Lesson Travel and Distance problems

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<H2>Travel and Distance problems</H2>
In this lesson simple typical word problems on &nbsp;<B><U>Travel and Distance</U></B>&nbsp; are presented  to show the approach and the methodology of their solutions.

<TABLE cellspacing="10"> 
  <TR>
  <TD>
<H3>Problem 1. Two objects moving toward each other</H3>

Two cars entered an Interstate highway at the same time and traveled toward each other&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
(see &nbsp;<B>Figure 1</B>). &nbsp;The initial distance between the cars was &nbsp;390 miles. &nbsp;The first car was running 
at the speed &nbsp;70 miles per hour, &nbsp;the second car was running at &nbsp;60 miles per hour.
How long will it take for the two cars to pass each other?
What distance each car will travel before passing?
  </TD>
  <TD>
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            line(-0.05,-0.03,  1.05, -0.03),

line( 0.0,  0.1,  0.0,  -0.1),
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        red(circle(1.0, 0.4, 0.005)),

blue(line( 0.0,  0.4,  0.12,  0.4)),
       blue(line(0.09,  0.47, 0.12,  0.4)),
       blue(line(0.09,  0.35, 0.12,  0.4)),
          locate(-0.08, 1.0, Car1_70_mph),

blue(line( 0.9,  0.4,  1.0,  0.4)),
       blue(line( 0.93, 0.47, 0.9,  0.4)),
       blue(line( 0.93, 0.35, 0.9,  0.4)),
          locate( 0.82, 1.0, 60_mph_Car2),

line( 0.0, -0.3,  0.0,  -0.5),
            line( 1.0, -0.3,  1.0,  -0.5),
            line( 0.0, -0.4,  0.35, -0.4),
            line( 0.65,-0.4,  1.0, -0.4),
            locate(0.4,-0.2, 390_miles),

line( 0.55, 0.9,  0.55, 0.7),
            line( 0.55, 0.6,  0.55, 0.4),
            line( 0.55, 0.3,  0.55, 0.1)
)}}}
<B>Figure 1. Two objects moving toward each other</B>
  </TD>
  </TR>
</TABLE> 
<B>Solution</B>

We will solve the problem by reducing it to the simple linear equation.
Let &nbsp;t&nbsp; be the unknown time in hours for the two cars to pass each other.
Since the rate of the first car is &nbsp;70 miles per hour, &nbsp;the distance it travels for time &nbsp;t&nbsp; is equal to &nbsp;70*t&nbsp; miles.
Since the rate of the second car is &nbsp;60 miles per hour, &nbsp;the distance it travels for time &nbsp;t&nbsp; is equal to &nbsp;60*t miles.
At the moment the two cars pass each other, &nbsp;the sum of these distances is equal to the initial distance between two cars, &nbsp;i.e. &nbsp;390 miles.

It gives an equation

&nbsp;&nbsp;&nbsp;&nbsp;70*t + 60*t = 390.

Simplify the left side of the equation using the distributive law:

&nbsp;&nbsp;&nbsp;&nbsp;70*t + 60*t = (70+60)*t = 130*t.

So, &nbsp;our equation takes the form

&nbsp;&nbsp;&nbsp;&nbsp;130*t = 390.

Divide both sides of the equation by &nbsp;130. &nbsp;You get

&nbsp;&nbsp;&nbsp;&nbsp;t = 390/130 = 3.

Thus, &nbsp;it will take &nbsp;3&nbsp; hours for the two cars to pass each other.

Now, &nbsp;check the solution.

The distance that the first car will travel for &nbsp;3&nbsp; hours is equal to &nbsp;70*3 = 210 &nbsp;miles.
The distance that the second car will travel for &nbsp;3&nbsp; hours is equal to &nbsp;60*3 = 180&nbsp;miles.
Since &nbsp;210 + 190 = 390, &nbsp;the solution is correct.

