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Here are a few recommended readings before getting started with this lesson.
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/s2 can be created following these two steps.
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.
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?
In the sentence Voyager 1 travels at 61000 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.
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 =Time intervalDistance traveled
From the above, three formulas can be derived relating speed, distance, and time. The triangular diagram below helps to set each formula.
Multiply the speed of the train by the time it took to go from one city to the other.
r=47.5mi/h, t=4h
Rewrite 47.5mi/h as h47.5mi
Cross out common factors
Cancel out common factors
1a=a
Multiply
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 d1 is given in meters, d2 is given in miles, and the total distance d1+d2 is required to be calculated.
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, d2 can be converted from miles to meters. In such instances, the conversion factor plays an important role.
Given Quantity | Conversion | Result |
---|---|---|
2 hours | 2 hours⋅1 hour60 minutes | 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 60 minutes1 hour. If the task was to convert 120 minutes to hours, 120 minutes would be multiplied by this conversion factor.
Given Quantity | Conversion | Result |
---|---|---|
120 hours | 120 minutes⋅60 minutes1 hour | 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⋅1 pound16 ounces | 48 ounces |
160 ounces | 160 ounces⋅16 ounces1 pound | 10 pounds |
1 mile | 1 mile⋅1 mile1760 yards | 1760 yards |
Some common conversions involve distance, mass, area, volume, time, and temperature.
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.
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.
Keep in mind that, despite the given quantity and the new quantity have different values, they represent the same amount.
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.
At his fastest, Tearrik can throw a fastball at 91miles per hour. How many seconds does it take for the ball to reach home plate? Round the answer to two decimal places.
Use the fact that 1 mile is 1609.34meters and 1hour is 3600seconds.
From the given information, speed and distance are given and time is required.
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.Multiply by 1609.34m/mi
Write as a fraction
Cross out common factors
Cancel out common factors
Multiply
Multiply by 3600s1h
Write as a fraction
Cross out common factors
Cancel out common factors
Calculate quotient
Round to 2 decimal place(s)
d=14.8m, r=40.68m/s
b/ca=ba⋅c
Cross out common factors
Calculate quotient
Round to 2 decimal place(s)
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.
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 | m2 | m2 |
Volume (V) | area × height | m3 | m3 |
Speed (r) | timedistance | m/s | m/s |
Acceleration (a) | timespeed | m/s2 | m/s2 |
Density (ρ) | length3mass | kg/m3 | kg/m3 |
Force (F) | mass × acceleration | kg⋅m/s2 | N (newton) |
Pressure (P) | areaforce | N/m2 | Pa (pascal) |
Energy (E) | force × distance | N⋅m | J (joule) |
Power (P) | timeforce×distance | J/s | W (watt) |
Frequency (F) | time1 | 1/s | Hz (hertz) |
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 =Time takenFinal speed−Initial speed
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.
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.
Final speed=86m/s, Initial speed=50m/s
Subtract terms
Change in speed=36m/s, Time taken=6s
Write as a fraction
ba/c=b⋅ca
Multiply
Calculate quotient
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.
Multiply by 1000g1kg
Cross out common factors
Cancel out common factors
Multiply
Calculate quotient
r=49m/s, t=4s
Write as a fraction
ba/c=b⋅ca
Multiply
Calculate quotient
m=0.145kg, a=12.25m/s2
Multiply
Round to 2 decimal place(s)
kg⋅m/s2=N
Saturday morning before baseball practice, Tearrik went out for a jog beginning at 10:00A.M. He was 1500 meters away from his starting location at 10:10A.M. Over the next 5 minutes he jogged another 500 meters. Tearrik then jogged another 1120 meters before finally stopping at 10:22A.M. Make a graph representing the time and Tearrik's speed during his jog.
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.
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) |
---|---|---|---|
10:00A.M. | 10:10A.M. | 10 | 1500 |
10:10A.M. | 10:15A.M. | 5 | 500 |
10:15A.M. | 10:22A.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 | 101500=150 |
5 | 500 | 5500=100 |
7 | 1120 | 71120=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 10:00A.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.
Next, translate the information from the table to the graph.
Time=SpeedDistance