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Total displacement of sail boat

  1. Sep 25, 2006 #1
    A sailboat tacking against the wind moves as follows:
    5.8 km at 45 degrees east of north, and then
    4.5 km at 50 degrees west of north.
    The entire motion takes 1 h 15 min
    What is the total displacement?
    What is the average velocity of this section of the trip?
    What is the speed, if it is assumed to be constant?

    Total Displacement
    /\x = 5.8*sin(45) + 4.5*sin(-50)= 6.116 km
    /\y = 5.8*cos(45) + 4.5*cos(-50)= 7.389 km
    /\d = sqrt(/\x^2 + /\y^2) = 9.592 km = 9592 m

    Average Velocity
    v= r/t

    How do I figure out the rest from this information? Any help would be appreciated.
     
  2. jcsd
  3. Sep 25, 2006 #2
    The speed (assuming it is constant through the trip) is just the total distance travelled divided by the time it took to travel that distance.

    Average velocity is displacement divided by the time.
     
  4. Sep 25, 2006 #3
    Average velocity=displacement/time
    Average velocity=9.592 km/1.25 h=7.623 km/h
     
  5. Sep 25, 2006 #4
    yeah that's right formula.

    However, looking at your calculation for displacement I don't think it's right. I don't think you're going to have a right angle triangle with those angles. I think that you will have to use the law of cosines to calculate displacement.
     
  6. Sep 25, 2006 #5
    hmmm .... Im new to vectors (and Physics as a whole) ..... but i did manage to do some calculations using the cosine law and it seems that big man is correct . The displacement should be around 10.3 Kms . As far as the average velocity is concerned ... well its about 2.3 ms^-1 . It says that the speed is assumed to be constant , so ... use this equation s = ut where s = total distance , t = total time and u = constant speed ....

    BTW which book are u going through ??
     
  7. Sep 25, 2006 #6
    sorry i think i got it wrong .... it said west of north damn !! anyways i think u get the picture right ?
     
  8. Sep 25, 2006 #7
    Welcome junior_j!!

    I actually calculated the displacement to be roughly 7 km.

    The angle opposite the displacement line is 85 degrees. Is that what you got?

    EDIT: haha that was a quick post. Disregard this post ;)
     
  9. Sep 25, 2006 #8
    now that u mentioned it im doing it again :)

    ..... yes those are what i got .... the angle opposite to the resultant vector is 85 degrees and the resultant displacement is 7km
     
  10. Sep 25, 2006 #9
    Alright here's another shot...

    Total Displacement
    180-angle A-angle B=angle C
    180-45-50=85
    v3^2=v1^2+v2^2-2(v1)(v2)cos(C)
    v3^2=5.8^2+4.5^2-(52.2)cos(85)
    v3^2=49.341
    v3=7.024 km

    Average velocity=displacement/time
    Average velocity=7.024 km/1.25 h=5.619 km/h

    How would I calculate the speed?
    Constant Acceleration
     
  11. Sep 25, 2006 #10
    speed = distance/time

    So 4.5 + 5.8 = 10.3 km. That is your total distance covered in the given time interval.

    Edit: Forgot to say that your above working is correct as well for the average velocity.
     
  12. Sep 25, 2006 #11
    speed=distance/time
    speed=4.5 km + 5.8 km/1.25 h= 8.24 km/h
     
  13. Sep 25, 2006 #12
    Yup that's it!!!
     
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