<B>Answer</B>. &nbsp;It will take &nbsp;3&nbsp; hours for the two cars to pass each other.
First car will travel &nbsp;210&nbsp; miles, &nbsp;and the second car will travel &nbsp;180 &nbsp;miles before passing.

<B>Note</B>. &nbsp;For those who prefer more &nbsp;<B>physical thinking</B>, &nbsp;the following alternative &nbsp;<B>solution</B>&nbsp; may seem more suitable.
The value of &nbsp;70+60 = 130&nbsp; miles per hour is the rate of decreasing the distance between two cars, &nbsp;so the time before passing is equal to &nbsp;t = 390/130 = 3 &nbsp;hours. &nbsp;Based on it, 
the distance traveled by the first car for this time is equal to &nbsp;3*70 = 210 &nbsp;miles, &nbsp;and the distance traveled by the second car for this time is equal to &nbsp;3*60 = 180 &nbsp;miles.
Surely, &nbsp;the answer is the same.

<TABLE cellspacing="10"> 
  <TR>
  <TD>
<H3>Problem 2. Two objects moving in the same direction</H3>

Two cars entered an Interstate highway at the same time at different locations and traveled&nbsp;&nbsp;
in the same direction as shown in &nbsp;<B>Figure 2</B>. &nbsp;The initial distance between cars was &nbsp;30&nbsp; miles. 
First car was running &nbsp;70&nbsp; miles per hour, &nbsp;the second car was running &nbsp;60&nbsp; miles per hour.
How long will it take for the first car to catch the second one?
What distance each car will travel before the first car catches the second one?
  </TD>
  <TD>
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line( 0.53, 0.0,   0.60,  0.0),
            line( 0.53, 0.03,  0.60,  0.03),
            line( 0.53,-0.03,  0.60, -0.03),

line( 0.63, 0.0,   0.70,  0.0),
            line( 0.63, 0.03,  0.70,  0.03),
            line( 0.63,-0.03,  0.70, -0.03),

line( 0.73, 0.0,   0.80,  0.0),
            line( 0.73, 0.03,  0.80,  0.03),
            line( 0.73,-0.03,  0.80, -0.03),

line( 0.83, 0.0,   1.05,  0.0),
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       blue(line(0.09,  0.47, 0.12,  0.4)),
       blue(line(0.09,  0.35, 0.12,  0.4)),
          locate(-0.08, 1.0, Car1_70_mph),

blue(line( 0.3,  0.4,  0.4,  0.4)),
       blue(line( 0.37, 0.47, 0.4,  0.4)),
       blue(line( 0.37, 0.35, 0.4,  0.4)),
          locate( 0.25,  1.0, Car2_60_mph),

line( 0.0, -0.3,  0.0,  -0.5),
            line( 0.3, -0.3,  0.3,  -0.5),
            line( 0.0, -0.4,  0.04, -0.4),
            line( 0.26,-0.4,  0.3,  -0.4),
            locate(0.05,-0.2, 30_miles),

line( 1.00, 0.9,  1.00, 0.7),
            line( 1.00, 0.6,  1.00, 0.4),
            line( 1.00, 0.3,  1.00, 0.1)
)}}}
<B>Figure 2. Two objects moving in the same direction</B>
  </TD>
  </TR>
</TABLE> 
<B>Solution</B>

I will show you first how to solve the problem by reducing it to the simple linear equation.
Let &nbsp;t &nbsp;be the unknown time in hours for the first car to catch the second one.
Since the rate of the first car is &nbsp;70&nbsp; miles per hour, &nbsp;the distance it travels for time &nbsp;t&nbsp; is equal to &nbsp;70*t&nbsp; miles.
Since the rate of the second car is &nbsp;60&nbsp; miles per hour, &nbsp;the distance it travels for time &nbsp;t&nbsp; is equal to &nbsp;60*t&nbsp; miles.
At the moment when the first car catches the second one, &nbsp;the difference of these distances is equal to the initial distance between two cars, &nbsp;i.e. &nbsp;30 miles.

