A car initially traveling at 30.1 m/s undergoes a constant negativeacceleration

Question

A car initially traveling at 30.1 m/s undergoes a constant negativeacceleration of magnitude 1.60 m/s2 after its brakesare applied. (a) How many revolutions does each tire make before the car comesto a stop, assuming the car does not skid and the tires have radiiof 0.330 m?
rev

(b) What is the angular speed of the wheels when the car hastraveled half the total distance?
rad/s (a) How many revolutions does each tire make before the car comesto a stop, assuming the car does not skid and the tires have radiiof 0.330 m?
rev

(b) What is the angular speed of the wheels when the car hastraveled half the total distance?
rad/s

Explanation / Answer

   Initial angular speed oftire   =   linear speed /radius       1   =   v/ r   =   30.1 /0.330   =   91.21   rad/s,          angularacceleration      =   a/ r   =   - 1.60 /0.330   =  - 48.48   rad/s2    a.   Final angularspeed   2   =   0          22   =   12   +   2* *        02   =     91.212   +   2* ( -48.48) *    angular displacement of tire before comingtohalt      =   8319.26/96.96   =   85.80   rad    1   rev   =   2   =   6.28   rad    =>      =   85.80/6.28   =   13.66   revs    b.   Over halfdisplacement      1/2   =   85.80/2   =   42.90   rad           1/22    =   12   +   2* * 1/2              1/22    =   91.212   +   2* ( - 48.48) * 42.90              1/2      =   4159.68   =   64.49   rad/s              1/22    =   91.212   +   2* ( - 48.48) * 42.90              1/2      =   4159.68   =   64.49   rad/s       

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