Question

A hotel elevator ascends 200 m with a maximum speed of 5.0 m/s. Its acceleration and deceleration both have a magnitude of 1.0 m/s². How far does the elevator move while accelerating to full speed from rest? how long does it take to make the complete trip from bottom to top?

2. MP HW P2.58 (Refer to example 2.10, p.47) A hotel elevator ascends 200m with maximum speed of 5m/s. Its acceleration and deceleration both have a magnitude of 1.0m/s a. How far does the elevator move while accelerating to full speed from rest? b. How long does it take to make the complete trip from bottom to top? A) i) Draw a motion diagram and show the velocity vectors and the acceleration vectors on the motion diagram; ii) Show the coordinate system and all variables you are going to use. iii) List known and unknown values in SI units below B) Sketch the position, velocity and acceleration versus time graphs. Clearly label the starting and ending instants of time of its acceleration (both speeding up and slowing) in each graph. Position-time graph Velocity-time graph Acceleration-time graph C) Find an equation to calculate the displacement the elevator moves while accelerating to full speed from rest. D) Calculate the time the elevator takes to make the complete trip from bottom to top

Answer 1

V^2 = u^2 + 2as (+ 5)^2 = (0)^2 + 2(+ 1)(S) 25/2 = S = > S = 1205 m V = 5 m/s u = 0 a = + 1 m/s^2 V = u + at 5 = 0 + (1)(t) t = 5 sec V^2 = u^2 + 2as 0^2 = (+ 5)^2 + 2(-1)S S = 25/2 = 12.5 m S = 12.5 m V = 0 u = 5 a = -1 = > 0 = 5 + (-1)t = > t = 5 sec how long it move = 12.5 m total journey time = 5 + 35 + 5 = 45 sec

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