This gives an equation

&nbsp;&nbsp;&nbsp;&nbsp;70*t - 60*t = 30.

Simplify the left side of the equation using the distributive law:

&nbsp;&nbsp;&nbsp;&nbsp;70*t - 60*t = (70-60)*t = 10*t.

So, &nbsp;our equation takes a form

&nbsp;&nbsp;&nbsp;&nbsp;10*t = 30.

Divide both sides of the equation by &nbsp;10. &nbsp;You get

&nbsp;&nbsp;&nbsp;&nbsp;t = 30/10 = 3.

Thus, &nbsp;it takes &nbsp;3&nbsp; hours for the first car to catch the second one.

You can check the solution.

The distance first car will travel for &nbsp;3&nbsp; hours is equal to &nbsp;70*3 = 210 miles.
The distance the second car will travels for &nbsp;3&nbsp; hours is equal to &nbsp;60*3 = 180 miles.

Since &nbsp;210 - 180 = 30, &nbsp;the solution is correct.

<B>Answer</B>. &nbsp;It will take &nbsp;3&nbsp; hours for the first car to catch the second one.
The first car will travel &nbsp;210 miles, &nbsp;and the second car will travel &nbsp;180 &nbsp;miles before the first car will catch the second.

<B>Note</B>. &nbsp;Again, &nbsp;there is alternative &nbsp;<B>physical</B> &nbsp;solution.
The value of &nbsp;70-60 = 10 &nbsp;miles per hour is the rate of decreasing the distance between two cars in this case, &nbsp;so the time before passing is equal to &nbsp;t = 30/10 = 3 &nbsp;hours. &nbsp;Based on it, 
the distance traveled by the first car for this time is equal to &nbsp;3*70 = 210&nbsp; miles, &nbsp;and the distance traveled by the second car for this time is equal to &nbsp;3*60 = 180 &nbsp;miles.
The answer the same.

<TABLE cellspacing="10"> 
  <TR>
  <TD>
<H3>Problem 3. Two objects moving toward each other</H3>

Two cars entered an Interstate highway at the same time and traveled toward each other&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
(see &nbsp;<B>Figure 3</B>). &nbsp;The initial distance between cars was &nbsp;390&nbsp; miles. 
The speed of the first car was in &nbsp;10&nbsp; miles per hour greater than that of the second car.
It took &nbsp;3&nbsp; hours for two cars to pass each other.
What was the speed of each car? &nbsp;What distance did each car travel before the cars passed each other?
  </TD>
  <TD>
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            line(-0.05,-0.03,  1.05, -0.03),

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        red(circle(1.0, 0.4, 0.005)),

blue(line( 0.0,  0.4,  0.12,  0.4)),
       blue(line(0.09,  0.47, 0.12,  0.4)),
       blue(line(0.09,  0.35, 0.12,  0.4)),
          locate(-0.08, 1.0, Car1_speed),

blue(line( 0.9,  0.4,  1.0,  0.4)),
       blue(line( 0.93, 0.47, 0.9,  0.4)),
       blue(line( 0.93, 0.35, 0.9,  0.4)),
          locate( 0.82, 1.0, speed_Car2),

line( 0.0, -0.3,  0.0,  -0.5),
            line( 1.0, -0.3,  1.0,  -0.5),
            line( 0.0, -0.4,  0.35, -0.4),
            line( 0.65,-0.4,  1.0, -0.4),
            locate(0.4,-0.2, 390_miles),

line( 0.55, 0.9,  0.55, 0.7),
            line( 0.55, 0.6,  0.55, 0.4),
            line( 0.55, 0.3,  0.55, 0.1)
)}}}
<B>Figure 3. Two objects moving toward each other</B>
  </TD>
  </TR>
</TABLE> 
<B>Solution</B>

