Question 1092990
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<pre>
Let D be the length of the track.

    (As you will see from the solution, the value of D does not participate in the solution; 
     so it does not matter even in which linear units we measure D).


Then  the speed of the faster track is {{{D/15}}} linear units per second,
while the speed of the slower track is {{{D/25}}} linear units per second.


The condition that the faster car will pass/overtake the slower is that the faster car will travel the distance 
    exactly D units longer than the slower track, or

{{{(D/15)*t}}} - {{{(D/25)*t}}} = D.


After canceling the factor D in both sides this equation takes the form

{{{t/15}}} - {{{t/25}}} = 1.


Simplify and solve for "t".  First step is to multiply both sides by 75:

5t - 3t = 75 ====>  2t  = 75  ====>  t = {{{75/2}}} = 37.5 seconds.


<U>Answer</U>.  The faster car will pass/overtake the smaller car in 37.5 seconds.
</pre>

Solved.



There is the lesson 

&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> 

in this site, where similar problem is solved and presented (Problem 2).



Also, &nbsp;you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;<A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lesson is the part of this textbook under the section "<U>Word problems</U>", &nbsp;the topic "<U>Travel and Distance problems</U>",

where you will find many other advanced and non-traditional Travel and Distance problems.



Save the link to this online textbook together with its description


Free of charge online textbook in ALGEBRA-I

https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson


to your archive and use it when it is needed.



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