The formula for the speed S of a moving object is S=\dfrac{D}{T}, where D is the distance covered, and T is the time taken.

For example, if a car travels a distance of 600\,\textrm{m} within a time of 30\,\textrm{s}, at constant speed, then we can find the speed of the car using the formula as follows:

\begin{align*} S &= \dfrac{D}{T} \\[5pt] &= \dfrac{600}{30} \\[5pt] &= 20 \end{align*}

But what is the unit of speed? We can find it by substituting the units for distance and time into the formula: \begin{align*} S &= \dfrac{D}{T} \\[5pt] &= \dfrac{\textrm{meters}}{\textrm{seconds}} \\[5pt] &= \dfrac{\textrm{m}}{\textrm{s}} \end{align*}

So, the speed of the car is 20\,\textrm{m}/\textrm{s}. When the unit has a quotient like this, we say "meters per second."

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The gravitational acceleration g of a body in free fall is given by the formula g = \dfrac{2h}{t^2}, where h is the distance, measured in meters, traveled by the body due to gravity in time t, measured in seconds. Determine an appropriate measurement unit for gravitational acceleration.

EXPLANATION

To find the unit of gravitational acceleration, we substitute the given units of distance and time into the formula:

\begin{align*} g & = \dfrac{2h}{t^2} \\[5pt] &= \dfrac{2\cdot \textrm{m}}{\textrm{s}^2}\\[5pt] &= 2\cdot \dfrac{\textrm{m}}{\textrm{s}^2} \end{align*}

Therefore the appropriate unit is \dfrac{\textrm{m}}{\textrm{s}^2}. Here, we ignore the unitless constant 2 since it does not affect the units.

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The momentum $p$ is related to mass $m$ and velocity $v$ by the equation $p = m v.$ If $m$ is measured in kilograms and $v$ is measured in $\dfrac{\textrm{m}}{\textrm{s}}$, what is an appropriate measurement unit for momentum?

a
$\dfrac{\textrm{m}}{\textrm{kg}}$
b
$\dfrac{\textrm{s}}{\textrm{kg} \, \textrm{m}}$
c
$\dfrac{\textrm{kg} \, \textrm{m}}{\textrm{s}}$
d
$\dfrac{\textrm{kg}}{\textrm{s}}$
e
$\dfrac{\textrm{kg} \, \textrm{m}}{\textrm{s}^2}$

The velocity of a body with uniform linear motion is given by the formula $v = \dfrac{d}{t},$ where $d$ is the distance traveled by the body in kilometers and $t$ is the time passed in hours.

Determine an appropriate measurement unit for velocity.

a
$\dfrac{\textrm{km}^2}{\textrm{hour}}$
b
$\dfrac{\textrm{km}}{\textrm{hour}}$
c
$\dfrac{\textrm{hour}^2}{\textrm{km}^2}$
d
$\dfrac{\textrm{km}^2}{\textrm{hour}^2}$
e
$\dfrac{\textrm{km}}{\textrm{hour}^2}$

The frequency of a wave is given by the formula f = \dfrac{v}{l}, where v represents the velocity of the wave in \dfrac{\textrm{m}}{\textrm{s}}, while l represents the length of the wave in meters. Find an appropriate measurement unit for frequency.

EXPLANATION

We can find the unit of frequency by substituting the given units of velocity and length into the formula and simplifying. \begin{align} \require{cancel} f & = \dfrac{v}{l}\\[5pt] & = \dfrac{\left( \dfrac{\textrm{m}}{\textrm{s}} \right)}{\textrm{m}}\\[5pt] & = \dfrac{1}{\textrm{m}} \cdot \left( \dfrac{\textrm{m}}{\textrm{s}} \right) \\[5pt] & = \dfrac{ \textrm{m} }{ \textrm{m} \cdot \textrm{s} } \\[5pt] & = \dfrac{ \cancel{\textrm{m}} }{ \cancel{\textrm{m}} \cdot \textrm{s} } \\[5pt] & = \dfrac{1}{\textrm{s}} \end{align}

Thus, the unit of frequency is \dfrac{1}{\textrm{s}}.

