A car initially traveling at 23.6 m/s undergoes aconstant negative acceleration

Question

A car initially traveling at 23.6 m/s undergoes aconstant negative acceleration of magnitude 1.70 m/s squared afterits brakes are applied.    How many revolutions does each tire makebefore the car comes to a stop, assuming the car does    not skid and the tires have a radi of .330m? ___________ rev     What is the angular speed of thewheels when the car has traveled half the totaldistance?______rad/s A car initially traveling at 23.6 m/s undergoes aconstant negative acceleration of magnitude 1.70 m/s squared afterits brakes are applied.    How many revolutions does each tire makebefore the car comes to a stop, assuming the car does    not skid and the tires have a radi of .330m? ___________ rev     What is the angular speed of thewheels when the car has traveled half the totaldistance?______rad/s

Explanation / Answer

let the distance to stop is S, vf = 0,vi = 23.6m/s, a = -1.70m/s2 we have: vf2 - vi2 =2aS =>S =( vf2 -vi2 ) / 2a =(0-(23.6m/s)2)/(2*(-1.70m/s2) ) =163.81m distance of one revolution is d1 = 2r = 2*3.14*0.330m =2.072 m total revs = 163.81m/2.072m = 79.059 Revs let the speed is vd at the half way, distance S1 =S/2 = 163.81m/2 = 81.905m vd2 - vi2 =2aS1 => vd = [2aS1+vi2]= [2*(-1.70m/s2)*81.905m + (23.6m/s)2]= 16.688m/s the angular speed at that moment is: = vd/r = 16.688m/s / 0.330m = 50.570rad/s we have: vf2 - vi2 =2aS =>S =( vf2 -vi2 ) / 2a =(0-(23.6m/s)2)/(2*(-1.70m/s2) ) =163.81m distance of one revolution is d1 = 2r = 2*3.14*0.330m =2.072 m total revs = 163.81m/2.072m = 79.059 Revs let the speed is vd at the half way, distance S1 =S/2 = 163.81m/2 = 81.905m vd2 - vi2 =2aS1 => vd = [2aS1+vi2]= [2*(-1.70m/s2)*81.905m + (23.6m/s)2]= 16.688m/s the angular speed at that moment is: = vd/r = 16.688m/s / 0.330m = 50.570rad/s

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