A plane is heading west at 200 mph. The wind is blowing S30°W at 25 mph.
1. What is the ground speed of the plane?
2. When you solve the triangle, what is the smallest angle?
3. What is the largest angle of the triangle?
4. What is the remaining angle?
5. What is the bearing of the plane?
A plane is is flying 240 mph heading N60°E. The wind is blowing S30°E at 30 mph.
6. What is ground speed of the plane?
7. What is the smallest angle in the triangle?
8. What is the biggest angle in the triangle?
9. What is the remaining angle in the triangle?
10. What is the bearing of the plane?
A plane is is flying 225 mph heading S25°W. The wind is blowing S80°E at 60 mph.
11. What is the bearing of the plane?
12. What is the smallest angle of the triangle?
13. What is the largest angle of the triangle?
14. What is the remaining angle of the triangle?
15. What is the ground speed of the plane?
16. Vector A has a magnitude of 3 in the leftward direction and B has a magnitude of 5 in the rightward direction. What is the value of 2A – B?
You can also answer other types of problems using vectors. Try this one:
Two forces are pushing on an object, one at 12 lbs of Force and one at 5.66 lbs of Force. The angle between them is 35° (each is 72.5 from horizontal, such that the forces make a v with the object in the center).
17. What is the total Force on the object?
18. What is the smallest angle of the triangle?
19. What is the largest angle in the triangle?
20. What is the remaining angle of the triangle?
I'll do a few, and you can use the method on the rest:
6. Draw the velocity triangle. The wind is blowing at right angles to the plane's air travel. So, you have a n air speed of 240 and a wind speed of 30, making the ground speed 30√65 = 241.87
7,8,9. The angles of the triangle are
smallest: arctan(1/8) = 7.125°
next: 90-7.125 = 88.875°
largest: 90°
10. Bearing is N67.125°E
oops. 90-7.125 = 82.875°
1. [S30oW] = 270-30 = 240o CCW.
Vpw = Vp + Vw = -200 + 25[240o] =
-200 + 25*Cos240 + i25*sin240 =
-200 - 12.5 - 21.7i = -212.5 - 21.7i =
-213.6[5.83o] = 213.6[185.83] CCW from
+X-axis. = [S84oW].
N60oE = 30o CCW from +X-axis.
S30oE = 300o CCW from +X-axis.
Vpw = Vp + Vw = 240[30o] + 30[300o] =
207.8 + 128i + 15 - 26i = 222.8 + 102i =
245mi/h[24.6o] CCW. = N65.4oE.
2. 6 Degrees.
3. 90 Degrees(84o= Largest acute angle)
4. 84 Degrees.
5. S84oW.
7. 24.6 Degrees.
8. 90 Degrees.
9. 65.4 Degrees.
10. N65.4oE.
11. S25oW = 245o CCW. S80E = 350o CCW.
Vpw = Vp + Vw=225[245o] + 60{350o] =
-95.1 - 204i + 59.1 - 10.4i = -36-214.4i
=-217.4[80.5o] = 217.4[80.5+180] =
217.4[260.5] CCW = 217.4mi/h[S9.5oW].
12. 9.5 Degrees.
13. 90 Degrees.
14. 90-9.5 = 80.5 Degrees.
15. 217.4 mi/h.
16. A = -3. B = 5.
2A - B = 2*(-3) - 5 = -11 = 11 Leftward.
17. F=F1 + F2 = 12[107.5o] + 5.66[72.5]=
12*Cos107.5 + i12*sin107.5 + 5.66*Cos72.5 + 5.66*sin72.5 =
-3.61 + 11.44i + 1.70 + 5.4i =
-1.91 + 16.9i =-17[-83.6o]=17[180-83.6]=
17Lbs[96.4o] CCW.= 17Lbs.[N6.4oW].
18. 6.4 Degrees.
19. 90 Degrees
20. 90-6.4 = 83.6 Degrees.
.
To solve these problems, we will use vector addition and trigonometry principles. Let's get started:
1. The ground speed of the plane can be found by considering the vector sum of the plane's velocity and the velocity of the wind. We can break down the wind velocity into its horizontal and vertical components. The horizontal component of the wind velocity is 25 mph * cos(S30°W), and the vertical component is 25 mph * sin(S30°W). Adding the horizontal component to the plane's velocity (westward) and the vertical component (southward), we get an overall ground speed vector. The magnitude of the ground speed is the square root of the sum of the squares of the horizontal and vertical components.
2. The smallest angle in the triangle can be found by considering the direction of the wind and the plane's heading. In this case, since the heading of the plane is directly opposite to the direction of the wind, the smallest angle will be between the wind direction and the plane's heading.
3. The largest angle of the triangle can be found by considering the direction of the wind and the plane's heading. In this case, the wind is blowing towards the south-west, and the plane is heading towards the west. The largest angle will be between the plane's heading and the wind direction.
4. The remaining angle of the triangle can be found by considering the fact that the sum of the angles in a triangle is always 180 degrees. Knowing the values of the smallest and largest angles, we can subtract them from 180 degrees to find the remaining angle.
5. The bearing of the plane can be determined by referencing a compass. The direction the plane is heading, in this case, will be the bearing angle. In this problem, the bearing is west.
For problems 6 to 10, follow the same steps as above but substitute the given values for the plane's heading and wind direction.
For problems 11 to 15, use the same process as above, but now we are given the plane's ground speed. To find the ground speed, we take the vector sum of the plane's velocity and the wind velocity. Using the Pythagorean theorem, we can find the magnitude of the ground speed vector.
For problem 16, "2A - B" implies multiplying vector A by 2 and subtracting vector B. We can multiply vector A by 2 by doubling its magnitude and keeping the direction constant. Subtracting vector B implies taking the opposite direction of vector B. Finally, we sum the two vectors obtained from the previous steps.
For problems 17 to 20, we use vector addition and trigonometry to solve the forces acting on the object. Break down the forces into horizontal and vertical components and add them accordingly to obtain the total force. Use the angles given to determine the smallest, largest, and remaining angles of the triangle formed by the forces.