Unformatted text preview: t 1 + t 2 = 8 . 33 s. The distances traveled during t 1 and t 2 are the same so that they total to 2(10 . 59) = 21 . 18 m. This implies that for a distance of 190 − 21 . 18 = 168 . 82 m, the elevator is traveling at constant velocity. This time of constant velocity motion is t 3 = 168 . 82 m 5 . 08 m / s = 33 . 21 s . Therefore, the total time is 8 . 33 + 33 . 21 ≈ 41 . 5 s....
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 Acceleration, Constant velocity motion

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