Physics quiz question: Mustang GT speeding-up distance

Physics 205A Quiz 2, fall semester 2018
Cuesta College, San Luis Obispo, CA

"2018 Ford Mustang GT 5.0.jpg"
Vauxford
commons.wikimedia.org/wiki/File:2018_Ford_Mustang_GT_5.0.jpg

According to Car and Driver magazine[*], a 2018 Ford Mustang GT is able to accelerate in top gear from 13 m/s to 22 m/s in 10.7 seconds. Assume that car traveled along a straight horizontal line, and that the acceleration was always pointed in the same direction as the velocity. The distance traveled by the car was:
(A) 48 m.
(B) 1.4⨉10² m.
(C) 1.9⨉10² m.
(D) 5.61⨉10² m.

[*] Eric Stafford, "2018 Ford Mustang" (May 2018), caranddriver.com/reviews/2018-ford-mustang-in-depth-review.

Correct answer (highlight to unhide): (C)

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

(x₀= 0 m),
(t₀= 0 s),
t= 10.7 s,
v_0x = +13 m/s,
v_x = +22 m/s.

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

v_x = v_0x + a_x·t,

x = (1/2)·(v_x + v_0x)·t,

x = v_0x·t + (1/2)·a_x·(t)²,

v_x² = v_0x² + 2·a_x·x.

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

x = (1/2)·(v_x + v_0x)·t,

x = (1/2)·((+22 m/s) + (+13 m/s))·(10.7 s) = 187.25 m,

Or to two significant figures, the magnitude of the acceleration would be 1.9⨉10² m.

(Response (A) is (1/2)·(v_x – v_0x)·t; response (B) is v_0x·t (which would be the distance traveled by the car if it maintained its initial speed throughout the entire 10.7 s time interval, instead of speeding up during that time); and response (D) is (1/2)·g·t² (the distance that this car would travel from being released from rest, and falling downwards for 10.7 s).)

Sections 70854, 70855
Exam code: quiz02HwRd
(A) : 4 students
(B) : 1 student
(C) : 47 students
(D) : 0 students

Success level: 90%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.20