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.

Question

nottim · Answer

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

nottim · Answer

I know that B takes 125s. 
CALC:
1. v=2.4-0.80=1.6m/s
2. t=200m/1.6m/s=125s

anonymous · Answer

note: 2.4 is speed relative to the earth?

nottim · Answer

nottim · Answer

@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 · Answer

can i use vectors to solve this?(it should be solved with vectors)

nottim · Answer

Wait....er. no. My friend's course probably doesn't use vectors.

anonymous · Answer

cos, the speed 2.4 is gonna have it's components...lol...without vectors....

nottim · Answer

well, go ahead and use vectors if you like. (is this vectors the basic kind, or from calculus and vectors)?

anonymous · Answer

|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