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

NotTim
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0note: 2.4 is speed relative to the earth?

NotTim
 3 years 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
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can i use vectors to solve this?(it should be solved with vectors)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0cos, the speed 2.4 is gonna have it's components...lol...without vectors....

NotTim
 3 years 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)?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw: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
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359314524260:dw yours is right. just missed a few lines

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmm...either it's wrong that i took 2.4 as the magnitude...what do you think?

NotTim
 3 years 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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oo, 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
 3 years ago
Best ResponseYou've already chosen the best response.0125s was an acceptable value too for some reason.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0nope.....hmm...am i overthinking this? for which reason?

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0alright. hopefully someone else will give it a go.
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