Getting the Job Done—Speed

Getting the Job Done—Speed exercise

• Find the distance travelled.

  Hints

  The equation relating distance, rate, and time is $d=r\cdot t$. Distance equals rate times time.

  For example: If you drive at $50~\frac{\text{mi}}{\text{hr}}$ for $2~\text{hr}$, you will travel $100~\text{mi}$:

  $d=50~\frac{\text{mi}}{\text{hr}} \cdot 2~\text{hr}$

  $d=100~\text{mi}$

  When using the equation $d=r\cdot t$, the values for $d$, $r$, and $t$ will have units. These units will need to be simplified by cancelling in order to determine the units for your solution.

  For example, in the equation:

  $d=5~\frac{\text{km}}{\text{hr}} \cdot 3~ \text{hr}$

  The hours cancel:

  $\frac{\text{km}}{\text{hr}} \cdot \text{hr} = \text{km}$

  Giving us:

  $d=15~\text{km}$

  Note that our solution was a distance, and distance can be measured in kilometers.

  A rate of speed is a distance covered per unit time. Some examples are:

  • miles per hour
  • kilometers per minute
  • feet per second

  Solution

  The distance travelled is the rate times the time. This relationship is expressed by the equation:

  $d=r\cdot t$

  $d$ represents the distance, $r$ represents the rate, and $t$ represents the time.

  Walter flew to the oasis at $20~\frac{\text{ft}}{\text{sec}}$. That's the rate. We recognize it's a rate because of its units, feet per second.

  Walter flew for $4~\text{sec}$. That's the time.

  The distance is unknown.

  To find the distance, we substitute the known values (with units) for $r$ and $t$ into the equation:

  $d=$ $20~\frac{\text{ft}}{\text{sec}}$ $\cdot$ $9~\text{sec}$

  First, let's simplify the units. The seconds cancel:

  $\frac{\text{ft}}{\text{sec}}\cdot\text{sec}=\frac{\text{ft}}{\text{sec}}\cdot\frac{\text{sec}}{1}=\text{ft}$

  This give us:

  $d=20~\text{ft}\cdot 9$

  Multiplying the values gives us:

  $d=180~\text{ft}$

  That's the distance Walter flew.

• Determine the time it took to travel.

  Hints

  The equation $d=r\cdot t$ features three variables. Typically, we are given the values for two of the variables, and we must find the third, unknown variable. This sometimes requires use of properties of equality to solve for the unknown.

  • If you want to know $d$, $d=r\cdot t$ is already solved for $d$.
  • If you want to know $t$, solve for $t$ by dividing.
  • If you want to know $r$, solve for $r$ by dividing.

  Once the general equation is solved for the unknown, you can then substitute to calculate the numerical value for the unknown.

  Dividing by a fraction is equivalent to multiplying by its reciprocal. This is useful when simplifying units in problems involving distance, rate, and time.

  Here is an example:

  $\frac{60~\text{km}}{30~\frac{\text{km}}{\text{hr}}} = \frac{60~\text{km}}{30} \cdot \frac{\text{hr}}{\text{km}} = 2~\text{hr}$

  The units of a quantity tell you if it's a distance, rate, or time. For example:

  • $30~\text{yd}$ is a distance.
  • $12~\frac{\text{yd}}{\text{min}}$ is a rate.
  • $7~\text{min}$ is a time.

  Solution

  We can use the equation $d=r\cdot t$.

  We know $d$ and $r$.

  • The distance to the oasis was given as $180~\text{ft}$.
  • The rate was given to us as $5~\frac{\text{ft}}{\text{sec}}$.

  We want to know $t$, the time it will take for them to get to the oasis.

  We solve our equation for $t$ by dividing by $r$:

  $\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline d=r\cdot t & \text{The equation for distance.}\\ \frac{d}{r}=\frac{r\cdot t}{r} & \text{Dividing both sides by}~r.\\ \frac{d}{r}=t & \text{Canceling}~r~\text{on the right.}\\ t=\frac{d}{r} & \text{Commutative property of equality.} \end{array}$

  The result is $t=\frac{d}{r}$.

