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6. Dimensional Analysis and Using Units
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6. 

Dimensional Analysis and Using Units

This lesson delves into the concept of dimensional analysis, a technique used to convert units and solve complex problems. It explains how conversion factors act as a bridge between different units of measurement, making it easier to perform calculations. Derived units are also discussed, which are new units created by combining basic units like meters and seconds. These derived units are particularly useful in calculating speed and distance. For example, if you know the speed of a train and the time it takes to travel between two stations, you can easily find the distance between those stations using derived units. The lesson is a valuable lesson for anyone looking to understand these mathematical tools and how they apply to everyday situations like travel, sports, and science experiments.

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Lesson Settings & Tools
13 Theory slides
12 Exercises - Grade E - A
Each lesson is meant to take 1-2 classroom sessions
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Dimensional Analysis and Using Units
Slide of 13
Some of the most common base units are meters, kilograms, and seconds. Apart from these, are there more units that are used in daily life? The answer is yes! Area, volume, and speed are some examples of quantities that can be found by combining base units.

Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.

Explore

Combining Base Units

The combination of base units creates new units of measure. Some of these combinations have their own names. Using the applet below, units like m/s^2 can be created following these two steps.

  • Click on s and hold the click.
  • Drag it to the denominator of the fraction and release the click.

To have a clean start, press the Reset button. Any of the units inside the boxes can be drag either to the numerator or denominator of the fraction at the right. Build new units and find out if they have a particular name.

Applet allowing different combination of base units
Explore

Distance to the Lightning

One rainy morning, Jordan saw lightning from her bedroom window. After a few seconds of time had passed, she heard the sound of thunder, which shook the photos on her desk. Could Jordan know the distance from her house to the lightning struck? What information does she need?

House, rain, lightning
Challenge

Voyager 1 Arrival Time

Voyager 1 is a space probe launched by NASA in 1977. Currently, it is the most distant human-made object from Earth. It takes more than 21 hours for a radio signal to get from Earth to Voyager 1 — traveling at the speed of light! The nearest star to Earth (apart from the Sun) is Proxima Centauri, which is about 4.24 light years, or 4.014* 10^(13) kilometers away.

Sun and Proxima Centauri with concentric circles around the sun
Assume Voyager 1 travels toward Proxima Centauri at a constant speed of 61 000 kilometers per hour. At 365 days per year, how many years will it take Voyager 1 to reach the star? Round the answer to the nearest integer.

Discussion

Combining Length and Time Units

In the sentence Voyager 1 travels at 61 000 kilometers per hour, both a unit of time and a unit of length are involved. In this case, the unit of length is distance traveled, and the unit of time is the amount of time it takes for the item to travel the given distance. Combining these units leads to a new type of unit that describes how fast an object is moving.

Concept

Speed

The rate or speed of a moving object, denoted by r, measures how fast the object is moving — that is, the rate of motion of the object. It is calculated by dividing the distance traveled by the amount of time spent traveling.

Speed = Distance traveled/Time interval

From the above, three formulas can be derived relating speed, distance, and time. The triangular diagram below helps to set each formula.

Triangle with distance, speed, and time
 The above formulas can be used in different daily life situations. For example, if the speed of a train and the time it takes to travel between two stations are known, the distance between those train stations can be calculated.
Example

Distance Between Train Stations

A train from New York departed at 10:30 A.M. and arrived in Boston at 2:30 P.M.

Train Traveling from New York to Boston
Assuming the train makes no stops between the cities and travels at a constant speed of 47.5 miles per hour, how far apart are the cities?

Hint
Multiply the speed of the train by the time it took to go from one city to the other.

Solution
Since the train left at 10:30 A.M. and arrived at 2:30 P.M., it took 4 hours to travel from one city to other. t = 4 h The speed of the train is 47.5 miles per hour. When the combined unit of miles per hour is multiplied by the base unit hours, the result will be in terms of miles. Therefore, to find the distance between the cities, multiply the speed by the time.

d = r* t
▼
Substitute values and simplify
SubstituteII

r= 47.5 mi/h, t= 4 h

d = 47.5 mi/h * 4 h
Rewrite

Rewrite 47.5 mi/h as 47.5 mih

d=47.5 mi/h* 4 h
CrossCommonFac

Cross out common factors

d = 47.5 mi/h* 4 h
CancelCommonFac

Cancel out common factors

d = 47.5 mi/1* 4
DivByOne

a/1=a

d = 47.5 mi* 4
Multiply

Multiply

d = 190 mi

In conclusion, the cities are 190 miles apart.