The situation is similar to that of the &nbsp;<B>Problem 1</B>, &nbsp;but the time to get passing is given and the speed is under the question in this case.
Again, &nbsp;we will solve the problem by reducing it to the simple linear equation.
Let v &nbsp; be the unknown speed &nbsp;(in miles per hour) &nbsp;of the first car.
Then the distance the first car traveled for &nbsp;3&nbsp; hours is equal to &nbsp;3*v&nbsp; miles.
Since the rate of the first car is in &nbsp;10&nbsp; miles per hour greater than that of the second car, &nbsp;the rate of the second car is equal to &nbsp;v-10 &nbsp;miles per hour.
Then the distance the second car traveled for &nbsp;3&nbsp; hours is equal to &nbsp;3*(v-10) &nbsp;miles.
At the moment two cars pass each other, &nbsp;the sum of these distances is equal to the initial distance between two cars, &nbsp;i.e. &nbsp;390&nbsp; miles.

It gives an equation

&nbsp;&nbsp;&nbsp;&nbsp;3*v + 3*(v-10) = 390.

Simplify the left side of the equation by opening brackets and collecting like terms:

&nbsp;&nbsp;&nbsp;&nbsp;3*v + 3*v - 3*10 = 6*v - 30.

So, &nbsp;our equation takes the form

&nbsp;&nbsp;&nbsp;&nbsp;6*v - 30 = 390.

Transfer the constant term from the left side to the right and collect like terms. &nbsp;You get

&nbsp;&nbsp;&nbsp;&nbsp;6*v = 390+30 = 420.

Divide both sides by &nbsp;6. &nbsp;You get

&nbsp;&nbsp;&nbsp;&nbsp;v = 420/6 = 70.

Thus, &nbsp;the speed of the first car was equal to &nbsp;70&nbsp; miles per hour.
Hence, &nbsp;the speed of the second car was equal to &nbsp;70-10 = 60 &nbsp;miles per hour. 
The distance the first car traveled for &nbsp;3&nbsp; hours was equal to &nbsp;70*3 = 210 &nbsp;miles.
The distance the second car traveled for &nbsp;3&nbsp; hours was equal to &nbsp;60*3 = 180 &nbsp;miles.
Since &nbsp;210 + 190 = 390, &nbsp;the solution is correct.

<B>Answer</B>. &nbsp;The speed of the first car was &nbsp;70&nbsp; miles per hour, &nbsp;the speed of the second car was &nbsp;60&nbsp; miles per hour.
The first car traveled &nbsp;210 &nbsp;miles, &nbsp;and the second car traveled &nbsp;180&nbsp; miles before the cars passed each other.

My other lessons on &nbsp;<B>Travel and Distance</B>&nbsp; problems in this site are

<TABLE>
  <TR>
  <TD>
- <A HREF=https://www.algebra.com/algebra/homework/word/travel/Travel-and-Distance-problems-for-two-bodies-moving-toward-each-other.lesson>Travel and Distance problems for two bodies moving in opposite directions</A>

- <A HREF=https://www.algebra.com/algebra/homework/word/travel/Typical-catching-up-Travel-and-Distance-problems.lesson>Travel and Distance problems for two bodies moving in the same direction (catching up)</A>

- <A HREF=http://www.algebra.com/algebra/homework/NumericFractions/Using-fractions-to-solve-Travel-problems.lesson>Using fractions to solve Travel problems</A>

- <A HREF=http://www.algebra.com/algebra/homework/word/travel/Wind-and-Current-problems.lesson>Wind and Current problems</A>

- <A HREF=http://www.algebra.com/algebra/homework/word/travel/More-problems-on-upstream-and-downstream-round-trips.lesson>More problems on upstream and downstream round trips</A>

- <A HREF=http://www.algebra.com/algebra/homework/word/travel/Wind-and-Current-problems-solvable-by-quadratic-equations.lesson>Wind and Current problems solvable by quadratic equations</A>

- <A HREF=https://www.algebra.com/algebra/homework/word/travel/Unpowered-raft-moving-downstream-along-a-river.lesson>Unpowered raft floating downstream along a river</A>

