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NotTim Group Title

2 canoeists, A & B, live on opposite shores of a 300.0 m wide river that flows east at 0.80 m/s. A lives on the north shore and B lives on the south shore. They both set out to visit a mutual friend X who lives on the north shore at a point 200.0m upstream from A and 200.0m downstream from B. both canoeists can propel their canoes at 2.4 m/s thru the water. How much time must canoeist A wait after canoeist B sets out so that they both arrive at X at the same time? Both canoeists make their respective trips by the most direct routes.

  • one year ago
  • one year ago

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  1. NotTim Group Title
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    This is for a friend, so I don't have much work shown.

    • one year ago
  2. NotTim Group Title
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    I know that B takes 125s. CALC: 1. v=2.4-0.80=1.6m/s 2. t=200m/1.6m/s=125s

    • one year ago
  3. Shadowys Group Title
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    note: 2.4 is speed relative to the earth?

    • one year ago
  4. NotTim Group Title
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    @Shadowys Yes, I believe so. Those links are images of an answer key I wrote down. Step 5 is indecipherable though. My friend had difficulties understanding the angle.

    • one year ago
  5. Shadowys Group Title
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    can i use vectors to solve this?(it should be solved with vectors)

    • one year ago
  6. NotTim Group Title
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    Wait....er. no. My friend's course probably doesn't use vectors.

    • one year ago
  7. Shadowys Group Title
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    cos, the speed 2.4 is gonna have it's components...lol...without vectors....

    • one year ago
  8. NotTim Group Title
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    well, go ahead and use vectors if you like. (is this vectors the basic kind, or from calculus and vectors)?

    • one year ago
  9. Shadowys Group Title
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    |dw:1359342186714:dw| the basic kind. where i=unit vector in x direction, j is unit vector in y direction. the velocity of A, \(\vec v_A = -2.4\vec i +0.8i=-1.6\vec i\) the displacement vector of A \(\vec S_A= -200\vec i\) so time taken for average journey = \(\frac{|\vec S_A|}{|\vec v_A|}=\frac{\sqrt{(-200)^2}}{\sqrt{(-1.6)^2}}=125s\) the angle of BX is \(tan^{-1} 3/2=0.9828 rad \) for B, velocity of B =\(\vec v_B = 2.4(cos 0.9828 \vec i+sin 0.9828 \vec j)+0.8\vec j=2.1313\vec i + 0.832\vec j\) displacement is \(\vec S_B= 200\vec i+300\vec j\) so time taken for average journey = \(\frac{|\vec S_B|}{|\vec v_B|}=\frac{\sqrt{(200)^2+(300)^2}}{\sqrt{(2.1313)^2 +(0.832)^2}}=157.35s\) so time needed to wait=157.34s-125s=32.35s

    • one year ago
  10. NotTim Group Title
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    answer key says 6.1s.

    • one year ago
  11. Shadowys Group Title
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    is my diagram wrong?

    • one year ago
  12. NotTim Group Title
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    |dw:1359314524260:dw| yours is right. just missed a few lines

    • one year ago
  13. NotTim Group Title
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    |dw:1359314590576:dw|

    • one year ago
  14. NotTim Group Title
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    Your part A is fine, its jsut part B

    • one year ago
  15. Shadowys Group Title
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    hmm...either it's wrong that i took 2.4 as the magnitude...what do you think?

    • one year ago
  16. NotTim Group Title
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    The first time you used it is fine, but I'ven ever seen the 2nd time 2.4 wasused (usign ctrl-F)

    • one year ago
  17. Shadowys Group Title
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    oo, right, i forgot to multiply the j component by 2.4. so \(\vec v_B =2.1313\vec i + 1.997 \vec j\) t=123.49s....

    • one year ago
  18. NotTim Group Title
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    Vb here was 131.15s?

    • one year ago
  19. NotTim Group Title
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    Sorry, Tb is 131.15s

    • one year ago
  20. NotTim Group Title
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    125s was an acceptable value too for some reason.

    • one year ago
  21. Shadowys Group Title
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    nope.....hmm...am i overthinking this? for which reason?

    • one year ago
  22. NotTim Group Title
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    125s was an acceptable solution for the time for B. And why? I do not know. I've looking over various different solutions and methods here (not understanding), and the values 131.15 and 125 are acceptable for some odd reason.

    • one year ago
  23. NotTim Group Title
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    of course, 131.15s is the on on the official answer key

    • one year ago
  24. NotTim Group Title
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    we can give it a rest now if you want.

    • one year ago
  25. Shadowys Group Title
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    alright. hopefully someone else will give it a go.

    • one year ago
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