|dt\bigr|}{t}$ — absolute values throughout, so errors accumulate no matter what the signs of the exponents. It is worth noting that this is the worst-case bound; if the errors are independent random quantities, the more realistic estimate is $\sqrt{0.2^{2}+0.4^{2}} \approx 0.45\%$.