Question 1109352
.
Let L be the "oval perimeter" length.


<pre>
    The condition says that L = 400 meters.
    But in reality, with all other given data, the solution below is VALID for any value of L: it DOES NOT DEPEND on the concrete value of L
</pre>

<U>1.  Short Physics solution</U>


<pre>
     Henry'   speed is  {{{L/2.5}}}  units of L per minute.

     Harriet' speed is  {{{L/2}}} units of L per minute.

     The relative speed is  {{{L/2 - L/2.5}}} = {{{L*(1/2 - 1/2.5)}}} = {{{2L*((1/4)-(1/5))}}} = {{{2L*(5/20-4/20)}}} = {{{(2L)*(1/20)}}} = {{{L/10}}} units of L per minute.


     It means that after start, the distance between Harriet (who runs faster) and Henry will increase at the rate of {{{L/10}}} per minute.


     Hence, Harriet will win one lap in 10 minutes.


     Notice that the solution does not depend on the concrete value of L.
</pre>

<U>2. - Algebra solution</U>


<pre>
     The distance along the oval from the starting point is

        D1 = {{{(L/2.5)*t}}} for Henry   (=speed*time)

        D2 = {{{(L/2)*t}}} for Harriet.


     The condition  D2 -D1 = L  is

        {{{(L/2)*t - (L/2.5)*t}}} = L,

        {{{(L/4 - L/5)*2t}}} = L,

        {{{(5L)/20 - (4L)/20)*2t}}} = L,

        {{{(L/20)*2t}}} = L,

        {{{(L/10)*t}}} = L

     and gives, finally, the same answer  t = {{{L/((L/10))}}} = 10 minutes.
</pre>

Solved.


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To see many other similar &nbsp;(or closely related) &nbsp;solved problems,  &nbsp;look into 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.