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28 August, 15:05

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?

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Answers (2)
  1. 28 August, 15:07
    0
    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
  2. 28 August, 15:35
    0
    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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