Cuesta College, San Luis Obispo, CA
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj1jtQMiMdKNxGi9qBJ6l4poFlYIqg7sg3P67v3P_f0wUZatdku5xQflLimqmBWGJ_EsegvHPmZxqsN7xTc_Uj8ZFw9DI1puw4wHVBuUyEp9-OsLqYe_lMdCwwn6f6BCOW8UHI03A/s1600/quiz02Cs1o-7.png)
(A) 0 s ≤ t ≤ 7 s.
(B) 7 s ≤ t ≤ 10 s.
(C) (There is a tie.)
(D) (Not enough information given.)
Correct answer (highlight to unhide): (C)
The displacement of the student is the area bounded by the student's vx(t) graph and the time axis. Since the student is always traveling in the forward direction (as all the velocity values are positive during this time interval), then the distance traveled is the same as its displacement.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgyLubS1AjLnfagavf0BVpsp7gVajujCh-2OceqaTVPBf653cYJg3kMyOs0CmPhjvAVvuUI9eezZNzmCjh6UwxUTW3R4Y_Wf9fr_vtsu1_GQORa6BHqb6tnu_wBIC2VReZD6tP5ag/s1600/quiz02Cs1o-7ab.png)
∆x = (1/2)·(5 s)·(2 m/s) + (2 s)·(2 m/s),
∆x = 5 m + 4 m = 9 m.
For 7 s ≤ t ≤ 10 s, the bounded area can be broken up into a rectangle with a triangle atop it:
∆x = (2 s)·(3 m/s) + (1/2)·(3 s)·(2 m/s),
∆x = 6 m + 3 m = 9 m.
Thus the student travels the same distance (9 m) during the 0 s ≤ t ≤ 7 s and the 7 s ≤ t ≤ 10 s time intervals.
Sections 70854, 70855
Exam code: quiz02Cs1o
(A) : 24 students
(B) : 6 students
(C) : 22 students
(D) : 1 student
Success level: 41%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.88
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