Acheetah starting from rest accelerates at a rate of 2.5m/s^2 to reach a speed of 45 m/s.how fare does the cheetah travel

405 m

Explanation:

Given:

v₀ = 0 m/s

a = 2.5 m/s²

v = 45 m/s

Find: Δx

v² = v₀² + 2aΔx

(45 m/s)² = (0 m/s)² + 2 (2.5 m/s²) Δx

Δx = 405 m

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Explanation:

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