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A bullet moving horizontally with a velocity of 40.0 m/s strikes a sandbag and penetrates a distance of 20.0 cm before coming to rest. what is the acceleration of the bullet in m/s?

Respuesta :

MathPhys

Answer:

-4000 m/s²

Explanation:

Given:

Δx = 0.200 m

v₀ = 40.0 m/s

v = 0 m/s

Find: a

v² = v₀² + 2aΔx

(0 m/s)² = (40.0 m/s)² + 2a (0.200 m)

a = -4000 m/s²

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