  Substituting $180~\text{ft}$ for $d$ and $5~\frac{\text{ft}}{\text{sec}}$ for $r$ gives us:

  $\begin{array}{l} t=\frac{180~\text{ft}}{5~\frac{\text{ft}}{\text{sec}}}\\ t=\frac{180~\text{ft}}{5}\cdot\frac{\text{sec}}{\text{ft}}\\ t=\frac{180}{5}~\text{sec}\\ t=36~\text{sec} \end{array}$

• Find the missing values.

  Hints

  Here is an example of how to simplify expressions involving division of units:

  $\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline \frac{150~{\text{m}}}{15\frac{\text{m}}{\text{sec}}} & \text{Given expression.}\\ \frac{150~{\text{m}}}{15}\cdot\frac{\text{sec}}{\text{m}} & \text{Dividing is multiplying by the reciprocal.}\\ \frac{150~\text{sec}}{15} & \text{Canceling units.}\\ 10~\text{sec} & \text{Division.} \end{array}$

  The units for distance are distances. Some examples are:

  • $\text{km}$
  • $\text{mi}$
  • $\text{m}$
  • $\text{ft}$

  The units for time are time units:

  • $\text{sec}$
  • $\text{min}$
  • $\text{hr}$

  The units for rates are distance units over time units:

  • $\frac{\text{km}}{\text{min}}$
  • $\frac{\text{m}}{\text{s}}$
  • $\frac{\text{mi}}{\text{hr}}$

  It is not always necessary to solve an equation to determine if a value is a solution. Given the following equation and the following potential solution, you can substitute and check to see if the value makes the equation true:

  \begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline \frac{x}{4}=12 & \text{Given equation.}\\ x=48 & \text{Potential solution.}\\ \frac{48}{4}=12 & \text{substituting to determine if }x=4 \text{ is a solution}\\ 12=12 &\text{Division confirms equality, so }x=4~ \text{is a solution.}\\ \end{array}

  Solution

  $\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline \frac{144~\text{m}}{48~\frac{\text{m}}{\text{min}}}=t & \text{Given equation.}\\ t=\frac{144~\text{m}}{48}\cdot \frac{\text{min}}{\text{m}} & \text{Dividing fractions is multiplying by reciprocal.}\\ t=3~\text{min} & \text{Dividing and canceling.}\\ \end{array}$

  $\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline 9~\text{min}\cdot\frac13~\frac{\text{yd}}{\text{min}}=d & \text{Given equation.}\\ d=9~\text{min}\cdot\frac13~\frac{\text{yd}}{\text{min}} & \text{Commutative property of equality.}\\ d=\frac{9}{3}~\text{yd} & \text{Canceling units.}\\ d=3~\text{yd} & \text{Dividing.} \end{array}$

  $\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline d=3~\frac{\text{ft}}{\text{sec}}\cdot 12~\text{sec} & \text{Given equation.}\\ d=3~\text{ft}\cdot 12 & \text{Canceling units.}\\ d =36~\text{ft} & \text{Multiplying.}\\ \end{array}$

  $\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline 36~\frac{\text{ft}}{\text{sec}}=\frac{d}{3~\text{sec}}& \text{Given equation.}\\ d=108~\text{ft} & \text{Choosing this potential solution for distance.}\\ 36~\frac{\text{ft}}{\text{sec}}=\frac{108~\text{ft}}{3~\text{sec}} & \text{Substituting the chosen value for }d.\\ 36~\frac{\text{ft}}{\text{sec}}=36~\frac{\text{ft}}{\text{sec}}& \text{Division.}\\ 36=36 & d=108~\text{ft confirmed as a solution.} \end{array}$

  $\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline t=\frac{60~\text{mi}}{5\frac{\text{mi}}{\text{sec}}} & \text{Given equation.}\\ t=\frac{60~\text{mi}}{5}\cdot\frac{\text{sec}}{\text{mi}} & \text{Dividing fractions is multiplying by the reciprocal.}\\ t=\frac{60}{5}~\text{sec} & \text{Canceling units.}\\ t=12~\text{sec} & \text{Division.}\\ \end{array}$

  $\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline r=\frac{144~{\text{km}}}{12~\text{sec}} & \text{Given equation.}\\ r=12~\frac{\text{km}}{\text{sec}} & \text{Division}\\ \end{array}$

• Comparing rates.

