PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 2, Problem 72P

(a)

To determine

Whether the passenger in the elevator is gone to a higher or lower floor.

(a)

Expert Solution
Check Mark

Answer to Problem 72P

The passenger in the elevator is gone to a higher floor.

Explanation of Solution

From the given graph the motion of the elevator is represented as an area under curve for the change in velocity. Each space along the t axis represents 2s.

A1=12bh (I)

Here, A1 is the area of the curve, b is the breadth, and h is the height.

Conclusion:

Substitute 8s for b and 0.2m/s2 for h in equation (I) for area A1.

A1=12(8s)(0.2m/s2)=0.8m/s

The elevator accelerates (ay>0) to 0.8m/s during the first 8s. Then, the elevator travels at 0.8m/s(ay=0) for next 8s.

Substitute 4s for b and 0.4m/s2 for h in equation (I) for area A2.

A2=12(4s)(0.4m/s2)=0.8m/s

The elevator slows down (ay<0) until it come to rest. Then it sits for the next 4s. Therefore, the passenger has gone to a higher floor.

(b)

To determine

Sketch the graph of the velocity Versus time.

(b)

Expert Solution
Check Mark

Answer to Problem 72P

The velocity vy of the elevator versus time is shown in the graph.

Explanation of Solution

Sketch the graph vy Versus t by plotting points at two seconds interval with vy determined from the ay versus t. Each rectangle represents (2s)(0.2m/s2)=0.4m/s.

t(s)vyt(s)vy
0The elevator is at rest, so vy=0120.8m/s
21/4(0.4m/s)=0.1m/s140.8m/s
41(0.4m/s)=0.4m/s160.8m/s
67/4(0.4m/s)=0.7m/s181(0.4m/s)+0.8m/s=0.4m/s
82(0.4m/s)=0.8m/s201(0.4m/s)+0.4m/s=0
100.8m/s, since ay=0220, since ay=0

Conclusion:

The graph shows the velocity vy of the elevator versus time taken.

PHYSICS, Chapter 2, Problem 72P , additional homework tip  1

(c)

To determine

Sketch the graph of the position Versus time.

(c)

Expert Solution
Check Mark

Answer to Problem 72P

The acceleration ay of the elevator versus time is shown in the graph.

Explanation of Solution

Each vertical space represents 0.1m/s and each horizontal space is 2. It is increasing until the height is stops.

Break the region under the graph from part (b) into six sections:

Section 1: (0, 4s)

Section (2):(4s, 8s)

Section (3):(8s, 16s)

Section (4):(16s, 18s)

Section (5):(18s, 20s)

Section (6):(20s, 24s)

The approximate distance of the elevator travels in each section by counting boxes under the curves each of them represents displacement (0.1m/s)(2s)=0.2s.

Section 1:3(0.2m)=0.6m

Section (2):13(0.2m)=2.6m

Section (3):32(0.2s)=6.4m

Section (4):6.5(0.2m)=1.3m

Section (5):1.5(0.2m)=0.3m

Section (6):0(0.2m)=0

The plotting data shows the position at the end of each time interval:

t(s)x(m)
00
40.6
83.2
169.6
1810.9
2011.2
2411.2

Conclusion:

The graph shows the position x of the elevator versus time taken.

PHYSICS, Chapter 2, Problem 72P , additional homework tip  2

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Chapter 2 Solutions

PHYSICS

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Position/Velocity/Acceleration Part 1: Definitions; Author: Professor Dave explains;https://www.youtube.com/watch?v=4dCrkp8qgLU;License: Standard YouTube License, CC-BY