Physics quiz question: Around the World in 80 Days average speed

Physics 205A Quiz 1, fall semester 2013
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 1.19, Comprehensive Problem 1.57(d)

Evaluate the following calculation, using an appropriate number of significant figures and/or decimal places.

24,901.55 mi ÷ 8.1×10¹ d = ?

(A) 3.1×10² mi/d.
(B) 3.07×10² mi/d.
(C) 307.4 mi/d
(D) 307.43 mi/d.

Correct answer (highlight to unhide): (A)

24,901.55 has seven significant figures, while 8.1×10¹ d has two significant figures (as denoted in green).

For multiplication and division operations, the term with the least number of significant figures determines the number of significant figures in the result, so the resulting calculation is limited to two significant figures:

24,901.55 ÷ 8.1×10¹ d = 307.42654321 mi/d,

and thus the result of this calculation should be rounded to show only two significant figures in scientific notation, as given by 3.1×10² mi/d.

(N.b.: this is a calculation for the average speed of circumnavigating the globe in 81 days.)

Sections 70854, 70855, 73320
Exam code: quiz01cH3E
(A) : 65 students
(B) : 7 students
(C) : 2 students
(D) : 2 students

Success level: 86%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.45