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What distance does a mobile travel that with a constant acceleration of 3m / s increases its speed from 6m / s to 12m / s

Respuesta :

Answer:

18 m

Explanation:

Given:

v₀ = 6 m/s

v = 12 m/s

a = 3 m/s²

Find: Δx

v² = v₀² + 2aΔx

(12 m/s)² = (6 m/s)² + 2 (3 m/s²) Δx

Δx = 18 m

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