During a very quick stop, a car decelerates at 7.4 m/s^2. Assume the forward mot

Question

During a very quick stop, a car decelerates at 7.4 m/s^2. Assume the forward motion of the car corresponds to a positive directin for the rotation of the tires (and that they do not slip the pavement).

a) What's the angular acceleration of it's tires in rad/s^2, assuming they have a radius of 0.29 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 90 rad/s

c) How long does it take the car to stop completely in seconds?

d) What distance does the car travel in this time in meters?

e) What's the car's initial speed in m/s?

Explanation / Answer

(A) alpha = a / r

= - 7.4 / 0.29

= - 25.5 rad/s^2  

(B) wf^2 - wi^2 =2 alpha theta

0^2 - 90^2 = 2(-25.5)(theta)

theta = 158.7 rad  

revolutions = theta / 2pi = 25.3 revolutions

(C) wf = wi + alpha t

0 = 90 - 25.5 t

t = 3.53 sec  

(D) distance = theta r = 46 m

(e) v = w r = 90 x 0.29 = 26.1 m/s  

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.