SOLUTION: Two railroad tracks intersect at right angles. At noon there is a train on each track approaching the crossing at 40 mi.per hour., one being 100 mi., the other 200 mi. distant. F

Question 1194695: Two railroad tracks intersect at right angles. At noon there is a train on each
track approaching the crossing at 40 mi.per hour., one being 100 mi., the
other 200 mi. distant. Find (a) when they will be nearest together, and
(b) what will be their minimum distance apart?
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

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)

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

Now the task is to solve
In other words, solve:

Because the denominator can never be zero, this forces the numerator to be zero.
Solve 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

which is approximate
Round this approximate value however your teacher instructs.

Answers:
(a) The two trains will be nearest together at exactly 3:45 PM
(b) The minimum distance is about 70.7107 miles

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