Answer:
about 4 minutes 55.5 seconds
Step-by-step explanation:
There are many possible ways to solve this. Perhaps one of the easiest is to make use of trigonometry to find the distance from JBLM to the point of closest approach.
The distance from JBLM to Mt. Rainier is found using the Pythagorean theorem:
d = √((56 km)² +(40 km)²) = 8√74 km
The bearing from JBLM to Mt. Rainier is ...
arctan(40/56) ≈ 35.538° . . . . . south of east
The heading of the airplane to its destination is ...
arctan(289/218) ≈ 52.972° . . . . . south of east
Then the difference between these angles is ...
52.972° -35.538° ≈ 17.434°
and the distance to the point of closest approach is ...
8√74×cos(17.434°) ≈ 65.657 . . . . km
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The relation between time, speed, and distance is ...
time = distance / speed
Then the time from JBLM to the point of closest approach is ...
time = (65.657 km)/(800 km/h) ≈ 0.0820715 h ≈ 4.9243 minutes
≈ 4 minutes 55.5 seconds
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There's a lot of math involved in this question no matter how you solve it.
Alternative approach 1
Define Mt. Rainier as the origin and write an equation for the line that defines the airplane's path. Use this equation to find the distance from the mountain to the point of closest approach, and use that distance to find the distance from JBLM to the point of closest approach. Figure time as above.
Using Mt. Rainier as the origin, JBLM has coordinates (-56, 40), and the line of travel of the airplane can be written as ...
289(x+56) +218(y -40) = 0
In general form, this is ...
289x +218y +7464 = 0
Using the formula for the distance from point (x, y) to line ax+by+c=0, we can find the distance to this line from Mt. Rainier:
d = |ax+by+c|/√(a²+b²) = |289·0 +218·0 +7464|/√(289²+218²)
d = 7464/√131045 ≈ 20.6187
This is the short leg of the right triangle that has the JBLM to Mt Rainier distance as its hypotenuse and the distance to closest approach as the other leg. In other words, the distance to closest approach (c) will be ...
c² = (56²+40²) -(20.6187)² = 4736 - 7464²/131045 ≈ 4310.869
c ≈ √4310.869 ≈ 65.657 . . . . km from JBLM to point of closest approach
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Alternative approach 2
Write an expression for the distance from the plane to Mt. Rainier as a function of time. Find the time that minimizes the value of this expression.
In order to minimize the use of annoying numbers, we will define the constants "e" and "s" to represent the easterly and southerly speed of the airplane.
The total distance to the undisclosed location is given by the Pythagorean theorem as ...
d = √(289² +218²) = √131045 km
Then the fraction of that distance that is in the easterly direction is ...
easterly fraction of distance = 218 km/√131045 km = 218/√131045
This is the same as the easterly fraction of speed, so the speed in the easterly direction is ...
e = (800 km/h)(218/√131045) = 174400/√131045 km/h
Similarly, the fraction of speed in the southerly direction is ...
s = (800 km/h)(289/√131045) = 231200/√131045 km/h
Now, the position P of the airplane at time t is ...
P = (et, -st)
And the distance from that position to Mt Rainier is found using the distance formula ...
d² = (x2-x1)² +(y2-y1)²
d² = (et -56)² +(-st -(-40))² = e²t² -112et +3136 +s²t² -80st +1600
= (e² +s²)t² -(112e +80s)t +4736
The expression for the square of the distance is a quadratic in t. The minimum of the quadratic ax²+bx+c is x=-b/(2a), so the value of t that minimizes the distance to the mountain is ...
time to closest approach = (112e +80s)/(2(e² +s²)) = (56e+40s)/(e² +s²)
We can fill in the above values for e and s to find the time as ...
((56·174400+40·231200)/√131045)/800² ≈ 0.0820715 . . . hours
