Question

High-speed elevators function under two limitations: (1) the maximum magnitude of vertical acceleration that a typical human body can experience without discomfort is about 1.2 m/s2, and (2) the typical maximum speed attainable is about 9.1 m/s . You board an elevator on a skyscraper's ground floor and are transported 230 mabove the ground level in three steps: acceleration of magnitude 1.2 m/s2 from rest to 9.1 m/s , followed by constant upward velocity of 9.1 m/s , then deceleration of magnitude 1.2 m/s2 from 9.1 m/s to rest.

A)

determine the elapsed time for each of these 3 stages.

Express your answers using two significant figures separated by commas.

tacc,tconstant,tdec

B)Determine the change in the magnitude of the normal force, expressed as a % of your normal weight during each stage.

Express your answers using two significant figures separated by commas.

ΔFN,accFN,ΔFN,constantFN,ΔFN,decFN

Answer 1

For the first stage of the trip, the appropriate equations are

v(t) = v(0) + at

s(t) = s(0) + v(0)t + ½at²

where

s(t) and v(t) are the position and speed, respectively at time t)

s(0) is the initial position (which we will take to be 0 here; the value doesn't matter, since we are concerned only with differences of position here)

v(0) is the initial speed (0 here)

a is the constant acceleration (1.2 here)

For the first stage, set v(t) = 9.1 and solve for t:

9.1 = 1.2t

7.58 = t

So it takes 7.58 sec to accelerate from rest to 9.1 m/s at a constant acceleration of 1.2 m/s²

During this time, the elevator travels s(7.58) m:

s(7.58) = ½(1.2)(7.58)² = 34.5 m

For the second stage, v is constant (9.1 m/s). Let's set our clock back to zero for the beginning of this stage; then for this stage

s(t) = 9.1t

s(t*) = 9.1t*

where t* is the time when we start to decelerate.

For the third stage, the appropriate equations are again

v(t) = v(0) + at

s(t) = s(0) + v(0)t + ½at²

where again we've reset the clock to 0 at the beginning of the deceleration. Now v(0) = 9.1 and a = -1.2, and we want t such that v(t) = 0

0 = 9.1 - 1.2t

7.58 = t

s(7.58) = 9.1(7.58) + ½(-1.2)(7.58)² = 34.5 m

These sounds familiar...

So the first and third stages of the trip take 7.58 seconds each, and account for 2*34.5 = 69 m of the total trip.

Since the total trip is to be 230 m, the length of the second stage is 230 - 69 = 161 m

From our stage 2 work, we found that s(t) = 9.1t, and we now know that s(t*) = 161. Solving for t*,

t* = 161/9.1 = 17.69 sec

So you accelerate for 7.58 sec, travel at constant speed for 17.69 sec, and decelerate for 7.58 sec.

ANS: 7.6s, 18s, 7.6s (2 significant figures-sf)

b)

During the first stage, the normal force is m(9.8 + 1.2) m/s²; during the third stage, it is m(9.8 - 1.2) m/s² (where m is the rider's mass in kg) So for the first stage, the change in magnitude is 1.2m/9.8m; during the third stage, it is -1.2m/9.8m (the m's cancel)

ANS: 12%, 0, -0.12%

IN CASE YOU HAVE PART C:

c) What fraction of the total transport time does the normal force not equal the person's weight?

The normal force equals the person's weight only when the elevator is not accelerating or decelerating; this is the case only during the second stage of the trip, so the fraction of transport time when this is not the case is (7.58 + 7.58) / (7.58 + 17.69 + 7.58) = 0.46 of the time.