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17 January, 14:31

A local elevator moves upward at a constant 2.1 mps passing a stopped express elevator. Precisely 2.7 seconds later the express elevator catches up with the local elevator where the velocity of the local with respect to the express elevator is - 9.5 m/s. Determine (a) the acceleration of the express elevator in m/s/s he distance traveled in m for the

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  1. 17 January, 15:32
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    4.3 m/s² (approx)

    Explanation:

    Velocity of local elevator with respect to express elevator initially is 2.1 mps

    After 2.7 seconds, the gap between them is

    2.7 x 2.1 = 5.67 m

    The express elevator begins with acceleration. At the time of catching up

    velocity of local with respect to express is - 9.5 m / s. That means velocity of

    express with respect to local will be 9.5 m/s.

    Relative velocity of express is 9.5 m/s

    Absolute velocity of express is 9.5 + 2.1 = 11.6 m/s

    Acceleration of express = change in velocity / time

    = (11.6 - 0) / 2.7 = 4.3 m/s² (approx)
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