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p02_039

# Fundamentals of Physics,Vol 1 (Chapters 1 - 20)

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39. We assume the periods of acceleration (duration t 1 ) and deceleration (duration t 2 ) are periods of constant a so that Table 2-1 can be used. Taking the direction of motion to be + x then a 1 =+1 . 22 m/s 2 and a 2 = 1 . 22 m/s 2 . We use SI units so the velocity at t = t 1 is v = 305 / 60 = 5 . 08 m/s. (a) We denote ∆ x as the distance moved during t 1 , and use Eq. 2-16: v 2 = v 2 0 +2 a 1 x = x = 5 . 08 2 2(1 . 22) which yields ∆ x =10 . 59 10 . 6m . (b) Using Eq. 2-11, we have t 1 = v v 0 a 1 = 5 . 08 1 . 22 =4 . 17 s . The deceleration time t 2
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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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