A plane flew at 210° from City A 2 hours at 280mph and then at 320° for 1.5 hours at 240°mph. In what direction must it fly to return directly back to City A.​

Respuesta :

Answer: 180° + angle.

Step-by-step explanation: The plane must fly at

12

0

120

 to return directly back to City A

1Calculate the total distance traveled in the first leg of the journey.

Given the speed of 280 mph and time of 2 hours, use the formula: distance = speed * time.

Distance = 280 mph * 2 hours = 560 miles.

2Calculate the total distance traveled in the second leg of the journey.

Given the speed of 240 mph and time of 1.5 hours, use the formula: distance = speed * time.

Distance = 240 mph * 1.5 hours = 360 miles.

3Draw a vector diagram to represent the two legs of the journey.

Let the first leg be represented by a vector of 560 miles at

21

0

210

 and the second leg by a vector of 360 miles at

32

0

320

.

4Find the resultant vector of the journey.

To find the resultant vector, add the two vectors using vector addition. This can be done by breaking the vectors into x and y components and then adding them up.

5Calculate the x-component of the resultant vector.

x-component = 560 miles * cos(

21

0

210

) + 360 miles * cos(

32

0

320

).

6Calculate the y-component of the resultant vector.

y-component = 560 miles * sin(

21

0

210

) + 360 miles * sin(

32

0

320

).

7Find the direction of the resultant vector.

Use the arctan function to find the angle of the resultant vector: angle = arctan(y-component / x-component).

8Convert the angle to the appropriate direction.

The direction of the plane to return directly back to City A is the direction opposite to the angle calculated in Step 7.

9Calculate the final direction.

Final direction = 180° + angle.