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A cargo ship initially traveling at 14.1 m/s slows down uniformly and
comes to rest over a distance of 1.60 km (1,600 m). The average
acceleration of the cargo ship is m/s.

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MathPhys

Answer:

-0.0621 m/s²

Explanation:

Given:

Δx = 1600 m

v₀ = 14.1 m/s

v = 0 m/s

Find: a

v² = v₀² + 2aΔx

(0 m/s)² = (14.1 m/s)² + 2a (1600 m)

a = -0.0621 m/s²

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