During a very quick stop, a car decelerates at 7.8 m/s 2 . Assume the forward mo

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

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---