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18 August, 07:03

An elevator is moving upward at 0.91 m/s when it experiences an acceleration 0.31 m/s2 downward, over a distance of 0.61 m. What will its final velocity be?

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  1. 18 August, 09:55
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    Given:

    u = 0.91 m/s

    a=0.31m/s∧2

    s = 0.61 m

    s = ut + 1/2 (at∧2)

    where s is the displacement of the object

    u is the initial velocity

    t is the time

    a is the acceleration

    Substituting the values

    0.61=0.91*t + (0.31 * t∧2) / 2

    0.61=0.91 t + 0.155 t∧2

    t=0.61 secs

    Consider the equation

    v=u + at

    where v is the initial velocity

    u is the initial velocity

    a is the acceleration

    t is the time

    Substituting the values we get

    v = 0.91 + (0.31*0.61)

    v = 1.099 m/s
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