Physics quiz question: braking car

Physics 205A Quiz 2, fall semester 2011
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Comprehensive Problem 2.57

A car traveling at 19 m/s is brought to a full stop in 3.2 s after the brakes are applied. While braking, the car travels:
(A) 5.9 m.
(B) 30 m.
(C) 61 m.
(D) 97 m.

Correct answer: (B)

The displacement is given by:

∆x = 0.5·(v_fx + v_ix)·∆t = 0.5·(0 + 19 m/s)·(3.2 s) = 30.4 m,

or 30 m with two significant figures.

Response (A) is ∆v/∆t, response (C) is ∆v·∆t, response (D) is 0.5·∆v·(∆t)²

Sections 70854, 70855
Exam code: quiz02p4iN
(A) : 6 students
(B) : 36 students
(C) : 10 students
(D) : 1 student

Success level: 68%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.56

Physics quiz archive: kinematics, free fall

Physics 205A Quiz 2, fall semester 2011
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855, version 1Sections 70854, 70855 results
Exam code: quiz02p4iN

 0- 6 : **  [low = 6]
 7-12 : ****
13-18 : ************
19-24 : *****************  [mean = 21.8 +/- 6.5]
25-30 : ******************  [high = 30]

Physics quiz question: position from velocity graph

Physics 205A Quiz 2, fall semester 2011
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 2.11

Consider this v_x(t) graph of an object traveling in a straight line. The object starts at x = 0 at t = 0. At t = 6 s, the object is located at:
(A) –6 m.
(B) –2 m.
(C) 0 m.
(D) +2 m.
(E) +6 m.

Correct answer: (E)

The displacement delta(x) is given by the area bounded by the v_x(t) graph from t = 0 to t = 6 s, which is a triangular area:

delta(x) = (1/2)*base*height = (1/2)*(2 m/s)*(6 s) = +6 m,

which is positive because velocity was positive during this time interval (the bounded area is above the time axis). Since the object started at x = 0 at t = 0, then at t = 6 s, the object is located at x = +6 m.

Sections 70854, 70855
Exam code: quiz02p4iN
(A) : 3 students
(B) : 3 students
(C) : 27 students
(D) : 3 students
(E) : 17 students

Success level: 32%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.72

Physics quiz question: speed and direction

Physics 205A Quiz 2, fall semester 2011
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 2.11

Consider this v_x(t) graph of an object traveling in a straight line. The object starts at x = 0 at t = 0. From t = 6 s to t = 10 s, the object __________ while moving in the __________ direction.
(A) slows down, positive.
(B) speeds up, positive.
(C) slows down, negative.
(D) speeds up, negative.

Correct answer: (D)

From t = 6 s to t = 10 s, speed increases in the negative direction, (with constant acceleration) with velocity changing from v_x = 0 to v_x = -4 m/s.

Sections 70854, 70855
Exam code: quiz02p4iN
(A) : 5 students
(B) : 1 student
(C) : 10 students
(D) : 37 students

Success level: 70%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.33

Physics quiz question: free fall, initial upwards velocity

Physics 205A Quiz 2, fall semester 2011
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 2.46(a)

A ball is launched vertically upwards from ground level. It reaches a highest point of 12.0 m above the ground. Neglect air resistance. Choose up to be the +y direction. The ball was launched from ground level with an initial speed of:
(A) 3.83 m/s.
(B) 15.3 m/s.
(C) 16.9 m/s.
(D) 118 m/s.

Correct answer: (B)

At the highest point, ∆y = +12.0 m, and v_fy = 0 m/s. So:

v_fy² - v_iy² = 2·a_y·∆y,

and solving for v_iy:

v_iy = sqrt(-2·a_y·∆y) = sqrt(-2*(-9.80 m/s²)·(+12.0 m)) = ±15.33623161 m/s, or 15.3 m/s, to three significant figures.

Response (A) is -a_y·∆t/2, response (C) is ∆y - a_y/2, response (D) is -a_y·∆y.

Sections 70854, 70855
Exam code: quiz02p4iN
(A) : 0 students
(B) : 47 students
(C) : 3 students
(D) : 2 students

Success level: 89%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.33