Vectors and Angles in Helicopter Flight

In summary, the problem involves a helicopter flying a path of 17 km at 63 degrees east of south, followed by 45 km at 73 degrees west of north, and then 34 km at 79 degrees west of north. To return to the launch pad, the displacement vector needs to be calculated by summing the north/south and east/west distances. The final leg of the journey will be in the opposite direction of the result.
  • #1
gric122

Homework Statement



Bob has a helicopter and from the launch pad he flies the following path. First he travels from the launch pad a distance of 17 kilometers at heading of 63 degrees East of South. Then he flies 45 kilometers heading 73 degrees West of North. After this he flies 34 kilometers heading 79 degrees West of North. Now he is ready to return to the launch pad.

What is the displacement vector that he needs to take to return directly to the launch pad from his present location (for the heading give the number degrees north of east - your answer may be greater than 90 degrees)?

Homework Equations


none

The Attempt at a Solution


So I've attempted to draw this problem out and then solve it with trig, but I'm stuck on where the angles point to. From what I've drawn, it looks like a slanted parallelogram, and I don't think I can trig it
 
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  • #2
I would just work out the displacement in the north/south and east/west direction for all the given distances, and then add those all up.
The final leg of the jouney will be in the opposite direction of the result.
 
  • #3
I calculated the horizontal distance and got -61.26 km, but for some reason my vertical is 8.66 which seems really low, am I doing somethign wrong? For x I did 17cos(333) + 45cos(163) + 34cos(169), and for y I just replaced cosine with sine
 
  • #4
gric122 said:
17cos(333) + 45cos(163) + 34cos(169), and for y I just replaced cosine with sine
When I do that I get 11.9.
 
  • #5
haruspex said:
When I do that I get 11.9.
How did you get that? Are u talking about the x or y value? I keep typing it in the calculator and x doesn't come out to be 11.9. I'm using degrees
 
  • #6
gric122 said:
How did you get that? Are u talking about the x or y value? I keep typing it in the calculator and x doesn't come out to be 11.9. I'm using degrees
I'm referring to the y value, obtained as you posted:
gric122 said:
17cos(333) + 45cos(163) + 34cos(169), and for y I just replaced cosine with sine
I.e. just replace all of those cos functions with sine functions. And, yes, angles in degrees.
 

FAQ: Vectors and Angles in Helicopter Flight

1. What is a vector with an angle?

A vector with an angle is a mathematical representation of a quantity that has both magnitude and direction. The angle represents the direction of the vector, while the magnitude represents the length of the vector.

2. How do you calculate the magnitude of a vector with an angle?

To calculate the magnitude of a vector with an angle, you can use the Pythagorean theorem. This states that the magnitude is equal to the square root of the sum of the squares of the components of the vector.

3. What is the difference between a vector with an angle and a scalar?

A vector with an angle has both magnitude and direction, while a scalar only has magnitude. This means that a vector with an angle can be represented by a line with an arrow, while a scalar can be represented by a single number.

4. How do you add or subtract two vectors with angles?

To add or subtract two vectors with angles, you can use the parallelogram rule. This involves drawing the two vectors as sides of a parallelogram, and the resultant vector is the diagonal of the parallelogram.

5. Can a vector with an angle have a negative magnitude?

No, a vector with an angle cannot have a negative magnitude. The magnitude of a vector is always a positive value, as it represents the length of the vector. However, the direction of the vector can be negative if it points in the opposite direction of a positive reference direction.

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