Question 1149914
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Let "t" be the time after second train started.


The we have for distances from the starting point


    First train :  x(t) = 40 + 40*t

    Second train : y(t) = 60*t


The distance between the trains is  z(t) = {{{sqrt((x(t))^2+ (y(t))^2)}}} = {{{sqrt((40+40*t)^2 + (60*t)^2)}}}


The rate of separating, which the problem is asking about, is


    {{{(dz)/(dt)}}} = {{{(1/2)*((2*(40+40*t)*40 + 2*60*t*60)/sqrt((40+40*t)^2 + (60*t)^2))}}} 


Now substitute  t= 2 hours into the last formula and calculate


    {{{(dz)/(dt)}}} = {{{(1/2)*((2*(40+40*2)*40 + 2*60*2*60)/sqrt((40+40*2)^2 + (60*2)^2))}}} = {{{(1/2)*(24000/sqrt(28800))}}} = 70.711 km/h.    <U>ANSWER</U>
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