Physics clicker question: average velocity vs. instantaneous velocity

Physics 205A, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Multiple-Choice Questions 2.2-2.3 (extended)

Students were asked the following clicker question (Classroom Performance System, einstruction.com) at the beginning of their learning cycle, after having average velocity and average speed defined, and two quick examples of calculating the average velocity and the average speed for the ball for the t = 0.9 s to t = 1.8 s time interval (v_av = -4.4 m/s, and average speed = 4.4 m/s).

A ball is thrown upwards, and follows the trajectory shown at right. What is the average velocity of the ball for the t = 0 to t = 1.8 s time interval?

(A) –4.4 m/s.
(B) –2.2 m/s.
(C) 0.0 m/s.
(D) +2.2 m/s.
(E) +4.4 m/s.
(F) (I'm lost, and don't know how to answer this.)

Sections 70854, 70855
(A) : 2 students
(B) : 13 students
(C) : 23 students
(D) : 2 students
(E) : 2 students
(F) : 1 student

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 70854, 70855
(A) : 5 students
(B) : 6 student
(C) : 31 students
(D) : 0 students
(E) : 2 students
(F) : 0 students

Correct answer: (C)

Average velocity is displacement divided by the elapsed time interval. Because the ball is at the same position for both the initial and final locations, then the displacement (the vector that points from initial to final locations) is zero, and thus the average velocity v_av = 0 for the t = 0 to t = 1.8 s time interval! However, average speed is the distance traveled divided by the interval, and is not "fooled" by a trajectory that doubles back on itself as shown here, and this average speed = 4.4 m/s for the same t = 0 to t = 1.8 s time interval.