Physics midterm question: Chrysler 300C braking acceleration magnitude

Physics 205A Midterm 1, fall semester 2012
Cuesta College, San Luis Obispo, CA

"2012 Chrysler 300 - First Drive"
NRMA Motoring and Services

The braking distance[*] for a 2012 Chrysler 300C to slow down from 31 m/s to a complete stop is 50.3 m. Assume that the acceleration is constant during this process. As the car slows down to a stop, the magnitude of the acceleration is:
(A) 9.6 m/s2.
(B) 19 m/s2.
(C) 38 m/s2.
(D) 41 m/s2.

[*] "Braking, 70–0 mph: 165 ft," Andrew Wendler, "2012 Chrysler 300C Long-Term Road Test Update," caranddriver.com/reviews/2012-chrysler-300c-long-term-update-review.

Correct answer (highlight to unhide): (A)

The following quantities are given (or assumed to be known):

(x0= 0 m),
(t0= 0 s),
x= +50.3 m,
v0x = +31 m/s,
vx = 0.

So in the equations for constant (average) acceleration motion in the horizontal direction, the following quantities are unknown, or are to be explicitly solved for:

vx = v0x + ax·t,

x = (1/2)·(vx + v0xt,

x = v0x·t + (1/2)·ax·(t)2,

vx2 = v0x2 + 2·ax·x.

With the unknown quantity ax to be solved for appearing in the fourth equation, with all other quantities given (or assumed to be known), then:

vx2 = v0x2 + 2·ax·x,

(vx2 - v0x2)/(2·x) = ax = (02 – (+31 m/s)2)/2·(+50.3 m) = –9.55268389662 m/s2,

Or to two significant figures, the magnitude of the acceleration would be 9.6 m/s2.

(Response (B) is (vx2v0x2)/x; response (C) is 2·(vx2v0x2)/x; response (D) is (x)2/(2·v0x2).)

Sections 70854, 70855
Exam code: midterm01sWFf
(A) : 24 students
(B) : 29 students
(C) : 3 students
(D) : 0 students

Success level: 42%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.35

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