Physics 205A Quiz 2, fall semester 2012
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 2.47
An Olympic diver jumps upwards off of a platform and spends 1.78 s in the air until diving into the water 10.0 m below the platform[*]. Neglect air resistance. Choose up to be the +y direction. As the diver travels from the platform to the water, which is the greater quantity?
(A) Average speed.
(B) Magnitude of average velocity.
(C) (There is a tie.)
(D) (Not enough information is given.)
[*] Source: Minute Physics, "Usain Bolt vs. Gravity," http://www.youtube.com/watch?v=9YUtFpLpGfk.
Correct answer: (A)
The straight-line (vertical) distance from the platform to the water is 10.0 m, which is the magnitude of the (vertical) displacement ∆y of the diver. However, the diver traveled upwards from the platform, reaching the highest point above the platform, then traveling downwards into the water, thus the (vertical) distance traveled is greater than the magnitude of the displacement.
Comparing average speed and magnitude of average velocity:
average speed = (distance traveled)/(time elapsed ∆t),
magnitude of average velocity |vav,y| = |displacement ∆y|/(time elapsed ∆t),
since the elapsed time ∆t is 1.78 s for both quantities, then the average speed will be greater than the magnitude of the average velocity. (If the diver had traveled downwards from the platform into the water, without changing direction, then the distance traveled would be the same as the magnitude of the displacement, and the average speed would be the same as the magnitude of average velocity.)
Sections 70854, 70855
Exam code: quiz02Bo74
(A) : 28 students
(B) : 15 students
(C) : 13 students
(D) : 2 students
Success level: 48%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.46
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