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9 October, 07:00

The position equation for a particle is s of t equals the square root of the quantity 2 times t plus 1 where s is measured in feet and t is measured in seconds. find the acceleration of the particle at 4 seconds. 3 ft/sec2 one third ft/sec2 negative 1 over 27 ft/sec2 none of these

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  1. 9 October, 07:58
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    Given that,

    s = √ (2t + 1)

    Time, t = 4 s

    Acceleration, a = ?

    Since,

    Acceleration = velocity / time

    Velocity = distance / time

    Acceleration = distance / time²

    s/t² = √ (2t+1) / t²

    putting t = 4 sec, we have

    a = √ (2*4+1) / 4²

    a = √ (5) / 16

    a = 0.139 ft/s²

    Therefore, acceleration of the given particle will be 0.139 feet / second².
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