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The pressure $P$ applied to a surface is related to the magnitude of the normal force $F$ and the area $A$ of the surface on contact by the equation $P=\dfrac{F}{A}.$ If $F$ is measured in Newtons $\left(\dfrac{\textrm{kg} \, \textrm{m}}{\textrm{s}^2}\right)$, and $A$ is measured in square meters, determine an appropriate measurement unit for pressure.

a
$\dfrac{\textrm{s}^2}{\textrm{kg} \, \textrm{m}}$
b
$\dfrac{\textrm{s}^2 \, \textrm{m}}{\textrm{kg}}$
c
$\dfrac{\textrm{kg} \, \textrm{s}^2}{m}$
d
$\dfrac{\textrm{kg} \, \textrm{m}}{\textrm{s}^2}$
e
$\dfrac{\textrm{kg}}{\textrm{m} \, \textrm{s}^2}$

The period of a wave is given by the formula $ p = \dfrac{l}{v}, $ where $l$ represents the length of the wave in meters, and $v$ represents the velocity of the wave in $\dfrac{\textrm{m}}{\textrm{s}}.$ Find an appropriate measurement unit for period.

a
$\dfrac{1}{\textrm{s}}$
b
$\dfrac{\textrm{s}}{\textrm{m}^2}$
c
$\textrm{m}$
d
$\textrm{s}$
e
$\dfrac{\textrm{m}^2}{\textrm{s}}$

Sometimes, we can be given a formula with values whose units are not the standard ones we know.

For instance, suppose the formula for a person's popularity P is given by P=\dfrac{2f}{s}, where f is the number of friends that person has, and s is the number of siblings they have.

If we want to find the unit of popularity, then we need to substitute the units for the number of friends ( f ) and the number of siblings ( s ) into the formula.

  • The units for the number of friends is ' \textrm{friends} ' since we say 'I have 5 friends'.

  • Similarly, the units for the number of siblings is ' \textrm{siblings}. '

Substituting these units into the formula, we get \begin{align*} P &= \dfrac{2f}{s} \\[5pt] &= \dfrac{2\cdot\textrm{friends}}{\textrm{siblings}} \\[5pt] &= 2\cdot\dfrac{\textrm{friends}}{\textrm{siblings}} \end{align*}

Therefore the appropriate unit is \dfrac{\textrm{friends}}{\textrm{siblings}} , or "friends per sibling." Here we ignore the constant 2 since it does not affect the units.

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A farm with 115 chickens produces 20\,000 eggs per year. To determine the number of eggs P produced by one chicken on average, the farmer uses the formula P=\dfrac{N}{C}, where N is the number of eggs produced by the farm during the year, and C is the number of chickens on the farm.

In words, what is the appropriate measurement unit for P?

EXPLANATION

We are solving for measurement units, so we can ignore the quantities 115 and 20\,000. They do not matter if we are just solving for units.

Since N represents \textrm{eggs} and C represents \textrm{chickens}, the appropriate measurement unit for P=\dfrac{N}{C} is \dfrac{\textrm{eggs}}{\textrm{chickens}}. In words, this is "eggs per chicken."

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$15$ children eat $400$ cookies monthly. If we want to determine the number of cookies that each child eats on average, we use the formula $A = \dfrac{n}{c},$ where $n$ represents the total of cookies while $c$ represents the total of children.

Find an appropriate measurement unit for $A.$

a
$\dfrac{\textrm{children}}{\textrm{cookie}}$
b
$\dfrac{\textrm{cookies}}{\textrm{child}^2}$
c
$\dfrac{\textrm{children}}{\textrm{cookie}^2}$
d
$\dfrac{\textrm{cookies}^3}{\textrm{child}^2}$
e
$\dfrac{\textrm{cookies}}{\textrm{child}}$

In a dog kennel, $20$ dogs eat $60$ treats each day. If we want to determine the number of treats that each dog eats on average, we use the formula $A = \dfrac{n}{d},$ where $n$ represents the total of treats while $d$ represents the total number of dogs.

Find an appropriate measurement unit for $A.$

a
$\dfrac{\textrm{treats}}{\textrm{dog}^2}$
b
$\dfrac{\textrm{treats}}{\textrm{dog}}$
c
$\dfrac{\textrm{dogs}}{\textrm{treat}^2}$
d
$\dfrac{\textrm{treats}^2}{\textrm{dog}^2}$
e
$\dfrac{\textrm{dogs}}{\textrm{treat}}$
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