a ball rolling down a hill was displaced 21.9 m while using uniformly accelerating from rest, If the final velocity was 7.14 m/s, what was t

Question

a ball rolling down a hill was displaced 21.9 m while using uniformly accelerating from rest, If the final velocity was 7.14 m/s, what was the rate of acceleration?

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Philomena 8 months 2021-07-24T18:50:10+00:00 1 Answers 19 views 0

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    2021-07-24T18:51:58+00:00

    Answer:

    a = 1.16 m/s²

    Explanation:

    In order to find the rate of acceleration of the ball, we will use third equation of motion, as follows:

    2as = Vf² – Vi²

    where,

    a = rate of acceleration = ?

    s = distance covered by the ball = 21.9 m

    Vf = Final Velocity of the ball = 7.14 m/s

    Vi = Initial Velocity of the ball = 0 m/s (Since, the ball started from rest)

    Therefore,

    2(a)(21.9 m) = (7.14 m/s)² – (0 m/s)²

    a = (50.97 m²/s²)/(43.8 m)

    a = 1.16 m/s²

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