Discussion

Converting Units of Measure

Sometimes, two quantities can be given using different units of measure. However, to operate with the quantities, they need to have the same unit. Consider the diagram below, where d_1 is given in meters, d_2 is given in miles, and the total distance d_1+d_2 is required to be calculated.

Two houses and a stadium in between. d1=300 meters and d2=0.25 miles

In such cases, one of the quantities needs to be converted into an equivalent quantity using the other quantity's unit of measure. For example, d_2 can be converted from miles to meters. In such instances, the conversion factor plays an important role.

Concept

Conversion Factor

A conversion factor is a fraction where the numerator and denominator represent the same quantity with different units. Example Conversion Factor [0.5em] 60 minutes/1 hour Recall that 1 hour and 60 minutes represent the same quantity. Multiplying a quantity by a conversion factor changes the quantity to an equivalent quantity in different units. Examine how to convert 2 hours to minutes using the above conversion factor.

Given Quantity Conversion Result
2 hours 2 hours * 60 minutes/1 hour 120 minutes

Although the final result is in minutes, both quantities represent the same amount of time. Note that the opposite conversion, from minutes to hours, has a conversion factor of 1 hour60 minutes. If the task was to convert 120 minutes to hours, 120 minutes would be multiplied by this conversion factor.

Given Quantity Conversion Result
120 minutes 120 minutes * 1 hour/60 minutes 2 hours

As shown in the examples above, the process of including units of measurement as factors is called dimensional analysis. Dimensional analysis can also be used when deciding which conversion factor will produce the desired units. In the table, some common conversion factors are used to convert the given measures.

Given Quantity Conversion Result
3 pounds 3 pounds * 16 ounces/1 pound 48 ounces
160 ounces 160 ounces * 1 pound/16 ounces 10 pounds
1 mile 1 mile * 1760 yards/1 mile 1760 yards

Some common conversions involve distance, mass, area, volume, time, and temperature.

Why

 The Reason That the Quantities Are Equivalent
The numerator and denominator of the conversion factor represent the same quantity. That means their quotient equals 1. Then, by the Identity Property of Multiplication, the amount of the given quantity does not change when multiplied by the conversion factor.

Given Amount * 1 = Given Amount

When converting from one unit to another, the desired unit needs to be in the numerator of the conversion factor while the given unit needs to be in the denominator. That way when the quantity is multiplied by the conversion factor, the given unit will cancel out and the desired unit will remain.

X Given Unit * (Y Desired Unit)/(Z Given Unit) = (XY/Z) Desired Unit

Keep in mind that, despite the given quantity and the new quantity have different values, they represent the same amount.

To convert d_2 from miles to meters, use the fact that 1 mile is the same as 1609.34 meters. 0.25 mi * 1609.34 m/1 mi = 402.335 m It has been shown that 0.25 miles is the same as 402.335 meters. Now that both d_1 and d_2 have the same unit of measure, d_1+d_2 can be found.

d_1+ d_2 &= 300 m + 402.335 m &= 702.335 m
Example

Fastball Time to Plate

Tearrik is studying about the theory of baseball during study hall. He reads that on a baseball field, the distance from the pitcher's mound to home plate is 14.8 meters.

Baseball field and data

At his fastest, Tearrik can throw a fastball at 91 miles per hour. How many seconds does it take for the ball to reach home plate? Round the answer to two decimal places.

Hint
Use the fact that 1 mile is 1609.34 meters and 1 hour is 3600 seconds.

Solution
From the given information, speed and distance are given and time is required.

Baseball field and data

Note that the speed of the ball is given in miles per hour, but the distance from the pitcher's mound to home plate is in meters. This suggests that a conversion from miles to meters is needed. Miles to Meters Conversion Factor = 1609.34 m/1 mi Multiplying the given speed by the conversion factor will convert the speed from miles per hour to meters per hour.

