Applications Problem Solving. 6/25/2013 Applications 2 Four-step Method 1. Define variables Name the quantities to be found Write these down Example:

Applications

Problem Solving

Applications 26/25/2013

Four-step Method

1. Define variables Name the quantities to be found

Write these down Example:

Let t = time in seconds Let V = velocity at time t in mph

Solution Methodology


Four-step Method

2. Write down relationships symbolically Example: V = 2t + 20

3. Substitute and solve algebraically Combine relationships if possible E.g., write some variables in terms

of others



Four-step Method

4. State and check solutions Check for reasonableness

Are values within physical limitations? Are negative values reasonable?

Write out response to initial question Does the result answer the question?



Four-step Method Review

1. Define variables

2. Write down relationships symbolically

3. Substitute and solve algebraically

4. State and check solutions

Solution Methodology: Review


Example: Linear Acceleration A vehicle traveling at 20 mph accelerates

by 2 mph per second for 10 seconds

Write an equation for velocity as a linear function of time, graph the function, find the velocity at time 6.5 seconds

Applications


Example: Linear Acceleration Identify variables

Let t = time in seconds

Let a = acceleration (2 mph per second)

Let V = velocity in mph at time t

Let V0 = initial velocity (20 mph)

Applications


Example: Linear Acceleration Relationships

Acceleration a is rate-of-change of velocity

So a is the slope of the graph of V as a linear function of t

Applications

20

40

60

5 10

V mph

tsecs

6.5

33 • •


Linear Acceleration (continued) Relationships (continued)

V Substitute and Solve

V = at + V0

= 2t + 20

At t = 6.5

V = 2(6.5) + 20 = 33

Applications

= at + V0

mph

20

40

60

5 10

V mph

tsecs

6.5

33 • •


Linear Acceleration (continued)

Applications

20

40

60

5 10

V mph

tsecs

6.5

33

State and Check Solution Velocity is 33 mph

after 6.5 seconds

Question:

Why does this t (6.5) work ?

Can we use any t between 0 and 10 ?

Why is 10 seconds not used ?

•


Rates of Completion

We think of some jobs as units of work

Examples: Filling a tank Painting a house Loading a truck

These all allow work sharing One job done by multiple agents, such as people, pipes, machines, etc


Rates of Completion

A certain job takes T minutes to complete How much of the job will be done in 1

minute?

So, per-minute rate of completion is

T1

of the job

T1

jobs/min


Rates of Completion

So, per-minute rate of completion is

Question:

How much work is done in 2T minutes?

T1

jobs/min

T1 jobs

min( ) 2T min( ) 2TT1( )•= min

jobsmin( )

= 2 jobs

*

* Note: A units accounting trick


Rates of Completion

For a given job, each of n agents have

individual completion times of ti where

i = 1, 2, 3, ... , n

Individual completion rates are then ti 1

In T minutes (hours, days, …) each

individual agent completes of the jobti T


Rates of Completion

In T minutes (hours, days, …) each

individual agent completes of the jobti T

E.g. , if the ith person can complete a job in ti = 4 hours, then that person’s rate of completion is one fourth of a job per hour

1ti

14

=

In T = 2 hours, that person completes half the job

Tti

24

=12

=


Rates of Completion

In T hours, n persons working together each complete a fraction of the job, so total job is

Tt1

Tt2

+ +Tt3

+ +Ttn

• • • = 1

= 11t1

1t2

+ +1t3

+ +1tn

• • •T ( )1T

=1t1

1t2

+ +1t3

+ +1tn

• • •

job

job


Rates of Completion

1T

=1t1

1t2

+ +1t3

+ +1tn

• • •

Total job rate of completion is sum of individual agent rates of completion

(Completion Rate) (Completion Time) ●

= One Job

=T1t1

1t2

+ +1t3

+ +1tn

• • •( ) T1T ( )

= 1


Example: A heavy-duty pump can empty a 60,000

gallon tank in 6 hours, while a light-duty pump can empty it in 18 hours

Emptying the tank is the JOB

Rates of completion

First pump:

Second pump:

Combined Rates of Completion

6 1 of the job per hour

18 1 of the job per hour


Example: (continued) With both pumps, overall completion

time is T hours Overall completion rate is


T1

=6 1

18 1+

9 2=

T2 9=

4.5= hours

=18 3

18 1+ =

18 4


Example: An auditorium has three exits of different sizes First exit can empty the room in

t1 = 10 minutes

Rate of Completion =

Second exit can empty the room in

t2 = 8 minutes



1t1

110

=

1t2

18

=


Third exit can empty the room in t3 = 5 minutes


The job : Clear the room in time T

using all three exits

Rate of completion


1t3

15

=

T1 1

t1 = + +

1t2

1t3

110

18= + +

15

The Auditorium



T1 1

t1 = + +

1t2

1t3

110

18= + +

15

4 + 5 + 840=

1740=

110

18

= + +15

44

55

18

Solving for T

T1740

= = 2.3592411 … minutes


Can we meet the fire code requirement for a 2-minute evacuation ?


T1740

= = 2.3592411

Room can be cleared in 2.359 minutes

Question:

Clearly not !

What can we do to meet the requirement ?

Alter the slowest door to clear the room in t minutes and force T to be under 2 minutes



Alter the slowest door to clear the room in t minutes and force T to be under 2 minutes

How fast would the altered door have to work to meet the code, i.e. what is t ?

Combined rates of completion are then

21

T1

=t1=

81

51+ +

t1 =

81

51–

21 –

The New Auditorium



t1 =

81

51–

21 –

t1 =

407

Can you show this ?

t =407

≈ 5.714 … minutes

The altered door must empty the room within 5.7 minutes

So alter to match fastest door (5 minutes)

WHY ?



Alter to match fastest door (5 minutes)

1T

8 + 5 + 840

= 2140

=

≈ 1.905 minutes

… with 5.7 seconds to spare !!

New completion rate is

15

= + +

18

15

T 4021

=

This meets the code requirement

The New Auditorium


Finding Averages

Add n values, divide by n

Call Avg the average of n values of x

Calculations with Data

Avg =x1 + x2 + x3 + ··· + xn

n

=n

n1

k = 1xk


Example

Average of test scores 73, 85, 14, 92

Works fine for small number of values


414

k = 1xk

= 66

= (73 + 85 + 14 + 92) 14

= (264) 14


Example Travel d miles in t hours For d = 200 miles and t = 3 hours

Works for large/infinite number of values


Average speed

=dt

=200 miles3 hours

= 66.67 mph



Finding percentages

A per-100 proportion

a is to b as P is to 100

a as a percentage of number b is

ab

(100) = P

… that is

ab

= 100

P



Example In a chain saw 12 ounces of oil are

added per gallon of gasoline What percentage of the mixture is oil ? Note that 1 gal = 128 oz

For every 100 ounces of mixture 8.571 ounces are oil

% oil ≈ 8.571 %

12

128 + 12=

(100) %



Finding percent change A variable p changes by amount ∆p

What percentage of p is ∆p ?

Percent change in p is

where p is the initial value

∆pp

(100) %



Example The price of gasoline increased from

$2.56 per gallon to $3.89 per gallon

Change is: ∆p = 3.89 – 2.56 = 1.33

Percent change is

∆pp (100) =

1.332.56

(100) ≈ 51.95 %


Think about it !

Documents

Applications Problem Solving. 6/25/2013 Applications 2 Four-step Method 1. Define variables Name the quantities to be found Write these down Example: