Question 1203173
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U><B>Physics solution</B></U>


<pre>
Their relative speed is 55 - 40 = 15 kilometers per hour.

In other words, the distance between them is 15*t kilometers, as a function of time.


The cars will be 10 km apart in  {{{10/15}}} = {{{2/3}}} of an hour,  or in 40 minutes.    <U>ANSWER</U>
</pre>


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U><B>Algebra solution</B></U>



<pre>
Fast car distance from the start is  {{{d[fast]}}} = 55*t kilometers.

Slow car distance from the start is  {{{d[slow]}}} = 40*t kilometers.


The distance between the cars

    {{{d[fast]}}} - {{{d[slow]}}} = 55t - 40t.


The distance equation

    55t - 40 t = 10  kilometers.


Simplify and find t

       15t     = 10

         t     = {{{10/15}}} = {{{2/3}}} of an hour = 40 minutes.


The same answer.
</pre>

Solved.


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For simple Travel & Distance problems, &nbsp;see introductory lessons 

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

&nbsp;&nbsp;&nbsp;&nbsp;- <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> 

&nbsp;&nbsp;&nbsp;&nbsp;- <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>

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