NotTim
  • NotTim
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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
NotTim
  • NotTim
This is for a friend, so I don't have much work shown.
NotTim
  • NotTim
I know that B takes 125s. CALC: 1. v=2.4-0.80=1.6m/s 2. t=200m/1.6m/s=125s
anonymous
  • anonymous
note: 2.4 is speed relative to the earth?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

NotTim
  • NotTim
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
NotTim
  • NotTim
@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.
anonymous
  • anonymous
can i use vectors to solve this?(it should be solved with vectors)
NotTim
  • NotTim
Wait....er. no. My friend's course probably doesn't use vectors.
anonymous
  • anonymous
cos, the speed 2.4 is gonna have it's components...lol...without vectors....
NotTim
  • NotTim
well, go ahead and use vectors if you like. (is this vectors the basic kind, or from calculus and vectors)?
anonymous
  • anonymous
|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
NotTim
  • NotTim
answer key says 6.1s.
anonymous
  • anonymous
is my diagram wrong?
NotTim
  • NotTim
|dw:1359314524260:dw| yours is right. just missed a few lines
NotTim
  • NotTim
|dw:1359314590576:dw|
NotTim
  • NotTim
Your part A is fine, its jsut part B
anonymous
  • anonymous
hmm...either it's wrong that i took 2.4 as the magnitude...what do you think?
NotTim
  • NotTim
The first time you used it is fine, but I'ven ever seen the 2nd time 2.4 wasused (usign ctrl-F)
anonymous
  • anonymous
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....
NotTim
  • NotTim
Vb here was 131.15s?
NotTim
  • NotTim
Sorry, Tb is 131.15s
NotTim
  • NotTim
125s was an acceptable value too for some reason.
anonymous
  • anonymous
nope.....hmm...am i overthinking this? for which reason?
NotTim
  • NotTim
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.
NotTim
  • NotTim
of course, 131.15s is the on on the official answer key
NotTim
  • NotTim
we can give it a rest now if you want.
anonymous
  • anonymous
alright. hopefully someone else will give it a go.

Looking for something else?

Not the answer you are looking for? Search for more explanations.