Question

3. During a very quick stop, a car decelerates at 6.2 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). a. What is the angular acceleration of its tires in rad/s2, assuming they have a radius of 0.28 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 92 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?

Answer 1

② a= 6.2 m/s2 8 = 0.28M a = o = 6.2 – 22.14 rat/s² 0.1% wi=g2rad/s 2=-22 14 rad /s² wr=0 v2 = uz taas wy² = w; ² + 2a0 o= (922 – 2 x 22.140 - 0= (32) ² 44.28 o = 191.14 yad 0 191.14 - o 30 revolutions 20 revela 2T wf-witat to ganat + = 92 = 4.15 sec 22:14 Vi= rw, = 0.28 x 92= 25.76 M/S a S = Uit & latt s= 25.76 x 4.15-1 x 6.2x (4.15) 2 s = is 106.goy - 53.389 = 53.51m Ni swi 25.76M/s.