Question

Starting position of an object is represented as \( x = 5.1 \pm 0.2 \text{ m} \) and the finishing position as \( y = 6.9 \pm 0.3 \text{ m} \). What will be the displacement and error in displacement?

• A. \( \text{Displacement = 1 m, Error = 0.5 m} \)
• B. \( \text{Displacement = 2 m, Error = 0.36 m} \)
• C. \( \text{Displacement = 1.8 m, Error = 0.36 m} \)
• D. \( \text{Displacement = 1.5 m, Error = 0.4 m} \)

Hint:

When subtracting or adding measurements with uncertainties, use the root-sum-square (RSS) method to find the total uncertainty:
\( \Delta Z = \sqrt{(\Delta A)^2 + (\Delta B)^2} \). This method assumes the uncertainties are independent and random, which is standard in experimental error analysis.

Answer 1

The Correct Option is C

Given:
Starting position, \( x = 5.1 \pm 0.2 \text{ m} \)
Finishing position, \( y = 6.9 \pm 0.3 \text{ m} \) We need to calculate the displacement and the error in displacement. 1. Calculate the Displacement:
Displacement \( \Delta P \) is the difference between the finishing and starting positions:
\[ \Delta P = y - x = 6.9 \, \text{m} - 5.1 \, \text{m} = 1.8 \, \text{m} \] 2. Calculate the Error in Displacement:
When two quantities with associated errors are subtracted (or added), the combined error is calculated using the root-sum-square method:
\[ \Delta E = \sqrt{(\Delta x)^2 + (\Delta y)^2} \] \[ \Delta E = \sqrt{(0.2)^2 + (0.3)^2} = \sqrt{0.04 + 0.09} = \sqrt{0.13} \approx 0.3606 \, \text{m} \] Rounding to two decimal places (consistent with the precision of the given uncertainties), we get:
\[ \Delta E \approx 0.36 \, \text{m} \] Thus, the displacement is \( 1.8 \, \text{m} \) and the error in displacement is \( 0.36 \, \text{m} \). Comparing with the given options:

• (A) Displacement = 1 m, Error = 0.5 m
• (B) Displacement = 2 m, Error = 0.36 m
• (C) Displacement = 1.8 m, Error = 0.36 m \quad (Correct)
• (D) Displacement = 1.5 m, Error = 0.4 m

Option (C) matches both the calculated displacement and error values.