Question 1194695
<font color=black size=3>
t = elapsed number of hours since 12:00 PM noon
example: t = 3 means it is 3 hours after noon, placing the time at 3:00 PM exactly


Draw an xy axis. 
The x and y axis are perpendicular to each other, i.e. they form a 90 degree angle, i.e. they intersect at right angles.
This will represent the two train tracks.


Label the two trains A and B. 
I'll place train A on the x axis, and train B on the y axis. 
Train A is 100 miles from the intersection. Meaning train A's location could be (-100,0). 
This places train A exactly 100 miles west of the intersection. Train A will head east at 40 mph.


Now form the distance equation for train A.
distance = rate*time
d = r*t
d = 40t


Train A is heading east in the positive x direction. 
We'll add the 40t onto the x coordinate to get the point (-100+40t, 0)
This describes train A's location at any given time t.


Example:
At 2 PM it will have the location of...
(-100+40t, 0) = (-100+40*2, 0) = (-20, 0)
It is currently 20 miles west of the intersection.


Meanwhile, train B will travel along the north/south train track (aka y axis). 
I'll place train B at (0,-200) which is 200 miles south of the intersection.
Its distance equation is also d = 40t because it travels the same speed of 40 mph
So we'll add 40t to the y coordinate of the starting position of the train.


The position of train B at time t is (0,-200+40t)
It travels north at a speed of 40 mph.


-----------------------------------------------------------------------------------


To recap:
Train A is at (-100+40t, 0)
Train B is at (0,-200+40t)
where t is the number of hours since 12:00 PM noon


Apply the distance formula to figure out how far apart the two trains are, at any given time t.


Train A = (x1,y1) = (-100+40t, 0) and Train B = (x2,y2) = (0,-200+40t)
{{{d = sqrt( (x[2]-x[1])^2+(y[2]-y[1])^2 )}}}


{{{d = sqrt( (0-(-100+40t))^2+(-200+40t-0)^2 )}}}


{{{d = sqrt( (0+100-40t)^2+(-200+40t-0)^2 )}}}


{{{d = sqrt( (100-40t)^2+(-200+40t)^2 )}}}


{{{d = sqrt( (1600t^2-8000t+10000)+(1600t^2-16000t+40000) )}}}


{{{d = sqrt( 3200t^2-24000t+50000 )}}}


{{{d = sqrt( 400(8t^2-60t+125) )}}}


{{{d = sqrt( 400)*sqrt(8t^2-60t+125)}}}


{{{d = 20*sqrt(8t^2-60t+125)}}}


You can use a graphing calculator to find the lowest point to be (3.75, 70.7106781186548)
The value 3.75 is exact; in contrast, the 70.7106781186548 is approximate


Or if you are in a calculus class, then you can use the derivative to determine the local minimum.
I'll use x in place of t, and y in place of d.
x = t = time
y = d = distance
{{{y = 20*sqrt(8x^2-60x+125)}}}


{{{(dy)/(dx) = 20*(1/(2*sqrt(8x^2-60x+125)))*expr(d/(dx))(8x^2-60x+125)}}}


{{{(dy)/(dx) = 20*(1/(2*sqrt(8x^2-60x+125)))*(16x-60)}}}


{{{(dy)/(dx) = (160x-600)/(sqrt(8x^2-60x+125))}}}


Now the task is to solve {{{(dy)/(dx) = 0}}}
In other words, solve: {{{(160x-600)/(sqrt(8x^2-60x+125)) = 0}}}


Because the denominator can never be zero, this forces the numerator to be zero.
Solve {{{160x-600 = 0}}} to get x = 600/160 = 3.75 exactly


This means t = x = 3.75 hours
3.75 hours = 3 hours + 0.75 hours
3.75 hours = 3 hours + (0.75 hours)*(60 min/1 hour)
3.75 hours = 3 hours + 45 min
The starting point was 12:00 PM noon. 3 hrs, 45 min later places us at 3:45 PM. This is the answer to part (a).


Now use this t value to find d
{{{d = 20*sqrt(8t^2-60t+125)}}}


{{{d = 20*sqrt(8(3.75)^2-60(3.75)+125)}}}


{{{d = 20*sqrt(12.5)}}}


{{{d = 70.7106781186548}}} which is approximate
Round this approximate value however your teacher instructs.



Answers:
(a) The two trains will be nearest together at exactly <font color=red>3:45 PM</font>
(b) The minimum distance is about <font color=red>70.7107 miles</font> 
</font>