Finding the Rescue Direction for a Plane Emergency Landing

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A plane departs from Galisteo, flying 170 km at 68 degrees east of north, then 230 km at 48 degrees south of east, before making an emergency landing. The calculations for the x and y components of each leg of the flight reveal errors in the signs and values used. After correcting these, the resultant vector is found to be approximately 330 km at 19 degrees south of east. The rescue crew should head in this direction to reach the plane. Drawing a sketch is recommended to visualize the vector components and ensure accuracy in direction.
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Homework Statement



A plane leaves the airport in Galisteo and flies 170km at 68 degrees east of north. And then flies 230 km at 48 degrees south of east. It makes an emergency landing at it's pasture. When the airport sends out a rescue crew, in which direction and how far should they go to fly directly to the plane?

Homework Equations


X and Y components, Vector addition.


The Attempt at a Solution



170(cos68deg) = x comp = 63.7km
170(sin68deg) = y comp = 157.8km

230(cos48deg)=x comp = 153.9km
230(sin48deg)=y comp = 171km

Rx=217.6km
Ry=328.6km
R= 394.1km, by calculating the magnitude

Ry/Rx (arctan) = 56.5 degrees

394.1km at 56.5 degrees West to get to the plane.
 
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alexhenderson said:

Homework Statement



A plane leaves the airport in Galisteo and flies 170km at 68 degrees east of north. And then flies 230 km at 48 degrees south of east. It makes an emergency landing at it's pasture. When the airport sends out a rescue crew, in which direction and how far should they go to fly directly to the plane?

Homework Equations


X and Y components, Vector addition.


The Attempt at a Solution



170(cos68deg) = x comp = 63.7km
170(sin68deg) = y comp = 157.8km

230(cos48deg)=x comp = 153.9km
OK!
230(sin48deg)=y comp = 171km
+ 171 or -171??
 
PhanthomJay said:
OK! + 171 or -171??
I'm thinking it would be +171.
 
Always draw a rough sketch on paper. In which direction does the y component of the 230 km vector point...up (positive) or down (negative)?
 
Negative, yeah, I think that's the mistake.
 
Yes, good. Now determine the resultant vector...magnitude and direction. Don't forget to draw a rough sketch of the problem...placing the tail of the 2nd vector on the head of the first...then sketching in the resultant will give you a graphical approximation of its direction.
 
Was my solution incorrect? Rather than forgetting the -171.
 
alexhenderson said:
Was my solution incorrect? Rather than forgetting the -171.
It was incorrect. If the y comp of the first vector is +63.7, and the y comp of the 2nd vector is -171, then Ry = ??
 
Ry= -103.7km. And then if I change Ry, and then do the magnitude and Ry/Rx arctan would I get the right answer?
 
  • #10
Oh, sorry, I'm messing you up, I misread the numbers. The y comp of the first vector is 157.8[/color] and the y comp of the 2nd vector is -171, and thus Ry = ?. Then find the magnitude and angle, and draw a sketch roughly to scale to make sure you get the direction correct.
 
  • #11
Ry= -13.2 so the magnitude is 217.2km, and would it be going 68 degrees east to the plane?
 
  • #12
Oh, let's start from the beginning again...you have made some errors and I have not been careful checking them, my apologies, so let's give it another try. Let the east direction be the positive x axis, and the north direction be the positive y axis.

Homework Statement



A plane leaves the airport in Galisteo and flies 170km at 68 degrees east of north. And then flies 230 km at 48 degrees south of east. It makes an emergency landing at it's pasture. When the airport sends out a rescue crew, in which direction and how far should they go to fly directly to the plane?

Homework Equations


X and Y components, Vector addition.


The Attempt at a Solution



170(cos68deg) = x comp = 63.7km
the plane initially flies 68 degrees east of north, which means it flies initially in a direction 22 degrees above the horizontal axis. So the horizontal x comp = 170 cos 22 = 157.6 km
170(sin68deg) = y comp = 157.8km
this should be 170 sin 22 = y comp = 63.7 km
230(cos48deg)=x comp = 153.9km
good
230(sin48deg)=y comp = 171km
should be -171
Rx=217.6km
Ry=328.6km
R= 394.1km, by calculating the magnitude
make necessary corrections
Ry/Rx (arctan) = 56.5 degrees
make correction
394.1km at 56.5 degrees West to get to the plane.
make correction, draw sketch
 
  • #13
292.4km 49 degrees south east? I did the Ry/Rx(arctan) and got 19 degrees and subtracted that from the 68 degrees.
 
  • #14
How are you getting this? There's probably too many numbers on the board.
Rx is the sum of the x comp = 157.6 + 153.9 = 311.5
Ry is the sum of the y comp = 63.7 - 171 = - 107.3

Thus you can get magnitude of the resultant (sq rt of sum of squares) and its angle with the x-axis (arc tan Ry/Rx), but be sure to draw a sketch to get the correct compass heading.
 
  • #15
I am doing those methods, but I think I am miscalculating, I did the square root sum of squares and got 329.5 km, then I did -107.3/311.5(arctan) and got 19 degrees.
 
  • #16
alexhenderson said:
I am doing those methods, but I think I am miscalculating, I did the square root sum of squares and got 329.5 km, then I did -107.3/311.5(arctan) and got 19 degrees.
That's very good, you are quite right, call it 330 km at 19 degrees, but the question is, 19 degrees with respect to which axis? I can tell you that it is with respect to the horizontal (east) axis, but the rescue crew must travel 330 km at a heading of (multiple choice)
a.) 19 degrees south of east ,
or is it
b.) 19 degrees north of east??

HINT: Draw a sketch roughly to scale! Place the tail of the 2nd vector on the head (arrow) of the first vector...then sketching in the resultant vector( R) by drawing a vector line from the tail of the first vector to the head of the 2nd vector will give you a graphical approximation of its direction.
 
  • #17
I drew a sketch and I got 19 degrees south of east.
 
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