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please see disrciption for my full question.

please expalin in deatil how the answer to question 1 and question 2 was acheived. Two planes leasve airpot at same time. Plane A flies at a speed of 400 miles per hour in a direction 60 degrees east of north. Plane B flies at a speed of 425 mph in a direction 40 degrees east of south. if after flying 3 hrs, the planes change direction and fly directly toward each other. 1. How long would they be able to fly along this path before a collision occurs? 2. what would be the distance travel by each plane be at the point of impact?

I wish WyzAnt would do something about the "pending review" issue. It seems silly to have 8 people give nearly identical answers because they are unaware of each other's answers.

Hi Jay! Here's an approximate approach until AW's solution clears ...

After 3hr => A is 1200mi 30 deg N of E ... 600mi N, 1030mi E ...

B is 1275mi 40 deg E of S ... 850mi E, 975mi S ... gaps to close ==> 1575N, 180E ...

direct-line gap => approx. 1600mi covered by combo speed 825mph in ~ 2 hrs less 3%.

A flies 800mi less 3% ~ 775mi ... B flies 850mi less 3% ~ 825mi ... Best regards, sir :)

Brad - what is 3% for ?  trying to understand the 1575n, 180e ??
Hey Jay -- since I don't use calculators, 1600/825 to me is 1600/800 ... the 3% comes
from the 25 offset out of the 825 ~ 25/825.

The N/S gap to cover is 600N + 975S = 1575N ... E/W gap is 1030E less 850E = 180E :)
In problems like this, I like to refer to the angles in terms of the regular cartesian plane.

Plane A flies at 400 mph in a direction of 30° east of north, which is the same as saying 30° (with respect to the +x axis.

Similarly, Plane B flies at 425 mph at an angle of 310° again with respect to the +x axis.

The magnitude of Plane A's vector is (400 mph)(3 hr) = 1200 miles

The magnitude of Plane B's vector is (425 mph)(3 hr) = 1275 miles

We thus have two sides of a triangle with the angle between them being 80°

The length,a, of the line the planes travel as they're heading towards each other is given by the law of cosines:

a2 = (1200 miles)2 + (1275 miles)2 - (2)(1200 miles)(1275 miles)cos(80°)

a2 = (1440000 mi.2) + (1625625 mi.2) - (531363 mi.2) = 2534261.58 mi.2

a = 1592 miles

a/[(400 mph) + 425 mph)] = 1.93 hours to collision

In 1.92 hours plane A flies (1.92 hours)*(400 mph) = 771.8 miles

In 1.92 hours plane A flies (1.92 hours)*(425 mph) = 820.1 miles

When a plane (A) flies for three hours at 400 mph it describes one leg of a triangle 1200 miles long.

When a second plane (B) leaves the same airport, flying for three hours at 425 mph, it describes another leg of a triangle 1275 miles long.

Plane A flies in a direction 60° east of north or 30° north of east.

Plane B flies in a direction 40° east of south or 50° south of east.

50° + 30° = 80°

So, imagine a triangle with one leg 1200 miles long and another leg 1275 miles long with the angle between them being 80 degrees.

You can find the magnitude (i.e., length of the opposite side of the triangle (the one that describes the "collision course" from the Law of Cosines (see my original answer).

The planes then head towards each other at a "combined speed) of 400 mph + 425 mph, or 825 mph.

The distance they have to travel to collide is the length of the third side of the triangle, the one I have called "a," and which has a length of 1592 miles.

That distance (in miles) divided by the combined speed (in mph) gives the time that will elapse until the planes collide (1.92 hours)

I can't make it any simpler than this.
x-y coordinates of plane A after 3hrs. (1200 * cos 30, 1200 * sin 30) ---->(1039.23, 600)
x-y coordinates of plane b after 3hrs. (1275 * cos -50, 1275 * sin -50) --->(819.55, -976.71)

we calculate distance between these two points:

√[(1039.23 - 819.55)2 + (600 - (-976.71))2] = √[48259.30 + 2486014.42] = 1591.94 miles

400t + 425t = 1591.94

825t = 1591.94

t ≈ 1.93 hours.

distances traveled before impact:

plane A: 1.93 * 400 = 771.85 miles

plane B: 1.93 * 425 = 820.25 miles

Let's first find out where the two planes will be when they change direction after 3 hours.

Plane 1 will have traveled 400mph*3hr=1200 miles at 60° E of N.
Plane 2 will have traveled 425mph*3hr=1275 miles at 40° E of S.

Draw a standard x-y-coordinate system, with "east" being the x-direction. Then the coordinates (in miles) of the planes at that time are
Plane 1: (1200 sin 60, 1200 cos 60) = (1039, 600)
Plane 2: (1275 sin 40, -1275 cos 40) = (820, -977)

Now find their distance at that time:
d= sqrt((1039-820)²+(600+977)²) = 1592 miles.