91 mi/h
Mult

Multiply by 1609.34 m/mi

91 mi/h * 1609.34 m/mi
▼
Simplify
WriteAsFrac

Write as a fraction

91 mi/1 h * 1609.34 m/1 mi
CrossCommonFac

Cross out common factors

91 mi/1 h * 1609.34 m/1 mi
CancelCommonFac

Cancel out common factors

91/1 h * 1609.34 m/1
Multiply

Multiply

146 449.94 m/h

Because the answer is required to be in seconds, a conversion from hours to seconds has to be made. To do so, the conversion factor needs to have a quantity in hours in the numerator and the equivalent quantity in seconds in the denominator. meters/hour * hour/seconds = meters/second Using the fact that that 1 hour is equivalent to 3600 seconds, the conversion factor can be set. Hours to Seconds Conversion Factor = 1 h/3600 s Having the conversion factor set, the conversion can be performed.

146 449.94 m/h
Mult

Multiply by 1 h/3600 s

146 449.94 m/h * 1 h/3600 s
▼
Simplify
WriteAsFrac

Write as a fraction

2 449.94 m/1 h* 1 h/3600 s
CrossCommonFac

Cross out common factors

2 449.94 m/1 h* 1 h/3600 s
CancelCommonFac

Cancel out common factors

146 449.94 m/3600 s
CalcQuot

Calculate quotient

40.680538... m/s
RoundDec

Round to 2 decimal place(s)

40.68 m/s

When units of distance are divided by units of speed, the units of distance cancel out, giving a result in units of time. meters/meterssecond = second Therefore, to compute the time it takes the ball to reach the home plate, the distance must be divided by the speed.

t = d/r
▼
Substitute values and simplify
SubstituteII

d= 14.8 m, r= 40.68 m/s

t = 14.8 m/40.68 m/s
DivByFracD

a/b/c= a * c/b

t = 14.8 m* s/40.68 m
CrossCommonFac

Cross out common factors

t = 14.8 m* s/40.68 m
CalcQuot

Calculate quotient

0.363815... s
RoundDec

Round to 2 decimal place(s)

t = 0.36 s

The ball reaches home plate in about 0.36 seconds, around one-third of a second. This is the same amount of time it takes to blink!

Discussion

Defining New Units

As these examples show, speed combines two different base units, distance and time, in a quotient. Similarly, different combinations of base units can lead to new units, which are called derived units.

Concept

Derived Units

Derived units are units of measurement formed by multiplying or dividing base units in different combinations.

Quantity (Symbol) Expression Base Units Unit Symbol (Name)
Area (A) length * width m^2 m^2
Volume (V) area * height m^3 m^3
Speed (r) distance/time .m /s. .m /s.
Acceleration (a) speed/time .m /s^2. .m /s^2.
Density ( ) mass/length^3 .kg /m^3. .kg /m^3.
Force (F) mass * acceleration .kg* m /s^2. N (newton)
Pressure (P) force/area .N /m^2. Pa (pascal)
Energy (E) force * distance N*m J (joule)
Power (P) force* distance/time .J /s. W (watt)
Frequency (F) 1/time .1 /s. Hz (hertz)
The derived units shown in this table are only some of the many possible combinations of base units.

One common example of a derived unit is the acceleration of an object, found by dividing the change in speed by the change in time. The change in speed is measured as the difference between the final and initial speeds.

Acceleration = Final speed-Initial speed/Time taken

Note that if the final speed is less than the initial speed, the acceleration is negative and the object is therefore decelerating. When the speed is constant, the acceleration is 0.
Example

Acceleration of a Formula 1 Car

When crossing the starting line, the speed of a Formula 1 car is 50 meters per second. After 6 seconds, its speed is 86 meters per second. Find the acceleration rate, in meters per second squared, of the car in that time interval.

Hint
The acceleration rate equals the change in speed divided by the time taken. The change in speed is the final speed minus the initial speed.

Solution
It is given that the initial speed of the Formula 1 car is 50 meters per second and the speed after 6 seconds is 86 meters per second. Since the final speed is greater, it means the car is accelerating.

