Please help!

XxGiornoxX

New member
Joined
Sep 2, 2022
Messages
1
You're flying from Joint Base Lewis-McChord (JBLM) to an undisclosed location 231 km south and 190 km
east. Mt. Rainier is located approximately 56 km east and 40 km south of JBLM. If you are flying at a
constant speed of 800 km/hr, how long after you depart JBLM will you be the closest to Mt. Rainier?
-minutes
 
Hello Giorno! This looks like a fun exercise that involves working with two perpendicular, intersecting lines on the xy-plane (that's not the aircraft, by the way ;)) – then finding their intersection point, determining the distance from the starting point to the intersection point, using the concept of "distance equals rate×time" and finally converting the answer from hours to minutes. How far did you get? If you're not sure how to begin, start by drawing a diagram with x-axis and y-axis. If you need help with that, then tell us specifically what you find confusing.

Please follow our posting guidelines and share your efforts, thoughts or questions. In case you've skipped the link provided to you earlier, a summary link appears below. Thank you!

[imath]\;[/imath]
 
You're flying from Joint Base Lewis-McChord (JBLM) to an undisclosed location 231 km south and 190 km east. Mt. Rainier is located approximately 56 km east and 40 km south of JBLM.
Hi. In case you're having issues posting your sketch, here's mine. Planes don't take off and land at 800 kilometers per hour (and such a path is not a straight line). So, I'm thinking as follows.

While flying at a fixed speed, the plane passes over both the military base (point B) and the undisclosed location (point C). The mountain is located at point M. When the aircraft is above point B, a timer starts.

310D4399-7BC3-491A-85E7-EEF23B37785E.jpeg

Next, we simply ignore the terrain and aircraft, and we model the aircraft's location as a point moving on a flat surface (the xy-plane) from B to C.

The shortest distance between point M and line BC occurs on a perpendicular line. In other words, on line BC, the closest location to point M is point L.

We're told that point M is 40 units lower on the xy-plane than point B. Calculate point M's y-coordinate.

Next, write the equation for line BC.

Please post your work for those two steps.

:)
[imath]\;[/imath]
 
Hi Giorno. Are you still interested?

Point M's y-coordinate is 40 less than point B's y-coordinate, so point M is located at (56,191).

The Slope-Intercept form of a line is:

y = m∙x + b

where symbol m is the slope and symbol b is the y-coordinate of the point where the line crosses the y-axis.

For line BC, we see that b = 231.

Slope is a ratio in fraction form: (change in y)/(change in x) as we move from any point on the line to another.

Moving from point B(0,231) to point C(190,0), we see the y-coordinate decreases 231 units, so (change in y) is -231. The x-coordinate increases from 0 to 190, so (change in x) is 190.

m = -231/190

The equation of line BC is: y = -(231/190)x + 231

The next step would be finding the coordinates of point L (in order to calculate the distance BL by formula). Do you have any thoughts to share about that?

:)
[imath]\;[/imath]
 
It seems like the OP has lost interest, so I'll finish this exercise as an example for future readers.

The slopes of a pair of perpendicular lines are negative reciprocals of each other. We know the slope of line BC is -231/190. Therefore, the slope of line LM is +190/231.

When we know a line's slope and the coordinates of a point on the line, then the Point-Slope Formula can be used to write the equation of a line.

For line LM, we know the slope is 190/231 and the coordinates of a point are (56,191). Therefore, the equation of line LM is:

y - 191 = (190/231)(x - 56)

In Slope-Intercept form, we now have a system of two linear equations:

Line BC: y = -(231/190)x + 231

Line LM: y = (190/231)x + 4783/33

The solution to this system gives the location of point L (the intersection point of lines BC and LM).

Using the Substitution Method for solving a system, we can replace symbol y in the second equation with its definition from the first equation (because point L lies on both lines):

-(231/190)x + 231 = (190/231)x + 4783/33

Solving this new equation yields the x-coordinate of point L. Then, substituting that solution for x in one of the original equations yields the corresponding y-coordinate.

We find the exact coordinates of point L are (3777200/89461, 16073211/89461).

Those improper fractions will not be nice to work with, so I'm going to proceed based on an assumption that the minutes (for the plane to travel from B to L) are to be reported in decimal form, rounded to the nearest tenth of a minute. Switching to decimal approximations for the remaining work, I'll round all intermediate results to three decimal places, to avoid round-off error in the final answer.

The rounded coordinates at point L are: (42.222, 179.667).

Next, the common relationship between distance traveled, speed (rate) and elapsed time is:

Distance = Rate × Time

Dividing each side of that equation by Rate yields a formula for Time:

Time = Distance/Rate

We know the plane's rate is 800 kph. We need the distance from B to L. We use the Distance Formula for that part.

DistanceBL = [(0 - 42.222)^2 + (231 - 179.667)^2]

The estimated distance is 66.466 kilometers.

The estimated Time is 66.466/800 (Distance/Rate) or 0.083 hours.

We convert the estimated time from hours to minutes.

Rounded to the nearest tenth of a minute, my answer is, "You will be closest to Mount Rainier 5.0 minutes after passing the base."

?
[imath]\;[/imath]
 
Top