Question

009 10.0 points The driver of a car traveling at 43.7 m/s ap- plies the brakes and undergoes a constant deceleration of 1.59 m/s. How many revolutions does each tire make before the car comes to a stop, assuming that the car does not skid and that the tires have radii of 0.22 m? Answer in units of rev.

Answer 1

initial Speed = deakceleration 43.7 m/s rate = 1.59 m)82 final velocity = m/s using equation . 4² - 4² = 295 U find speed - u= initial speed s = distance travelled at aculleration (in this case it is deaubert on substituting 0²-(43.712 = = 2(-1.59) 21-1.59) s I Ove sign for decuelesation of 2 :,-1909169 = -4X1059 XS = 1909.69 2x1'59 600.53m Now radil of tire = 0.gam circumferene = ex īX 0.29 = 2X3.14% 0.22 =_103816 m Total no. of revolutions done by tire = total distance travelled _cireumference of the 600.63m 1.3816 m 434.66 revolutions