SOLUTION: The Krakow airport is 3 km west and 5 km north of the city center. At 1 pm, Zuza took off in a Cessna 730. Every six minutes, the plane�s position changed by 9 km east and 7 km n

Question 1074231: The Krakow airport is 3 km west and 5 km north of the city center. At 1 pm, Zuza
took off in a Cessna 730. Every six minutes, the plane�s position changed by 9 km east and
7 km north. At 2:30 pm, Zuza was flying over the town of Jozefow. In relation to the center
of Krakow, (a) where is Jozefow? (b) where was Zuza after t hours of flying?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
let x = the number of km east of krakow center.
let y = the number of km north of krakow center.

when t = 0, the position of the plane is 3km west and 5 km north of krakow center.

3 km west is the same as -3 km east.

if we graph the location of the plane, using x = number of km east of krakow center and y = number of km north of krakow center, then, if we assume krakow center is at point (0,0), the plane starts off at the point (-3,5).

x = -3 means the plane is -3 km east which is the same as 3 km west.
y = 5 means the plane is 5 km north.

if we let t = hours, then every 6 minute increment will be .1 * t.

for every 6 minute increment, the plane travels 1 * 9 km east and 1 * 7 km north.

therefore, for every 60 minute increment, the plane travels 10 * 9 km east and 10 * 7 km north.

simplify this to get:

the plane travels 90km east and 70 km north every 60 minutes.

since 60 minutes is equal to 1 hour, then you get:

the plane travels 90 km east and 70 km north every hour.

your equation for y and x in terms of t becomes:

x = 90 * t + b and y = 70 * t + b.

b represents the value of x or y when t = 0.

since the plane starts at x = -3 and y = 4, then your equation for x and y in terms of t becomes:

x = 90 * t - 3 and y = 70 * t + 5

when t = 0, you get x = -3 and y = 5
when t = 1, you get x = 87 and y = 75
when t = 2, you get x = 177 and y = 145

after t hours, the plane is 90 * t - 3 km east of krakow center and 70 * t + 5 km north of krakow center.

since the plane starts at 1:00 pm and is over jozefow at 2:30 pm, then t = 1.5 and the plane is 90 * 1.5 - 3 km east of krakow center and 70 * 1.5 + 5 km north of krakow center.

that puts the plane at 132 km east of krakow center and 110 km north of krakow center.

that means jozefow is 132 km east of krakow center and 110 km north of krakow center.

i could be wrong, but i believe that i did it right and that the above answers are the solutions you are looking for.

your solutions should be:

x = 90 * t - 3
y = 70 * t + 5
jozefow is 132 km east and 110 km north of krakow center.

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