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Direction of the explorer's displacement

  1. Sep 1, 2004 #1
    Getting Back. An explorer in the dense jungles of equatorial Africa leaves his hut. He takes 37.0 steps at an angle 45.0 Degrees north of east, then 82.0 steps at an angle 60.0 Degrees north of west, then 52.0 steps due south. Assume his steps all have equal length. Save him from becoming hopelessly lost in the jungle by giving him the displacement, calculated using the method of components, that will return him to his hut.

    What the magnitude of the explorer's displacement? 47.6 steps

    What the direction of the explorer's displacement? <--- ok for this one, i did arctan(y/x) which is -15.164 but it's the wrong answer. am i doing something incorrectly?

    my work:


    37cos(45)i +37sin(45)j
    -82cos(30)i+82sin(30)j
    + (-55)j
    -------------------------------------
    -44.85i+12.16j

    arctan (12.16/-44.85) =-15.164 Degrees

    can anyone tell me why i am getting the incorrect answer?
     
  2. jcsd
  3. Sep 1, 2004 #2
    try 120 degrees for the second batch of steps (60 degrees north of west)
     
  4. Sep 1, 2004 #3
    What's the correct answer?
     
  5. Sep 1, 2004 #4
    i dont know what the answer is, i have to submit it thru an online server and it checks your answer.


    sorry i forgot to say that the answer must be ______ East of South .

    what do you mean bt trying 120 Degrees for the second batch? my setup is correct cause i got the first answer correct, but im having trouble with the second question
     
  6. Sep 1, 2004 #5
    just use vector addition for the direction (head to tail addition)
     
  7. Sep 1, 2004 #6
    what do you mean? can you give me an example?
     
  8. Sep 3, 2004 #7
    can someone help me with this one? my work is shown in my first post, im just reposting so it will go to the top so other people will be able to see.

    What the direction of the explorer's displacement? <--- this is the one i need help on. (the answer must be ______ East of South)

    im asking agian, because the homework is due in 7 hours. and this is the only question im missing. I went to the physics tutoring center today at school and asked the TA for help, he helped me and told me the incorrect answer. i thought he was correct so i went home to my computer and went online and submitted my hw and it was incorrect. so you guys are my only help. im trying to do the question over and over agian, but i cannot get it correct, i really need help. thanks in advance
     
  9. Sep 3, 2004 #8

    Doc Al

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    Staff: Mentor

    Where did the 30 degrees come from? Your problem said "60 degrees N of W".
     
  10. Sep 3, 2004 #9
    Man, if you only gave the correct information, you wouldn't be wasting so much time, waiting. You probably also gave the wrong info to the TA.
     
  11. Sep 3, 2004 #10
    no i didnt give you guys or the TA the incorrect time, i gave the TA a print out. i took the angle next to the 60 degrees, which is 30. 90-60 = 30 Degrees, i just did the problem differently.is my setup correct? my friend helped me set it up at school, and maybe i just got lucky and got the first answer correctly. is the correct setup? and how would i find the answer to my second question?

    i seen other students at the tutoring center set it up this way:

    37cos(45)i +37sin(45)j
    -82cos(60)i+82sin(60)j
    + (-55)j
    -------------------------------------
    and they get an incorrect answer when they submit their answer. this is probably the way you guys were thinking. students that set it up my way got the correct answer to the first question, so i dont know...
     

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  12. Sep 3, 2004 #11

    Doc Al

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    It's certainly true that 90-60 = 30, but so what? If you want the x-component of a displacement you'd better be using [itex]cos\theta[/itex], where [itex]\theta[/itex] is the angle from the x-axis. That angle is 60 degrees, not 30. (Your diagram is not draw to scale.)
     
  13. Sep 3, 2004 #12

    Doc Al

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    Staff: Mentor

    This is the correct way to find the displacement of the explorer from the starting point. To find the displacement that will return him to the camp, take the negative of that sum.
     
  14. Sep 3, 2004 #13
    37cos(45)i +37sin(45)j
    -82cos(60)i+82sin(60)j
    + (-55)j
    -------------------------------------
    -14.83i + 42.17j

    what do you mean by taking the negative sum?


    (-14.83i + 42.17j) <-- original sum

    14.83i^2 - 42.17j^2 = C^2 <--- is that what you mean? instead of adding, i should substract? can i get an example? isnt the answer suppose to be in degrees?

    im sorry, im confused. im really not good at physics, that's why im in the tutoring center everyday. :(
     
  15. Sep 3, 2004 #14
    well i did it this way,

    sin45*37+sin60*82-52 = y

    cos60*82 -cos45*37= x

    tan^-1(y/x) = theta

    however you want east of south, so 90 - theta will be the degrees east of south. However though, this direction is to get back towards the original point, not it's displacement direction.
     
  16. Sep 3, 2004 #15

    Doc Al

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    Staff: Mentor

    Yes.
    The direction can be expressed as an angle. Figure it out by drawing the vector and finding [itex]tan\theta[/itex], like you were doing in the first post.
     
  17. Sep 3, 2004 #16

    Doc Al

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    Staff: Mentor

    Correction: 52, not 55

    That last piece should be 52 steps, not 55. You'll have to redo the sum.
     
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