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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.**

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.

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?

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.

Assume Voyager $1$ travels toward Proxima Centauri at a constant speed of $61000$ 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.

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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.

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

### Solution

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?

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

Since the train left at $10:30A.M.$ and arrived at $2:30P.M.,$ it took $4$ hours to travel from one city to other.
In conclusion, the cities are $190$ miles apart.

$t=4h $

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

$r=47.5mi/h$, $t=4h$

$d=47.5mi/h⋅4h$

Rewrite

Rewrite $47.5mi/h$ as $h47.5mi $

$d=h47.5mi ⋅4h$

CrossCommonFac

Cross out common factors

$d=h47.5mi ⋅4h$

CancelCommonFac

Cancel out common factors

$d=147.5mi ⋅4$

DivByOne

$1a =a$

$d=47.5mi⋅4$

Multiply

Multiply

$d=190mi$

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.

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.

A conversion factor is a fraction where the numerator and denominator represent the *same* quantity with different units.

### Why

The Reason That the Quantities Are Equivalent

$Example Conversion Factor1hour60minutes $

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 |
---|---|---|

$2$ $hours$ | $2hours ⋅1hour 60minutes $ | $120$ $minutes$ |

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 $60minutes1hour .$ If the task was to convert $120$ minutes to hours, $120$ minutes would be multiplied by this conversion factor.

Given Quantity | Conversion | Result |
---|---|---|

$120$ $hours$ | $120minutes ⋅60minutes 1hour $ | $2$ $hours$ |

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 |
---|---|---|

$3$ $pounds$ | $3pounds ⋅1pound 16ounces $ | $48$ $ounces$ |

$160$ $ounces$ | $160ounces ⋅16ounces 1pound $ | $10$ $pounds$ |

$1$ $mile$ | $1mile ⋅1mile 1760yards $ | $1760$ $yards$ |

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.

$0.25mi⋅1mi1609.34m =402.335m $

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} =300m+402.335m=702.335m $

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.

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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.$Miles to Meters Conversion Factor=1mi1609.34m $

Multiplying the given speed by the conversion factor will convert the speed from miles per hour to meters per hour.
$91mi/h$

Mult

Multiply by $1609.34m/mi$

$91mi/h⋅1609.34m/mi$

Simplify

WriteAsFrac

Write as a fraction

$1h91mi ⋅1mi1609.34m $

CrossCommonFac

Cross out common factors

$1h91mi ⋅1mi1609.34m $

CancelCommonFac

Cancel out common factors

$1h91 ⋅11609.34m $

Multiply

Multiply

$146449.94m/h$

$hour meters ⋅secondshour =secondmeters $

Using the fact that that $1$ hour is equivalent to $3600$ seconds, the conversion factor can be set.
$Hours to Seconds Conversion Factor=3600s1h $

Having the conversion factor set, the conversion can be performed.
$146449.94m/h$

Mult

Multiply by $3600s1h $

$146449.94m/h⋅3600s1h $

Simplify

WriteAsFrac

Write as a fraction

$1h2449.94m ⋅3600s1h $

CrossCommonFac

Cross out common factors

$1h2449.94m ⋅3600s1h $

CancelCommonFac

Cancel out common factors

$3600s146449.94m $

CalcQuot

Calculate quotient

$40.680538…m/s$

RoundDec

Round to $2$ decimal place(s)

$40.68m/s$

$secondmeters meters =second $

Therefore, to compute the time it takes the ball to reach the home plate, the distance must be divided by the speed.
$t=rd $

Substitute values and simplify

SubstituteII

$d=14.8m$, $r=40.68m/s$

$t=40.68m/s14.8m $

DivByFracD

$b/ca =ba⋅c $

$t=40.68m14.8m⋅s $

CrossCommonFac

Cross out common factors

$t=40.68m14.8m⋅s $

CalcQuot

Calculate quotient

$0.363815…s$

RoundDec

Round to $2$ decimal place(s)

$t=0.36s$

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 | $m_{2}$ | $m_{2}$ |

Volume $(V)$ | area $×$ height | $m_{3}$ | $m_{3}$ |

Speed $(r)$ | $timedistance $ | $m/s$ | $m/s$ |

Acceleration $(a)$ | $timespeed $ | $m/s_{2}$ | $m/s_{2}$ |

Density $(ρ)$ | $length_{3}mass $ | $kg/m_{3}$ | $kg/m_{3}$ |

Force $(F)$ | mass $×$ acceleration | $kg⋅m/s_{2}$ | N (newton) |

Pressure $(P)$ | $areaforce $ | $N/m_{2}$ | 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.

