Question

Part A During a very quick stop, a car decelerates at 7.40 m/s. What is the angular acceleration of its 0.280-m-radius tires, assuming they do not slip on the pavement? Submit Answer Incompatible units. No conversion found been rpn' and the required units. Tries 0/5 Previous Tries Part B How many revolutions do the tires make before coming to rest, given their initial angular velocity is 90.0 rad/s. (You do not need to enter any units.) Submit Answer Tries 0/5 Part 6 How long does the car take to stop completely? Subrnit Answer Tries 0/5 Part D What distance does the car travel in this time? Submit Answer Tries 0/5 Part E What was the car's initial velocity? (Do the values obtained seem reasonable, considering that this stop happens very quickly?) Submit Answer Tries 0/5

Answer 1

Peltas we know that, id=0 The angular acceleration of the car is, de-a ४ =- 7.40mis 2 0.280m :-26.42 radls? = PastB;. u know that, Wf² =0,² +220 Given that, wf = Oradis W1 = 9 Oradis -(0-1902) 26-26:42 = 153.29 rad = 153129 2XT o = 24,40 rev Partc! Wp = cotat t = wf-wi (0-90) -26:42 t = 3.40 sec

Part Ei The initial Speed, Vi= 60; 7 Vi = 90% 0.280 25.2 misec Posto: The distance toavelled , d = Vitt hat? = 2512 X3.40 +0.547.40X (3,4032 -85.68 +420772 d - 128.45m