Change in speed = Final speed-Initial speed
SubstituteII

Final speed= 86 m/s, Initial speed= 50 m/s

Change in speed = 86 m/s - 50 m/s
SubTerms

Subtract terms

Change in speed = 36 m/s

Now that the change in speed is known, the acceleration rate of the Formula 1 car can be computed.

a = Change in speed/Time taken
SubstituteII

Change in speed= 36 m/s, Time taken= 6 s

a = 36 m/s/6 s
WriteAsFrac

Write as a fraction

a =36 m1 s/6 s
DivFracD

a/c/b= a/b* c

a = 36 m/6 s * s
Multiply

Multiply

a = 36 m/6 s^2
CalcQuot

Calculate quotient

a = 6 m/s^2

In conclusion, the acceleration rate of the car during those 6 seconds was 6 meters per second squared.

Sometimes information is not provided explicitly, but it can be inferred from context.
Example

Impact Force

During baseball practice, a ball hit by Tearrik traveled 4 seconds before hitting the center field wall at a speed of 49 meters per second. Tearrik wondered about the force with which the ball impacted the wall. Since force is mass times acceleration, to help him, the coach told Tearrik that a baseball has a mass of 145 grams.

Baseball field, animation pitching and batting a baseball
If the speed of the ball when Tearrik hit it was 0 meters per second, find the impact force with which the ball hit the wall. Round the answer to two decimal places.

Hint
From the derived units table, the base units for newtons are kilograms multiplied by meters per second squared. Therefore, the mass of the ball needs to be converted from grams to kilograms. Acceleration equals the change in speed divided by the change in time.

Solution
According to the derived units table, force is obtained by multiplying the mass of the object, in kilograms, by its acceleration in meters per second squared. F = mass_(kg)* acceleration_(m/s^2) First, convert the mass of the baseball from grams to kilograms using the fact that 1 kilogram is 1000 grams.

145 g
Mult

Multiply by 1 kg/1000 g

145 g * 1 kg/1000 g
▼
Simplify
CrossCommonFac

Cross out common factors

145 g * 1 kg/1000 g
CancelCommonFac

Cancel out common factors

145 * 1 kg/1000
Multiply

Multiply

145 kg/1000
CalcQuot

Calculate quotient

0.145 kg

It is given that the ball hit the center field wall at a speed of 49 meters per second. This is the final speed of the ball. Also, it is said that the speed of the ball at the moment it was hit — the initial speed — was at 0 meters per second. Change in speed = 49 m/s - 0 m/s ⇓ Change in speed = 49 m/s Using this speed and the fact that the ball traveled 4 seconds before hitting the wall, its acceleration can be found.

a = r/t
▼
Substitute values and evaluate
SubstituteII

r= 49 m/s, t= 4 s

a = 49 m/s/4 s
WriteAsFrac

Write as a fraction

a = 49 m1 s/4 s
DivFracD

a/c/b= a/b* c

a = 49 m/4 s*s
Multiply

Multiply

a = 49 m/4 s^2
CalcQuot

Calculate quotient

a = 12.25 m/s^2

Finally, the force with which the ball impacted the center field wall can be calculated.

F = m * a
▼
Substitute values and simplify
SubstituteII

m= 0.145 kg, a= 12.25 m/s^2

F = 0.145 kg * 12.25 m/s^2
Multiply

Multiply

F = 1.77625 kg*m/s^2
RoundDec

Round to 2 decimal place(s)

F ≈ 1.78 kg*m/s^2
Substitute

kg*m/s^2= N

F ≈ 1.78 N

The ball impacted the wall with about 1.78 netwons of force.

Graphs can also be used to represent the relationship between units in a given situation. When time is involved, it usually is placed on the horizontal axis.
Example

Time vs. Speed Graph

Saturday morning before baseball practice, Tearrik went out for a jog beginning at 10:00 A.M. He was 1500 meters away from his starting location at 10:10 A.M. Over the next 5 minutes he jogged another 500 meters. Tearrik then jogged another 1120 meters before finally stopping at 10:22 A.M. Make a graph representing the time and Tearrik's speed during his jog.

Answer
Time vs Speed graph

Hint
Place time on the horizontal axis and speed on the vertical axis. Find the speed in each time interval by dividing the distance covered by the time interval.

Solution
Start by organizing the given information in a table. Note that information for three time intervals is given.

Start Time End Time Time Interval (minutes) Distance Covered (meters)
10:00 A.M. 10:10 A.M. 10 1500
10:10 A.M. 10:15 A.M. 5 500
10:15 A.M. 10:22 A.M. 7 1120

To find the speed in each interval, divide the distance by the time interval.

