A car initially traveling at 26.0 m/s undergoes a constant negative acceleration of magnitude 1.50 m/s2 after its brakes are applied.
(a) How many revolutions does each tire make before the car comes to a stop, assuming the car does not skid and the tires have radii of 0.330 m?
rev
(b) What is the angular speed of the wheels when the car has traveled half the total distance?
rad/s
(a) How many revolutions does each tire make before the car comes to a stop, assuming the car does not skid and the tires have radii of 0.330 m?
rev
(b) What is the angular speed of the wheels when the car has traveled half the total distance?
rad/s
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Use kinematics formula v^2 - u^2 = 2*a*s
(a) Final speed v = 0, initial speed u = 26m/s, a = -1.5m/s^2
Linear Distance travelled s = -26^2 / (2*-1.5) m = 225.33m
Circumference of wheel = 2pi*r = 2pi*0.33m = 2.07m
So number of revs = 225.33 / 2.07 = 108.7
(b) When distance s = 225.33/2 = 112.67m,
2*-1.5*112.67 = v^2 - 26^2
Final speed v = sqrt(676-338) m/s = 18.38 m/s
So angular speed w = v/r = 18.38 / 0.33 rad/s = 55.7 rad/s
(a) Final speed v = 0, initial speed u = 26m/s, a = -1.5m/s^2
Linear Distance travelled s = -26^2 / (2*-1.5) m = 225.33m
Circumference of wheel = 2pi*r = 2pi*0.33m = 2.07m
So number of revs = 225.33 / 2.07 = 108.7
(b) When distance s = 225.33/2 = 112.67m,
2*-1.5*112.67 = v^2 - 26^2
Final speed v = sqrt(676-338) m/s = 18.38 m/s
So angular speed w = v/r = 18.38 / 0.33 rad/s = 55.7 rad/s