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 one year ago
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
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.

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NotTim
 one year ago
Best ResponseYou've already chosen the best response.0This is for a friend, so I don't have much work shown.

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0I know that B takes 125s. CALC: 1. v=2.40.80=1.6m/s 2. t=200m/1.6m/s=125s

Shadowys
 one year ago
Best ResponseYou've already chosen the best response.1note: 2.4 is speed relative to the earth?

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0https://fbcdnsphotosaa.akamaihd.net/hphotosakprn1/533567_154212501399299_1017958671_n.jpg https://fbcdnsphotosha.akamaihd.net/hphotosaksnc7/421635_154212521399297_25972720_n.jpg https://fbcdnsphotosfa.akamaihd.net/hphotosakprn1/76140_154212481399301_1414230843_n.jpg https://fbcdnsphotosda.akamaihd.net/hphotosakash4/387012_154212484732634_845307988_n.jpg https://fbcdnsphotosga.akamaihd.net/hphotosakash4/428325_154212474732635_582085865_n.jpg https://fbcdnsphotosba.akamaihd.net/hphotosakprn1/150879_154212514732631_1274474888_n.jpg https://fbcdnsphotosda.akamaihd.net/hphotosakprn1/554471_154214558065760_366131359_n.jpg https://fbcdnsphotosha.akamaihd.net/hphotosakash3/44769_154214568065759_779969484_n.jpg https://fbcdnsphotosfa.akamaihd.net/hphotosaksnc7/486007_154214574732425_1940920149_n.jpg https://fbcdnsphotosba.akamaihd.net/hphotosakash3/537274_154214594732423_1396018523_n.jpg https://fbcdnsphotosda.akamaihd.net/hphotosakprn1/45480_154216528065563_1014789527_n.jpg

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0@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.

Shadowys
 one year ago
Best ResponseYou've already chosen the best response.1can i use vectors to solve this?(it should be solved with vectors)

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0Wait....er. no. My friend's course probably doesn't use vectors.

Shadowys
 one year ago
Best ResponseYou've already chosen the best response.1cos, the speed 2.4 is gonna have it's components...lol...without vectors....

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0well, go ahead and use vectors if you like. (is this vectors the basic kind, or from calculus and vectors)?

Shadowys
 one year ago
Best ResponseYou've already chosen the best response.1dw: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=157.34s125s=32.35s

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0dw:1359314524260:dw yours is right. just missed a few lines

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0Your part A is fine, its jsut part B

Shadowys
 one year ago
Best ResponseYou've already chosen the best response.1hmm...either it's wrong that i took 2.4 as the magnitude...what do you think?

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0The first time you used it is fine, but I'ven ever seen the 2nd time 2.4 wasused (usign ctrlF)

Shadowys
 one year ago
Best ResponseYou've already chosen the best response.1oo, 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
 one year ago
Best ResponseYou've already chosen the best response.0125s was an acceptable value too for some reason.

Shadowys
 one year ago
Best ResponseYou've already chosen the best response.1nope.....hmm...am i overthinking this? for which reason?

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0125s 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
 one year ago
Best ResponseYou've already chosen the best response.0of course, 131.15s is the on on the official answer key

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0we can give it a rest now if you want.

Shadowys
 one year ago
Best ResponseYou've already chosen the best response.1alright. hopefully someone else will give it a go.
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