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8 November, 21:48

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

+5
Answers (1)
  1. 9 November, 00:55
    0
    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
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