During a very quick stop, a car decelerates at 7.8 m/s 2 . Assume the forward mo
ID: 1399319 • Letter: D
Question
During a very quick stop, a car decelerates at 7.8 m/s2. Assume the forward motion of the car corresponds to a positive direction for the rotation of the tires (and that they do not slip on the pavement).Randomized Variablesat = 7.8 m/s2
r = 0.26 m
0 = 95 rad/s (a) What is the angular acceleration of its tires in rad/s2, assuming they have a radius of 0.26 m and do not slip on the pavement?
(b) How many revolutions do the tires make before coming to rest, given their initial angular velocity is 95 rad/s ?
(c) How long does the car take to stop completely in seconds?
(d) What distance does the car travel in this time in meters? (e) What was the car’s initial speed in m/s? During a very quick stop, a car decelerates at 7.8 m/s2. Assume the forward motion of the car corresponds to a positive direction for the rotation of the tires (and that they do not slip on the pavement).
Randomized Variablesat = 7.8 m/s2
r = 0.26 m
0 = 95 rad/s (a) What is the angular acceleration of its tires in rad/s2, assuming they have a radius of 0.26 m and do not slip on the pavement?
(b) How many revolutions do the tires make before coming to rest, given their initial angular velocity is 95 rad/s ?
(c) How long does the car take to stop completely in seconds?
(d) What distance does the car travel in this time in meters? (e) What was the car’s initial speed in m/s?
Explanation / Answer
given a = 7.8 m/s^2
a = alpha*r
alpha = a/r
alpha = 30 rad/s^2
part b )
wf^2 = wi^2 - 2alpha*theta
wf = 0
theta = wi^2/2alpha
wi = 95 rad/s ; alpha = 30 rad/s^2
theta = 150.4 = 150 radian
rev = 150/2pi = 23.87 rev
part c )
wf = wi + alpha*t
wf = 0
alpha = -30 rad/s^2
t = wi/alpha
t = 3.17 sec
part d )
s = theta *r
s = 150 * 0.26
s = 39 m
part e )
v = wr
w = 95
r = 0.26
v = 95 x 0.26
v = 24.7 m/s
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