Distance - Rate - Time – Different Directions

Distance - Rate - Time – Different Directions exercise

• Determine when and where the car will crash.

  Hints

  Fill in the table:

  $\begin{array}{|l|c|c|c|} \hline &r&t&d\\ \hline \text{car 1}&50&t&?\\ \hline \text{car 2}&70&t&?\\ \hline \end{array}$

  Don't forget the mnemonic device.

  Determine the unknown quantity, and use the appropriate equation:

  • $d=r\times t$
  • $r=\frac dt$
  • $t=\frac dr$

  The sum of the distances is equal to $1.2$ (miles).

  Solution

  The red car travels at $50$ mph, while the blue car travels in the opposite direction at $70$ mph.

  This is a distance-rate-time-problem with different directions.

  Let's start with a table:

  $\begin{array}{|l|c|c|c|} \hline &r&t&d\\ \hline \text{car 1}&50&t&50t\\ \hline \text{car 2}&70&t&70t\\ \hline \end{array}$

  Use the equation $d=r\times t$.

  Since the car cannot travel a distance greater than $1.2$ miles, we write the equation:

  $\begin{array}{rclcl} 50t+70t & = & 1.2 &&|\text{Combine Like Terms} \\ 120t&=&1.2 \\ \color{#669900}{\div120} && \color{#669900}{\div120} &&|\text{Opposite Operations} \\ t &=& 0.01 \end{array}$

  $t=0.01$ is the time in hours. If we want to know how many seconds this is, we can calculate the conversion. There are $60$ minutes in an hour and $60$ seconds for every minute. Multiply... there are $3,600$ seconds in one hour.

  $\begin{array}{rcll} \dfrac{60 ~\text{min.}}{1 ~\text{hr.}} \times \dfrac{60 ~\text{sec.}}{1 ~\text{min.}} &=& \dfrac{3,600 ~\text{sec.}}{1 ~\text{hr.}} \end{array}$

  $~$

  $\dfrac{3,600 ~\text{ sec.}}{1 ~\text{ hr.}}\times 0.01 ~\text{ hr.} = 36 ~\text{ sec.}$

  Now we know that the crash will happen after $36$ seconds.

  To find out where the crash will occur, we use the equation $d=r\times t$. Plugging in the known values gives us:

  $d = \dfrac{50 ~\text{mi.}} {\text{hr.}} \times 0.01 ~\text{hr.} = 0.5~\text{mi.}$ for the red car

  $d = \dfrac{70 ~\text{mi.}} {\text{hr.}} \times 0.01 ~\text{hr.} = 0.7~\text{mi.}$ for the blue car

  To document the crash, Johnny has to set up the camera $0.5$ miles far away from Point A.

• Calculate the rate at which the car travels.

  Hints

  Keep this triangle in mind - it's magic.

  If it takes $0.5$ hours to travel a distance of $30$ miles, the rate is $60$ mph.

  Solution

  You can remember the formulas by using the Distance Rate Time Triangle.

  Here we know

  • Time: $t=0.01$ hours
  • Distance: $0.4$ miles

  We need to know the rate of the blue car, so we choose the formula in the middle $r=\frac dt$.

  By plugging in the known values, we get:

  $r=\frac{0.4~\text{mi.}}{0.01~\text{hr.}}=40 ~\text{mph.}$

• Evaluate the time, if the rate changes.

  Hints

  First gather all given information:

  • $r=80$
  • $t=0.01$

  Decide which formula to use.

  If you already know the rate and the time, use $d=r\times t$ to find the distance.

  For example:

  The given information:

  • $r=90$
  • $t=0.01$

  To calculate the distance, we can calculate:

  $d = rt$ $d=90\times 0.01=0.9$ miles

  Solution

  The speed of the blue car is increased to $80$ mph.

  Johnny's a bit lazy and doesn't want to change the camera position. He can't change the rate or the time, but he can change the distance: To figure this out, use the equation: $d=r\times t$.

  Given:

  • Rate: $80$ mph
  • Time: $0.01$ hrs.

  $d=80\times 0.01=0.8$

  So the blue car must be a distance of $0.8$ miles from the crash point. The distance between the two cars at the start is $0.5+0.8=1.3$ miles.

• Find out when and where the two brothers will meet.