Now they travel towards each other with a relative speed of
v=400+425=825 mph.

They will collide at 1592/825= 1.93 hr after they changed direction (total time 1.93+3=4.93 hr).

Plane 1 will have traveled 400*1.93=772 miles plus the original 1200 miles for a total of 1972 miles.
Plane 2 will have traveled 425*1.93=820(=1592-772) miles plus the original 1275 miles for a total of 2095 miles.

Draw Plane A and plane B route on a North South East West horizontal plane.
Plane A travels 1200  miles 60 degree east of north which is same as 30 degrees north of east
Plane b travels 1275 miles 40 degree east of south which is same as 60 degrees south of east
Join the ends of two lines.
This becomes a triangle with two sides of 1200 miles and 1275 miles with 80 degrees in between
Now we need to find third side of the triangle.
Apply cosine law
third side ^2 = first side ^2 + second side^2 - (2xfirst sidexsecond side)Cos80degrees
third side ^2 = 1200^2 + 1275^2 - 2x1200x1275Cos80
third side ^2 = 1440000 + 1625625 - 531363.4
third side ^2 = 2534261.6
third side = square root 2534261.6 = 1592 miles
Relative speed of airplanes travelling directly towards each other = 400+475=875 mph
Time for air planes to travel third side = 1592/875=1.82 hrs
Time taken by planes to collide = 3 + 1.82 = 4.82 Hrs
Distance travelled by Plane A to point of impact = 1200 + (1.82x400) = 1928 miles
Distance travelled by Plane B to point of impact = 1275 + (1.82x475) = 2139.5 miles

Let us take northern direction as positive y, and eastern direction as positive x. Then the position of the first plane is given by the coordinates:
x1(miles)=400(mph)*sin(60)*t(hours)
y1(miles)=400(mph)*cos(60)*t(hours)

The northern component of the velocity of the first plane is 400*cos(60), the eastern one is 400*sin(6).

For the second plane, the eastern component of the velocity is given by 425*sin(40), northern component is given by -425*cos(40). Minus sign indicates that it is directed southward, as it should be according to the problem's statement. Thus, coordinates of the second plane are:

x2(miles)=425(mph)*sin(40)*t(hours)
y2(miles)=-425(mph)*cos(40)*t(hours)

In three hours, the first plane will be in the point A with coordinates (1039.23; 600), the second plane will be at the point B (819.55; -976.71). The distance between those points can be found by the formula:

D=√[(x2-x1)2+(y2-y1)2]≈1592 miles

If two planes fly towards each other, then the speed of approach is the sum of two speeds, that is 825 mph. Divide the distance D by 825 to obtain the time until collision, tc=1592/825≈1.93 hours or 1 hour 55 minutes and 47 seconds. This answers the first question.

The answer to the second question can be obtained by multiplying the time found in the first question, tc, by the velocity of each plane. So we obtain that the first plane will travel 400*1.93=772 miles, the second plane will travel 425*1.93=820 miles. These are the distances traveled from the point of turn to the collision point.
It's a long process, but I managed to get (with rounding), an answer for 1) of 1.93 hours.

After 3 hours, plane A flies 400*3 = 1200 miles along the "northeast hypotenuse".
After 3 hours, plane B flies 425*3 = 1275 miles along the "southeast hypotenuse".

I have the coordinates of the plane A at (600*sqrt(3), 600)
or (1200cos 30, 1200sin 30)

I have the coordinates of the plane B at (819.55, -976.71) or (1275cos 50, 1275sin 50).

*Remember plane B went southeast so it's on the other side of the due East direction from plane A.

sqrt(     (1039.23 - 819.55)2  +   (600 - (-976.71)))

Puts the distance between the two planes at 1591.94 miles.

400t + 425t = 1591.94

825t = 1591.94

t = 1.93 hours

Plane A would fly 400(1.93) = 772 miles
Plane B would fly 425(1.93) = 820.25 miles

772
820.25
--------
1592.25 miles = approx.  1591.94 miles...

Please note there's quite a bit of room for error here with rounding.  Hopefully, we'll get someone to check this answer.

•
3*400=       /  |
/    |
1200 mi  ↑       |
↑         ↓
/           ↓
/             |
• ) 100o     |
\              |   "x" miles
\            ↑
\          ↑
↓         |
↓       |
3*425=      \     |
1275 mi       \   |
•

Cosine Rule:

x2 = 12002 + 12752 - 2 * 1200 * 1275 * cos 100o

1220000 + 1625625 - 3060000 * cos 100o ≈ 2534261.37634

x ≈ 1592 mi

1592 / (400 + 425) ≈ 1.93 hours before  collision.

2.
400 * 1.93 ≈ 772 mi

1592 - 772 = 820 mi