Cuesta College, San Luis Obispo, CA
The vx(t) graph of an object traveling in a straight line is shown at right. The object starts at x = 0 at t = 0. From t = 0 to t = 8 s, the average speed of the object is: (A) 0 m/s.
(B) 0.5 m/s.
(C) 1 m/s.
(D) 3 m/s.
(E) 4 m/s.
Correct answer (highlight to unhide): (D)
Compare the definitions for the average speed, and the magnitude of average velocity:
average speed = (distance traveled)/(time elapsed ∆t),
magnitude of average velocity |vav,x| = |displacement ∆x|/(time elapsed ∆t).
Since the object travels from t = 0 to t = 8 s 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.
The displacement ∆x is the area "under" the vx(t) graph (more specifically, the area bounded by the trace of the vx(t) curve, and the time axis), which is the square area from t = 0 to t = 4 s, added to the triangular area from t = 4 to t = 8 s:
∆x = (+4 m/s)·(4 s) + (1/2)·(+4 m/s)·(4 s) = (+16 m) + (+8 m) = +24 m.
Then the average velocity from t = 0 to t = 8 s is then:
vav,x = ∆x/∆t = (+24 m)/(8 s) = +3 m/s,
of which the magnitude, 3 m/s, is the average speed for this time interval.
Sections 70854, 70855
Exam code: quiz02Bo74
(A) : 2 students
(B) : 30 students
(C) : 4 students
(D) : 20 students
(E) : 2 students
Success level: 34%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.24
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