How Many Revolutions Do Tires Make Before Stopping?

AI Thread Summary
A car traveling at 33.8 m/s experiences a constant negative acceleration of 1.80 m/s² after braking. The distance traveled before stopping is calculated to be 9.39 meters, resulting in approximately 1.49 revolutions of each tire, assuming a tire radius of 0.330 m. An error was noted in the initial calculations regarding the initial velocity, which should be squared. The discussion emphasizes the importance of accurate calculations in physics problems. The final results depend on correctly applying the equations of motion.
mandy9008
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Homework Statement


A car initially traveling at 33.8 m/s undergoes a constant negative acceleration of magnitude 1.80 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? (b) What is the angular speed of the wheels when the car has traveled half the total distance?


The Attempt at a Solution


v2=vo2 + aΔx
0= 33.8 m/s + 2(-1.8 m/s2)Δx
Δx=9.39
Δө=9.39 rad / 2π = 1.49 rev
 
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mandy9008 said:

Homework Statement


A car initially traveling at 33.8 m/s undergoes a constant negative acceleration of magnitude 1.80 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? (b) What is the angular speed of the wheels when the car has traveled half the total distance?


The Attempt at a Solution


v2=vo2 + aΔx
0= 33.8 m/s + 2(-1.8 m/s2)Δx
Δx=9.39
Δө=9.39 rad / 2π = 1.49 rev

You forgot to do square for v0. So calculate again.
 

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