20111014

Physics midterm question: position from velocity graph

Physics 205A Midterm 1, fall semester 2011
Cuesta College, San Luis Obispo, CA

At right is a vx(t) graph of an object traveling in a straight line. The object starts at x = 0 at t = 0. At t = 10 s, the object is located at:
(A) –10 m.
(B) –5 m.
(C) –0.4 m.
(D) 0 m.
(E) +0.4 m.
(F) +5 m.
(G) +10 m.

Correct answer (highlight to unhide): (G)

The (positive) displacement for t = 0 to t = 10 s is equal to the area bounded by the graph and the t axis, such that:

x = (1/2)·base·height = (1/2)·(10 s)·(+2 m/s) = +10 m,

such that the object is +10 m at the end of the time interval. Responses (E) and (C) are the slopes (and thus accelerations) of the first and second half of the 10 s time interval.

Sections 70854, 70855
Exam code: midterm01w4Sh
(A) : 0 students
(B) : 0 students
(C) : 1 student
(D) : 14 students
(E) : 2 students
(F) : 4 students
(G) : 32 students

Success level: 60%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.45

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