  Hints

  This is a distance, rate, and time problem with different directions.

  Use the formula $d=r\times t$.

  Keep in mind that the sum of the distances traveled is $3$ miles.

  Use the opposite operations:

  • Multiplication ($\times~\longleftrightarrow~\div$)
  • Division ($\div~\longleftrightarrow~\times$)
  • Addition ($+~\longleftrightarrow~-$)
  • Subtraction ($-~\longleftrightarrow~+$)

  You have to calculate the time first, and then you can calculate the distance.

  Solution

  Eugene is riding his bike at a rate of $10$ mph, while Shane is running at a rate of $2$ mph.

  Here we have a distance rate time problem involving different directions.

  First let's consider the distance. We know that $d=r\times t$, and the time $t$ is unknown.

  • $d=10t$ for Eugene
  • $d=2t$ for Shane

  But, we do know the total distance, so we can write the equation:

  $10t+2t=3$

  Combine the like terms: $12t=3$. Finally, divide by $12$: $t=\frac3{12}=\frac14=0.25$ hours.

  So the brothers will meet after $0.25$ hours or $15$ minutes.

  To determine the distance each boy will travel, we use the formula $d=r\times t$ once again:

  • Eugene: $d=10\times 0.25=2.5$ miles
  • Shane: $d=2\times 0.25=0.5$ miles

  In summary:

  • We know that the brothers will meet after 15 minutes.
  • Eugene starts $2.5$ miles away from the meeting point.
  • Shane starts $0.5$ miles away from the meeting point.

• Summarize the distance, rate, and time problem.

  Hints

  Use the magic triangle:

  • $d$ over $r$ as well as $d$ over $t$
  • $r$ and $t$ on the same level

  Remember what the variables stand for:

  • Distance
  • Rate
  • Time

  Solution

  To solve problems, use the Distance Rate Time triangle.

  • Distance
  • Rate
  • Time

  Starting at the top, if we want to calculate distance, $d$, we can see that $r$ and $t$ are on the same level. So our equation is modified: $d=r\times t$. To calculate the rate, $r$, we look at the other two variables, $d$ and $t$. $d$ stands over $t$ in the triangle, so we have $r=\frac dt$. To find time, $t$, we notice that $d$ is over $r$, so our final equation is: $t=\frac dr$.

• Determine when Jim has to leave his home to make it to his appointments on time.

  Hints

  First determine the time, $t$, Jim needs to travel each distance. You should use the following equation:

  $t=\frac dr$

  Next, calculate the time in hours. Convert to minutes if necessary.

  Subtract your answer from the time Jim plans to arrive at each destination. Remember, there are $60$ minutes in an hour.

  Solution

  What do we know?

  • The distance from Jim's house to each destination
  • Jim's walking rate is $2$ mph

  To calculate the time it takes Jim to travel to each of these destinations, we use the formula $t=\frac dr$.

  School Distance = $0.8$ miles

  $t=\frac{0.8}{2}=0.4$ hours. Multiply by 60: $t=0.4\times 60=24$ minutes.

  Subtract $24$ minutes from the arrival time:

  $8:00-0:24=7:36$ a.m.

  To get to school on time, Jim has to leave his house by $7:36$ a.m.

  Grandma's House Distance = $1$ mile

  $t=\frac{1}{2}=0.5$ hours. Again, multiply by $60$: $t=0.5\times 60=30$ minutes.

  Subtracting $30$ minutes from the arrival time gives us:

  $10:00-0:30=9:30$ a.m.

  Jim has to leave by $9:30$ a.m. to get to his grandma's house by $10:00$ a.m.

  Jack Distance = $3$ miles

  $t=\frac{3}{2}=1.5$ hours or, $1$ hour and $30$ minutes.

  Subtract this time from the arrival time (We used the $24$-hour clock for convenience.):

  $14:00-1:30=12:30$, or 12:30 pm.

  Jim has to leave by 12:30 p.m. to meet Jack at $2$ p.m.

  Cinema Distance = $4$ miles

  Oh, what a day. Jim deserves a little bit of time to kick back and relax. To get to the movies on time, he has to start $t=\frac42=2$ hours before the film starts playing. So he should leave his house at $6$ p.m. He walks $2$ hours to watch a movie!? I hope it's a good movie!