A car initially traveling at 28.5 m/s undergoes a constantnegative acceleration

Question

A car initially traveling at 28.5 m/s undergoes a constantnegative acceleration of magnitude 1.90 m/s^2 after its brakes areapplied. a.) How many revolutions does each tire make before the carcomes to a stop, assuming the car does not skid and the tires haveradii of 0.330m? b.) What is the angular speed of the wheels when the car hastraveled half the total distance? A car initially traveling at 28.5 m/s undergoes a constantnegative acceleration of magnitude 1.90 m/s^2 after its brakes areapplied. a.) How many revolutions does each tire make before the carcomes to a stop, assuming the car does not skid and the tires haveradii of 0.330m? b.) What is the angular speed of the wheels when the car hastraveled half the total distance?

Explanation / Answer

= aR (t) =0  - t ; where0=vo/r ( acceleration isnegative) (t) =vo/r  - (a/R) t (t)= 0 -0.5t2 since  0 =0and substituning other variables (t)=   0.5(a/R)t2 where t=0/=(vo/R)/(a/R) ;t=  vo/a (t)=  0.5(a/R)[vo/a]2 = 0.5 vo2/[Ra] a) number of revolutions = (t)/(2) N=0.5 vo2 / [2Ra] = vo2 / [4Ra] . b) for half the distance 1/2= /2 =( 0.5vo2/[Ra]) /2 = 1/2 = vo2/(4Ra) t= (2 1/2 / ) t= ( 2 vo2/(4Ra) / (a/R) ); t= ( vo2/(2R2))= vo /(2R) since =t = (a/R)vo /(2R) =a vo/(2R2) .

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