## 20160917

### Physics quiz question: motorcycle drag racing acceleration

Physics 205A Quiz 2, fall semester 2016
Cuesta College, San Luis Obispo, CA

"Motorbike drag racing"
Chris Jones
flic.kr/p/9WqjMY

A sample drag racing time slip for a street motorcycle[*] shows that midway through its race it had a speed of 36.8 m/s in the forward direction, and after traveling 201 m past this point to reach the finish line, the motorcycle then had a speed of 28.7 m/s traveling in the same direction. Assume that the racing strip is a straight horizontal line, and that the acceleration of the motorcycle is constant and always points in the opposite direction as its velocity. The acceleration of the motorcycle while traveling this 201 m distance was:
(A) –1.5 m/s2.
(B) –1.32 m/s2.
(C) –0.780 m/s2.
(D) –0.040 m/s2.

[*] tropiczoneracing.com/images/Sample%20timeslip.jpg.

Correct answer (highlight to unhide): (B)

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

(x0 = 0 m),
x = +201 m,
(t0 = 0 s),
v0x = +36.8 m/s,
vx = +28.7 m/s.

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,

(+28.7 m/s)2 = (+36.8 m/s)2 + 2·ax·(+201 m),

ax = ((+28.7 m/s)2 – (+36.8 m/s)2)/(2·(+201 m)) = –1.319776119 m/s2,

which to two significant figures is –1.3 m/s2.

(Response (A) is the calculation of ax = ∆vx/∆t, but finding the elapsed time from x/v0x; response (C) is vx/v0x; response (D) is ∆vx/∆x.)

Sections 70854, 70855, 73320
Exam code: quiz02SL1p
(A) : 1 student
(B) : 48 students
(C) : 4 students
(D) : 6 students

Success level: 81%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.54