Time Interval (minutes) Distance Covered (meters) Speed (meters per minute)
10 1500 1500/10 = 150
5 500 500/5 = 100
7 1120 1120/7 = 160

Next, place time on the horizontal axis and speed on the vertical axis. Each number to the right of the origin corresponds to the minutes after 10:00 A.M. Based on the table, it is convenient to divide the horizontal axis into 5-minute intervals and the vertical axis into intervals of 50 meters per minute.

Time vs Speed graph

Next, translate the information from the table to the graph.

  • The first 10 minutes, Tearrik's speed was at 150 meters per minute. This information is represented by a horizontal line from (0,150) to (10,150).
  • The next 5 minutes, Tearrik's speed was at 100 meters per minute. A line from (10,100) to (15,100) represents this information.
  • The last 7 minutes, Tearrik's speed was 160 meters per minute. This is represented by a line from (15,160) to (22,160).

Consequently, the graph representing the time and speed during Tearrik's jog is represented as follows.

Time vs Speed graph
Closure

Voyager 1 Arrival Year

With all this new information learned, the time it takes Voyager 1 to reach Proxima Centauri can now be calculated. Recall that Voyager 1 travels at a constant speed of 61 000 kilometers per hour and that the distance from Earth to Proxima Centauri is 4.24 light years, which is the same as 4.014* 10^(13) kilometers. Speed &= 61 000 km/h Distance &= 4.014* 10^(13) km To determine when Voyager 1 will reach Proxima Centauri, use the formula below.

Time= Distance/Speed

Next, substitute the values and simplify. Remember, r stands for the speed (rate).

t = d/r
▼
Substitute values and simplify
SubstituteII

d= 4.014* 10^(13) km, r= 61 000 km/h

t = 4.014* 10^(13) km/61 000 km/h
DivByFracD

a/b/c= a * c/b

t = 4.014* 10^(13) km* h/61 000 km
CrossCommonFac

Cross out common factors

t = 4.014* 10^(13) km* h/61 000 km
CancelCommonFac

Cancel out common factors

t = 4.014* 10^(13) h/61 000
Since the time is asked in years, a unit conversion is needed. To do this conversion, use the fact that 1 year has 365 days and each day has 24 hours.
4.014* 10^(13) h/61 000
Mult

Multiply by 1 day/24 h*1 year/365 days

4.014* 10^(13) h/61 000* 1 day/24 h*1 year/365 days
▼
Simplify
CrossCommonFac

Cross out common factors

4.014* 10^(13) h/61 000* 1 day/24 h*1 year/365 days
MultFrac

Multiply fractions

4.014* 10^(13) years/61 000* 24 * 365
Multiply

Multiply

4.014* 10^(13) years/534 360 000
CalcQuot

Calculate quotient

75 117.898046... years
RoundInt

Round to nearest integer

75 118 years
Consequently, Voyager 1 will take about 75 118 years to reach Proxima Centauri.

Trip of Voyager 1 from Earth to Proxima Centauri


Level 1
Level 2
Level 3
Dimensional Analysis and Using Units
Exercises
Carry out the following conversions. Rewrite 7ft 4in. to inches.

Rewrite 908kg to pounds. Use the fact that 1 pound is approximately 0.454 kilograms.

Rewrite 3.5h to minutes.

Rewrite 200cm to meters.

Converting between feet (ft) and inches (in.) will involve using a conversion factor. Since 12 inches equals 1 foot, the conversion factor becomes the following. 12in./1 ft Multiplying 7 ft by this conversion factor will convert it to inches.

By adding the original 4 inches to 84 inches, the length can be converted to inches only. 7 ft 4 in. =84+4= 88in.

Converting from kilograms (kg) to pounds (lbs) will involve using a conversion factor. As 1 pound is approximately the same as 0.454 kilograms, the conversion factor becomes the following. 1lb/0.454kg Multiplying 908 kilograms by this conversion factor will convert it to pounds.


Converting from hours (h) to minutes (min) involves using a conversion factor. Because in 1 hour there are 60 minutes, the conversion factor becomes the following. 60min/1h Multiplying 3.5 hours by this conversion factor will convert it to minutes.


