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A man roller skates up a hill at 12.5 m/s but slows down, comes to a stop and then accelerates
back down the hill until he is going 13.7 m/s the other direction. Assuming the acceleration is
constant and this take 18.3 s, what is the man's displacement?

Respuesta :

MathPhys

Answer:

-11 m

Explanation:

Given:

v₀ = 12.5 m/s

v = -13.7 m/s

t = 18.3 s

Find: Δx

Δx = ½ (v + v₀) t

Δx = ½ (-13.7 m/s + 12.5 m/s) (18.3 s)

Δx = -11 m

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