- <A HREF=https://www.algebra.com/algebra/homework/word/travel/Selected-problems-from-the-archive-on-a-boat-floating-Upstream-and-Downstream.lesson>Selected problems from the archive on the boat floating Upstream and Downstream</A>

- <A HREF=https://www.algebra.com/algebra/homework/word/travel/Selected-problems-from-the-archive-on-a-plane-flying-with-and-against-the-wind.lesson>Selected problems from the archive on a plane flying with and against the wind</A>

- <A HREF=https://www.algebra.com/algebra/homework/word/travel/Selected-Travel-and-Distance-problems-from-the-archive.lesson>Selected Travel and Distance problems from the archive</A>

- <A HREF=https://www.algebra.com/algebra/homework/word/travel/Had-a-car-move-faster-it-would-arrive-quicker.lesson>Had a car move faster it would arrive sooner</A>

- <A HREF=http://www.algebra.com/algebra/homework/word/travel/How-far-do-you-live-from-school.lesson>How far do you live from school?</A>

- <A HREF=http://www.algebra.com/algebra/homework/word/travel/One-unusual-Travel-problem.lesson>One unusual Travel problem</A>

- <A HREF=http://www.algebra.com/algebra/homework/word/travel/Another-unusual-Travel-problem.lesson>Another unusual Travel problem (Arnold's problem on two walking old women)</A>

- <A HREF=http://www.algebra.com/algebra/homework/word/travel/Travel-problem-on-a-messenger-moving-back-and-forth-along-the-marching-army-column.lesson>Travel problem on a messenger moving back and forth along the marching army's column</A>

- <A HREF=https://www.algebra.com/algebra/homework/word/travel/A-person-walking-along-the-street-and-the-buses-traveling-in-the-same-and-opposite-directions.lesson>A person walking along the street and buses traveling in the same and opposite directions</A>

</TD>
  <TD>&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/travel/Calculating-an-average-speed.lesson>Calculating an average speed: a train going from A to B and back</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/travel/One-more-problem-on-calculating-an-average-speed.lesson>One more problem on calculating an average speed</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/travel/Clock-problems.lesson>Clock problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/travel/Advanced-clock-problems.lesson>Advanced clock problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/travel/Problems-on-bodies-moving-on-a-circle.lesson>Problems on bodies moving on a circle</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/travel/A-train-passing-a-telegraph-post-and-passing-a-bridge.lesson>A train passing a telegraph post and passing a bridge</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/travel/A-train-passing-a-platform.lesson>A train passing a platform</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/travel/A-train-passing-a-tunnel.lesson>A train passing through a tunnel</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/travel/A-light-rail-train-passing-a-walking-person.lesson>A light-rail train passing a walking person</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/travel/A-train-passing-another-train.lesson>A train passing another train</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/travel/A-man-crossing-a-bridge-when-a-train-comes-from-behind.lesson>A man crossing a bridge and a train coming from behind</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/travel/27-A-problem-on-a-rower-going-on-a-river-who-missed-the-bottle-of-whiskey-under-a-bridge.lesson>A rower going on a river who missed the bottle of whiskey under a bridge</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/travel/Non-traditional-Travel-and-Distance-problems.lesson>Non-traditional Travel and Distance problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/travel/The-distance-covered-by-a-free-falling-body-during-last-second-of-the-fall.lesson>The distance covered by a free falling body during last second of the fall</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/travel/The-Doppler-shift.lesson>The Doppler Shift</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/travel/Entertainment-Travel-and-Distance-problems.lesson>Entertainment Travel and Distance problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/travel/OVERVIEW-of-lessons-on-Travel-and-Distance.lesson>OVERVIEW of lessons on Travel and Distance</A>

</TD>
 </TR>
</TABLE>

Use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>&nbsp; to navigate over all topics and lessons of the online textbook &nbsp;ALGEBRA-I.