# Thread: [SOLVED] boat travels accross a river...

1. ## [SOLVED] boat travels accross a river...

A guy traveling 5m/s accross a 40m wide river (banks are parallel). The water speed is fastest in the middle. The speed for the water is given as: (3/400)(x)(40-x) where x is the meters from shore. (for ease of communication assue the river goes North and the guy is traveling West to East).

1) How far down does he travel?
2) At what angle South (upstream) does he need to travel to end up directly accross the river from where he started.

My work: It takes him 8 seconds to cross the river because his speed is constant. From here I am not sure where to begin. I am pretty sure that there is an integral involved to "add up" all the downward drif but i cannot get it to work out.
As for the second part this should (i think) be easy because given that it is 40 meter accross and "Z" feet downstream is what he ended up. I can just use tangent to figure out that angle and say to go upstream.

Thanks....and sorry for spelling errors it clears my text as i found out

2. Originally Posted by snaes
A guy traveling 5m/s accross a 40m wide river (banks are parallel). The water speed is fastest in the middle. The speed for the water is given as: (3/400)(x)(40-x) where x is the meters from shore. (for ease of communication assue the river goes North and the guy is traveling West to East).

1) How far down does he travel?
2) At what angle South (upstream) does he need to travel to end up directly accross the river from where he started.

My work: It takes him 8 seconds to cross the river because his speed is constant. From here I am not sure where to begin. I am pretty sure that there is an integral involved to "add up" all the downward drif but i cannot get it to work out.
As for the second part this should (i think) be easy because given that it is 40 meter accross and "Z" feet downstream is what he ended up. I can just use tangent to figure out that angle and say to go upstream.
$v_x= 5$ m/s , so $x = 5t$

sub $5t$ for $x$ in the water speed expression ...

$v_y = \frac{3(5t)}{400}(40-5t)$

to get the distance downstream ...

$\int_0^8 v_y \, dt$

for the second part, I don't think the angle steered upstream is a constant ... I'll have to think on that.

3. Thanks that makes sense now. I actually treid that in my calculator only typed in wrong . Thanks, the explanation helped my understanding more though.

For number 2) if this helps you think: (I do the homework for comprehension not for a grade so take your time...)

Answer in back:It's approxonately 23.6 deg upstream and it shows a graph that has has xmin0 to xmax40 then line goes down that back up (like a -sin(x) but with period of 40) symetrically along the x axis. It has a zero at 20 and probably (just guessing by looking at graph) has a max of about 2. So the equation has to be somewhat close to -2sin[.05*pi*x].

You are correct that the angle would not be constant: Because the inverse tangent of (16/40)=21.8deg which obvioulsy is close but wrong.