The driver of a car traveling at 32 m/s applies

the brakes and undergoes a constant deceleration of 2.25 m/s^2.

How many revolutions does each tire make

before the car comes to a stop, assuming that

the car does not skid and that the tires have

radii of 0.32 m?

Answer in units of rev

Question

Kael

Physics

27 November, 15:40

The driver of a car traveling at 32 m/s applies

the brakes and undergoes a constant deceleration of 2.25 m/s^2.

How many revolutions does each tire make

before the car comes to a stop, assuming that

the car does not skid and that the tires have

radii of 0.32 m?

Answer in units of rev

+2

Answers (1)

Know the Answer?

Derrick Valentine · Answer 1

110 rev

Explanation:

Given:

v₀ = 32 m/s

v = 0 m/s

a = - 2.25 m/s²

Find: Δx

v² = v₀² + 2aΔx

(0 m/s) ² = (32 m/s) ² + 2 (-2.25 m/s²) Δx

Δx = 228 m

The circumference of the tires is 2π * 0.32 m = 2.01 m. Therefore:

228 m * (1 rev / 2.01 m) = 110 rev

The driver of a car traveling at 32 m/s appliesthe brakes and undergoes a constant deceleration of 2.25 m/s^2.How many revolutions does each tire makebefore the car comes to a stop, assuming thatthe car does not skid and that the tires haveradii of 0.32 m?Answer in units of rev

The driver of a car traveling at 32 m/s applies

the brakes and undergoes a constant deceleration of 2.25 m/s^2.

How many revolutions does each tire make

before the car comes to a stop, assuming that

the car does not skid and that the tires have

radii of 0.32 m?

Answer in units of rev