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A horse running at 4.0 m/s accelerates uniformly to a velocity of
18 m/s in 4.0 s. What is its displacement during the 4.0 s time
interval?

Respuesta :

MathPhys

Answer:

44 m

Explanation:

Given:

v₀ = 4.0 m/s

v = 18 m/s

t = 4.0 s

Find: Δx

Displacement is the average velocity times time:

Δx = ½ (v + v₀) t

Δx = ½ (18 m/s + 4.0 m/s) (4.0 s)

Δx = 44 m

aabdulsamir

Answer: 44m

Explanation:

Initial velocity (u) = 4.0m/s

Final velocity (v) = 18m/s

Time (t) = 4.0s

Acceleration (a) = ?

Displacement (s) =?

From equation of motion,

v = u + at

18 = 4 + a*4

18 = 4 + 4a

a = 14/4 ; a = 3.5 m/s²

From the final equation of motion

V² = U² + 2as

S = (v² - u²) / 2a

S = (18² - 4²) / 2*3.5

S = 44m

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