  Hints

  Here are some helpful conversion factors:

  • $3600~\text{sec}=1~\text{hr}$
  • $5280~\text{ft}=1~\text{mi}$
  • $1760~\text{yd}=1~\text{mi}$
  • $3~\text{ft}=1~\text{yd}$

  When comparing rates with different units, select a single unit that you can easily convert all the rates to, and which is intelligible to you. Some rates that may make sense to you are:

  • miles per hour
  • feet per second

  Here's an example of converting from $\frac{\text{ft}}{\text{min}}$ to $\frac{\text{mi}}{\text{hr}}$:

  $\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline r=\frac{21,120~\text{ft}}{10~\text{min}} & \text{Given equation.}\\ r=\frac{2112~\text{ft}}{\text{min}} & \text{Division.}\\ r=\frac{2112~\text{ft}}{1~\text{min}}\cdot \frac{1~\text{mi}}{5280~\text{ft}} & \text{From feet to miles.}\\ r=\frac{0.4~\text{mi}}{1~\text{min}}\cdot \frac{60~\text{min}}{1~\text{hr}} & \text{From minutes to hours.}\\ r=24~\frac{\text{mi}}{\text{hr}} & \text{Simplifying.} \end{array}$

  Solution

  $\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline \frac{21~\text{mi}}{3~\text{min}}=r & \text{Given equation.}\\ r=7~\frac{\text{mi}}{\text{min}} & \text{Dividing.}\\ r=\frac{7~\text{mi}}{\text{min}}\cdot \frac{60~\text{min}}{1~\text{hr}} & \text{From minutes to hours.}\\ r=420~\frac{\text{mi}}{\text{hr}} & \text{Simplifying.} \end{array}$

  $\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline 300~\text{ft}=r\cdot 6~\text{sec} & \text{Given equation.}\\ r=\frac{300~\text{ft}}{6~\text{sec}} & \text{Division property of equality.}\\ r=\frac{50~\text{ft}}{\text{sec}} & \text{Division.}\\ r=\frac{50~\text{ft}}{\text{sec}}\cdot \frac{3600~\text{sec}}{1~\text{hr}} & \text{From seconds to hours.}\\ r=180,000~\frac{\text{ft}}{\text{hr}} & \text{Canceling units and simplifying.}\\ r=180,000~\frac{\text{ft}}{\text{hr}} \cdot \frac{1~\text{mi}}{5280~\text{ft}} & \text{From feet to miles.}\\ r=34.\overline{09}~\frac{\text{mi}}{\text{hr}}& \text{Canceling units and simplifying.}\\ \end{array}$

  $\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline \frac{325~\text{yd}}{25~\text{sec}} =r& \text{Given equation.}\\ r=\frac{13~\text{yd}}{1~\text{sec}} & \text{Division.}\\ r=\frac{13~\text{yd}}{\text{sec}}\cdot \frac{3600~\text{sec}}{~\text{hr}} & \text{From seconds to hours.}\\ r=\frac{46,800~\text{yd}}{\text{hr}}\cdot \frac{1~\text{mi}}{1760~\text{yd}} & \text{From yards to miles.}\\ r=26\frac{13}{22}~\frac{\text{mi}}{\text{hr}} & \text{Simplifying.} \end{array}$

  $\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline r=\frac{3~\text{mi}}{8~\text{min}} & \text{Given equation.}\\ r=\frac{3~\text{mi}}{8~\text{min}}\cdot \frac{60~\text{min}}{~\text{hr}} & \text{From minutes to hours.}\\ r=22.5~\frac{\text{mi}}{\text{hr}} & \text{Simplifying.} \end{array}$

  $\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline 20,000~\text{yd} =r\cdot5~\text{hr}& \text{Given equation.}\\ r=\frac{20,000~\text{yd}}{5~\text{hr}} & \text{Division property of equality.}\\ r=\frac{4000~\text{yd}}{\text{hr}} & \text{Division.}\\ r=\frac{4000~\text{yd}}{\text{hr}}\cdot \frac{1~\text{mi}}{1760~\text{yd}} & \text{From yards to miles.}\\ r=2\frac{3}{11}~\frac{\text{mi}}{\text{hr}} & \text{Simplifying.} \end{array}$

• Identify the distance formula.

