please see disrciption for my full question.

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?

Answer 1

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

 

 

 

 

Answer 2

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.

 

Answer 3

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

Cosine Rule: 

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

1220000 + 1625625 - 3060000 * cos 100^o ≈ 2534261.37634 

x ≈ 1592 mi 

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

2.      
400 * 1.93 ≈ 772 mi 

1592 - 772 = 820 mi 

Answer 4

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

 

Answer 5

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, t_c=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, t_c, 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.

Answer 6

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))² )

 

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...

 

Answer 7

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

Answer 8

 

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 :)