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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.

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.
Now that the change in speed is known, the acceleration rate of the Formula $1$ car can be computed.
In conclusion, the acceleration rate of the car during those $6$ seconds was $6$ meters per second squared.

$Change in speed=Final speed−Initial speed$

SubstituteII

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

$Change in speed=86m/s−50m/s$

SubTerms

Subtract terms

$Change in speed=36m/s$

$a=Time takenChange in speed $

SubstituteII

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

$a=6s36m/s $

WriteAsFrac

Write as a fraction

$a=6s1s36m $

DivFracD

$ba/c =b⋅ca $

$a=6s⋅s36m $

Multiply

Multiply

$a=6s_{2}36m $

CalcQuot

Calculate quotient

$a=6m/s_{2}$

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.
### Hint

### Solution

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.

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.

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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.

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.
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.
Finally, the force with which the ball impacted the center field wall can be calculated.
The ball impacted the wall with about $1.78netwons$ of force.

$F=kgmass ⋅m/s_{2}acceleration $

First, convert the mass of the baseball from grams to kilograms using the fact that $1$ kilogram is $1000$ grams.
$145g$

Mult

Multiply by $1000g1kg $

$145g⋅1000g1kg $

Simplify

CrossCommonFac

Cross out common factors

$145g ⋅1000g 1kg $

CancelCommonFac

Cancel out common factors

$145⋅10001kg $

Multiply

Multiply

$1000145kg $

CalcQuot

Calculate quotient

$0.145kg$

$Change in speed=49m/s−0m/s⇓Change in speed=49m/s $

Using this speed and the fact that the ball traveled $4$ seconds before hitting the wall, its acceleration can be found.
$a=tr $

Substitute values and evaluate

SubstituteII

$r=49m/s$, $t=4s$

$a=4s49m/s $

WriteAsFrac

Write as a fraction

$a=4s1s49m $

DivFracD

$ba/c =b⋅ca $

$a=4s⋅s49m $

Multiply

Multiply

$a=4s_{2}49m $

CalcQuot

Calculate quotient

$a=12.25m/s_{2}$

$F=m⋅a$

Substitute values and simplify

SubstituteII

$m=0.145kg$, $a=12.25m/s_{2}$

$F=0.145kg⋅12.25m/s_{2}$

Multiply

Multiply

$F=1.77625kg⋅m/s_{2}$

RoundDec

Round to $2$ decimal place(s)

$F≈1.78kg⋅m/s_{2}$

Substitute

$kg⋅m/s_{2}=N$

$F≈1.78N$

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.

- 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 $(22,160).$

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 $61000$ 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.
Next, substitute the values and simplify. Remember, $r$ stands for the speed (rate).
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.
Consequently, Voyager $1$ will take about $75118$ years to reach Proxima Centauri.

$SpeedDistance =61000km/h=4.014×10_{13}km $

To determine when Voyager $1$ will reach Proxima Centauri, use the formula below. $Time=SpeedDistance $

$t=rd $

Substitute values and simplify

SubstituteII

$d=4.014×10_{13}km$, $r=61000km/h$

$t=61000km/h4.014×10_{13}km $

DivByFracD

$b/ca =ba⋅c $

$t=61000km4.014×10_{13}km⋅h $

CrossCommonFac

Cross out common factors

$t=61000km4.014×10_{13}km⋅h $

CancelCommonFac

Cancel out common factors

$t=610004.014×10_{13}h $

$610004.014×10_{13}h $

Mult

Multiply by $24h1day ⋅365days1year $

$610004.014×10_{13}h ⋅24h1day ⋅365days1year $

Simplify

CrossCommonFac

Cross out common factors

$610004.014×10_{13}h ⋅24h1day ⋅365days 1year $

MultFrac

Multiply fractions

$61000⋅24⋅3654.014×10_{13}years $

Multiply

Multiply

$5343600004.014×10_{13}years $

CalcQuot

Calculate quotient

$75117.898046…years$

RoundInt

Round to nearest integer

$75118years$