Applications Problem Solving. 6/25/2013 Applications 2 Four-step Method 1. Define variables Name the...

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

Applications 36/25/2013

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

Solution Methodology

Applications 46/25/2013

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?

Solution Methodology

Applications 56/25/2013

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

Applications 66/25/2013

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

Applications 76/25/2013

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

Applications 86/25/2013

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 • •

Applications 96/25/2013

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 • •

Applications 106/25/2013

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 ?

•

Applications 116/25/2013

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

Applications 126/25/2013

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

Applications 136/25/2013

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

Applications 146/25/2013

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

Applications 156/25/2013

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

=

Applications 166/25/2013

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

Applications 176/25/2013

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

Applications 186/25/2013

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

Applications 196/25/2013

Example: (continued) With both pumps, overall completion

time is T hours Overall completion rate is

Combined Rates of Completion

T1

=6 1

18 1+

9 2=

T2 9=

4.5= hours

=18 3

18 1+ =

18 4

Applications 206/25/2013

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

Rate of Completion =

Combined Rates of Completion

1t1

110

=

1t2

18

=

Applications 216/25/2013

Third exit can empty the room in t3 = 5 minutes

Rate of Completion =

The job : Clear the room in time T

using all three exits

Rate of completion

Combined Rates of Completion

1t3

15

=

T1 1

t1 = + +

1t2

1t3

110

18= + +

15

The Auditorium

Applications 226/25/2013

Combined Rates of Completion

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

Applications 236/25/2013

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

Combined Rates of Completion

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

Applications 246/25/2013

Combined Rates of Completion

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

Applications 256/25/2013

Combined Rates of Completion

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 ?

Applications 266/25/2013

Combined Rates of Completion

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

Applications 276/25/2013

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

Applications 286/25/2013

Example

Average of test scores 73, 85, 14, 92

Works fine for small number of values

Calculations with Data

414

k = 1xk

= 66

= (73 + 85 + 14 + 92) 14

= (264) 14

Applications 296/25/2013

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

Works for large/infinite number of values

Calculations with Data

Average speed

=dt

=200 miles3 hours

= 66.67 mph

Applications 306/25/2013

Calculations with Data

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

Applications 316/25/2013

Calculations with Data

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) %

Applications 326/25/2013

Calculations with Data

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) %

Applications 336/25/2013

Calculations with Data

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 %

Applications 346/25/2013

Think about it !