# Chapter 2 - Kinematics in One Dimension - Exercises and Problems - Page 66: 52

(a) The elevator moves 12.5 meters while accelerating. (b) It takes 45 seconds to make a complete trip from bottom to top.

#### Work Step by Step

(a) With an acceleration of $1.0~m/s^2$, the elevator takes 5.0 seconds to accelerate to a speed of 5.0 m/s. We can find the distance the elevator moves in the acceleration period. $y_a = \frac{1}{2}at^2$ $y_a = \frac{1}{2}(1.0~m/s^2)(5.0~s)^2$ $y_a = 12.5~m$ The elevator moves 12.5 meters while accelerating. (b) During the deceleration period, the elevator also moves a distance of 12.5 meters in 5.0 seconds. We can find the time it takes the elevator to move the remaining 175 meters at a speed of 5.0 m/s: $t = \frac{d}{v} = \frac{175~m}{5.0~m/s}$ $t = 35~s$ The total time is 5.0 s + 35 s + 5.0 s which is 45 seconds. It takes 45 seconds to make a complete trip from bottom to top.

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