  Hints

  Multiplication is commutative. That means:

  $a\cdot b = b \cdot a$

  Equality is commutative. That means:

  If $a=b$ then $b=a$.

  Below is an example of the process for dividing both sides of an equation to solve for a variable. In this example, we will solve for $t$:

  $\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline d=r\cdot t & \text{The equation for distance.}\\ \frac{d}{r}=\frac{r\cdot t}{r} & \text{Dividing both sides by}~r.\\ \frac{d}{r}=t & \text{Canceling}~r~\text{on the right.}\\ t=\frac{d}{r} & \text{Commutative property of equality.} \end{array}$

  Solution

  • $d=r\cdot t$ True This is the most common way of writing the equation relating distance, rate, and time.

  • $r=d\cdot t$ False The correct method for solving for $r$ is shown below.

  $\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline d=r\cdot t & \text{The equation for distance.}\\ \frac{d}{t}=\frac{r\cdot t}{t} & \text{Dividing both sides by}~t.\\ \frac{d}{t}=r & \text{Canceling}~t~\text{on the right.}\\ r=\frac{d}{t} & \text{Commutative property of equality.} \end{array}$

  • $r=\frac{d}{t}$ True See above.

  • $t=\frac{r}{d}$ False The correct method for solving for $t$ is shown below:

  $\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline d=r\cdot t & \text{The equation for distance.}\\ \frac{d}{r}=\frac{r\cdot t}{r} & \text{Dividing both sides by}~r.\\ \frac{d}{r}=t & \text{Canceling}~r~\text{on the right.}\\ t=\frac{d}{r} & \text{Commutative property of equality.} \end{array}$

  • $\frac{d}{r}=t$ True See above. Look at the second to last line.

  • $t\cdot r =d$ True See below.

  $\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline d=r\cdot t & \text{The equation for distance.}\\ r\cdot t =d & \text{Commutative property of equality.}\\ t\cdot r =d & \text{Commutative property of multiplication.} \end{array}$

  • $r=d\cdot t$ False See above for the correct method for solving for $r$.

  • $\frac{t}{r}=d$ False $d=r\cdot t$.

• Calculate the missing rate.

  Hints

  First, find the distance to the waterfall.

  If one rate is $r$, then twice that rate is $2r$.

  Create one expression for the distance traveled by Tatu and Princess during the day, and another for the distance traveled during the night.

  The sum of the distance traveled during the day and the distance traveled during the night is the distance to the waterfall.

  Solution

  Walter flew at $20$ miles per hour for $1$ day to arrive at the waterfall:

  $\begin{array}{l} r=20~\text{mi}\\ t=24~\text{hr} \end{array}$

  Distance is rate times time:

  $\begin{array}{l} d=20~\frac{\text{mi}}{\text{hr}}\cdot 24~\text{hr}\\ d=480~\text{mi}\\ \end{array}$

  The waterfall is $480~\text{mi}$ away. For Tatu and Princess, there are two different rates they travel, a rate during the day, $r$, and a rate at night, $2r$.

  The travel each rate for $12$ hours, accomplishing one distance during the day and another distance at night:

  $\begin{array}{l} d_\text{day}=r\cdot12~\text{hr}=12r\\ d_\text{night}=2r\cdot12~\text{hr}=24r\\ \end{array}$

  If they are going to make it to the waterfall in a day, the sum of these two distances must be $480~\text{mi}$:

  $\begin{array}{l} 12r+24r=480\\ 36r=480\\ r=13\frac13~\frac{\text{mi}}{\text{hr}}\\ \end{array}$

  That’s $r$, their day rate. Their night rate is $2r$, that is, $26\frac23~\frac{\text{mi}}{\text{hr}}$.