A plane is flying through the air at a velocity 88km/hr

  • #1

Homework Statement



A plane is flying through the air at a velocity 88km/hr direction N37'E on a day when there is no wind.
a) After 50 minutes, how far has the plane gone?
b) How far east is it from it's starting point?
c) How far north is it from its starting point?

If there is a wind blowing FROM due south at 17mk/hour, what would be the answers to the question above?

Homework Equations




time=distance/speed

The Attempt at a Solution



50 minutes = distance/ 88 km/h
 

Answers and Replies

  • #2
tiny-tim
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welcome to pf!

hi jacknjersey! welcome to pf! :smile:
time=distance/speed

50 minutes = distance/ 88 km/h

ok, now convert minutes to hours …

(you need everything to be in the same units! :wink:)

the distance is … ? :smile:

(time=distance/speed, speed=distance/time, distance=speed*time)
 
  • #3


50 minutes is 0.83 hours

so now I have distance = speed x time

distance = 0.83 x 88
73.04km

I'm not sure what to use to get b and c - how far north and east? argh.
 
  • #4
tiny-tim
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I'm not sure what to use to get b and c - how far north and east? argh.

well, you know the angle (37° to the right of north) …

so draw a triangle, and use cos and sin :wink:
 
  • #6


cos theta = adjacent/ hypotenuse
cos 37'=x/ 73.04
x=55.90584235
x=55.91km

sin theta = opposite/hypotenuse
sin 37'= y/73.04
y=73.04 sin 37
y=43.96km

If the wind was blwoing from the south at 17km/hour, the wind is blowing due north.

therefore:

after 50 minutes at 17 km p/h, plane has travelled
distance = 0.83 x 17
distance = 14.11km

how far east is it fromit's starting point?


how far north is it from it's starting point?
 
  • #7
tiny-tim
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you can regard the wind as being separate from the engine …

just ad the displacement due to the wind to the displacement due to the engine :wink:
 
  • #8


simply put, 17+88?
 
  • #9
tiny-tim
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no!!

add the displacements (the changes in position, the distances)
 
  • #10


the plane went 73.04km. add the wind of 17.
= 90.04
 
  • #11
tiny-tim
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haven't you been taught vector addition?

to add displacements, you have to add the x and y amounts (the components) separately

(try drawing it)
 
  • #12


do i have all the coordinates?
 
  • #13
tiny-tim
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  • #14


Ive done a drawing and to get the resultant vector, i need to draw a tail from the head of the second to the tail of the first and find that length. the length of this vector is the magnitude of the resultant vector.

Should i paste in my picture?
 
  • #15
tiny-tim
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hi jacknjersey! :smile:
Ive done a drawing and to get the resultant vector, i need to draw a tail from the head of the second to the tail of the first and find that length. the length of this vector is the magnitude of the resultant vector.

that's correct :smile:

but do b) and c) first (how far east and north), then you can use pythagoras to find the magnitude :wink:
 
  • #16


how far east? 73.04
how far north from starting point? 17 + 43.96 = 60.96km
how far has the plan gone? x^2=73.04^2 + 17^2 = 74.99km
 
  • #17
tiny-tim
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how far north from starting point? 17 + 43.96 = 60.96km

that's right! :smile:
how far east? 73.04

no 73.04 was the total distance from the engine, you need to use the east distance (and add 0 :wink:)
how far has the plan gone? x^2=73.04^2 + 17^2 = 74.99km

no, for the total magnitude, use pythagoras on the total north and total east
 

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