During a very quick stop, a car decelerates at 6.80 m/s2. (a) What is the angula

Question

During a very quick stop, a car decelerates at 6.80 m/s2.

(a) What is the angular acceleration (in rad/s2) of its 0.250 m radius tires, assuming they do not slip on the pavement? (Indicate the direction with the sign of your answer. Assume the tires initially rotated in the positive direction)

(b) How many revolutions do the tires make before coming to rest, given their initial angular velocity is 94.0 rad/s?

(c) How long (in s) does the car take to stop completely?

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

(e) What was the car's initial velocity (in m/s)? (Indicate the direction with the sign of your answer.)

(f) Do the values obtained seem reasonable, considering that this stop happens very quickly?

Explanation / Answer

A.

Angular acceleration = acceleration/radius

alpha = a/R = -6.8/0.25 = -27.2 rad/sec^2

B.

wf^2 = wi^2 + 2*alpha*theta

wf = 0

theta = wi^2/(2*alpha) = 94^2/(2*27.2)

theta = 162.43 rad

C.

w = alpha*t

t = w/alpha = 94/27.2 = 3.45 sec

D.

S = theta*R = 162.43*0.25 = 40.61 m

E.

V = w*r = 94*0.25 = 23.5 m/sec

F.

yes these values are reasonable.

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