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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.

Physics
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This is for a friend, so I don't have much work shown.
I know that B takes 125s. CALC: 1. v=2.4-0.80=1.6m/s 2. t=200m/1.6m/s=125s
note: 2.4 is speed relative to the earth?

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https://fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-prn1/533567_154212501399299_1017958671_n.jpg https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-snc7/421635_154212521399297_25972720_n.jpg https://fbcdn-sphotos-f-a.akamaihd.net/hphotos-ak-prn1/76140_154212481399301_1414230843_n.jpg https://fbcdn-sphotos-d-a.akamaihd.net/hphotos-ak-ash4/387012_154212484732634_845307988_n.jpg https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-ash4/428325_154212474732635_582085865_n.jpg https://fbcdn-sphotos-b-a.akamaihd.net/hphotos-ak-prn1/150879_154212514732631_1274474888_n.jpg https://fbcdn-sphotos-d-a.akamaihd.net/hphotos-ak-prn1/554471_154214558065760_366131359_n.jpg https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-ash3/44769_154214568065759_779969484_n.jpg https://fbcdn-sphotos-f-a.akamaihd.net/hphotos-ak-snc7/486007_154214574732425_1940920149_n.jpg https://fbcdn-sphotos-b-a.akamaihd.net/hphotos-ak-ash3/537274_154214594732423_1396018523_n.jpg https://fbcdn-sphotos-d-a.akamaihd.net/hphotos-ak-prn1/45480_154216528065563_1014789527_n.jpg
@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.
can i use vectors to solve this?(it should be solved with vectors)
Wait....er. no. My friend's course probably doesn't use vectors.
cos, the speed 2.4 is gonna have it's components...lol...without vectors....
well, go ahead and use vectors if you like. (is this vectors the basic kind, or from calculus and vectors)?
|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
answer key says 6.1s.
is my diagram wrong?
|dw:1359314524260:dw| yours is right. just missed a few lines
|dw:1359314590576:dw|
Your part A is fine, its jsut part B
hmm...either it's wrong that i took 2.4 as the magnitude...what do you think?
The first time you used it is fine, but I'ven ever seen the 2nd time 2.4 wasused (usign ctrl-F)
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....
Vb here was 131.15s?
Sorry, Tb is 131.15s
125s was an acceptable value too for some reason.
nope.....hmm...am i overthinking this? for which reason?
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.
of course, 131.15s is the on on the official answer key
we can give it a rest now if you want.
alright. hopefully someone else will give it a go.

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