Question

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

Answer 1

110 rev

Step-by-step 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