To convert centimeters to meters, a conversion factor is required. Since 100 centimeters equals 1 meter, the conversion factor becomes the following. 1 m/100 cm Multiplying 200 centimeters by this conversion factor will convert it to meters.


E
Hint & Answer
S
Solution

Which of the recipes requires the least amount of flour per loaf of bread?

  • Recipe A: 5 loaves, 1 cup of flour.
  • Recipe B: 12 loaves, 3 cups of flour.
  • Recipe C: 20 loaves, 6 cups of flour.
  • Recipe D: 21 loaves, 4 cups of flour.
  • Recipe E: 32 loaves, 8 cups of flour.

By determining the unit rate of each recipe, the recipe that requires the least amount of flour per loaf of bread can be determined. Unit Rate: # cups/1loaf Let's calculate the unit rate for each recipe.

Recipe # cups/# loaves # cups/1 loaf
A 1 cup/5 loaves 0.2 cups/1 loaf
B \dfrac {3 \text{ cups}}{12 \text{ loaves}} \dfrac{0.25 \text{ cups}} {1 \text{ loaf}}
C 6 cups/20 loaves 0.3 cups/1 loaf
D 4 cups/21 loaves ≈0.19 cups/1 loaf
E 8 cups/32 loaves 0.25 cups/1 loaf

Recipe D requires the least amount of flour per loaf.

E
Hint & Answer
S
Solution

Maya, Mark, and Dylan are driving from Los Angeles to New York. Their dad tells them that he did the same road trip in his younger years using the same car. Then, it had taken almost 50 hours to drive the 3200 miles. What is the average speed the siblings must drive, in miles per hour, to reach New York 50 hours after leaving Los Angeles? Assume they are constantly driving.

When the distance covered by an object and the time it takes are known, the speed can be calculated using the following formula. Distance=Speed * Time By substituting the time and the distance between the two cities, we can determine the speed at which the siblings need to drive. According to their dad, the distance between New York and Los Angeles is 3200 miles and took him 50 hours.

The siblings must drive at an average speed of 64 miles per hour to reach New York 50 hours after leaving Los Angeles.

E
Hint & Answer
S
Solution

Suppose we want to convert kilometers to miles. What unit should be in the numerator of the conversion factor? How about the denominator?

We know that 1 mile is about 1.609 kilometers. 1 mi ≈ 1.609 km For a conversion factor to work, we need to position these units so that the unit we do not want will cancel out. Since the original quantity is in kilometers, we should place kilometers in the denominator of our conversion factor. 1 mi/1.609 km When the original quantity is multiplied by the conversion factor, this allows for kilometers to cancel out. The miles unit remains. \begin{aligned} \left( ^{\,\text{Original}}_\text{Quantity} \right) {\color{#FF0000}{\cancel{{\color{#0000FF}{\text{km}}}}}} \cdot \dfrac{1 {\color{#009600}{\ \text{mi}}}}{1.609 \ {\color{#FF0000}{\cancel{{\color{#0000FF}{\text{km}}}}}}} = \underbrace{\left(\dfrac{^{\,\text{Original}}_\text{Quantity}}{1.609} \right)}_{\stackrel{\scriptsize{\text{converted}}}{\text{quantity}}} \ {\color{#009600}{\text{mi} }} \end{aligned} Therefore, the desired resulting units, miles, will be in the numerator of the conversion factor.

E
Hint & Answer
S
Solution

Diego ran 15 kilometers in 3 hours, LaShay ran 8 kilometers in 2 hours, and Kriz ran 9 kilometers in 3 hours. Pair each runner to their relative average speed compared to the other runners.

To determine the runner with the fastest average speed, we should calculate the average speed of each person. To do so, we will use the following formula. Speed=Distance/Time By substituting distance and time, we can determine the speed of each runner.

Runner Distance & Time Distance/Time Speed
Diego d= 15
t= 3 		15/3 5 km/h
LaShay d= 8
t= 2 		8/2 4 km/h
Kriz d= 9
t= 3 		9/3 3 km/h

As we can observe, Diego was the fastest, Kriz was the slowest, and LaShay was in the middle.

E
Hint & Answer
S
Solution

Dimensional Analysis and Using Units

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