A straight river is 20m wide. The velocity of the river at (x,y) is v=[-(3x(20-x))/100]j m/min, 0<=x<=20. A boat leaves the shore at (0,0) and travels through the water with a constant velocity. It arrives at the opposite shore at (20,0). The speed of the boat is always radical 20 m/min.
Find the velocity of the boat.
Find the location of the boat at time t.
Please help! I am really stuck.
I multiplied dy/dt out to = q-[(60pt-3pt^2)/100]. Then I tried to integrate and got [(q^2)/2]-[((30pt^2)-(pt^3))/100] but wasn't able to get q=2 so I know I'm doing something wrong.
Would the position of the boat just be the position vector 4ti+2tj?
Im sorry, im still not sure how you were able to get q=2 with both q and p being unknown variables. As you can tell, I am very confused. Thank you for all of your help.
Nevermind! I got that part. So the position would be 4txi+2tyj? or just 4tx+2ty?
The velocity , which we now know is , is the velocity of the boat relative to the water. (See the start of my original post.) In other words, if we are in a boat ourselves that is not moving through the water - so to us, the water appears to be still - the first boat's velocity appears to be . Since the units here are m/min, that means that the first boat is moving at m/min faster across the river than we are, and m/min faster upstream (since and are unit vectors across and upstream, respectively).
Originally Posted by antman
But the water is, of course, moving relative to the bank of the river, and will carry the boat (and us, in our imaginary boat) with it. This is why we need the vector velocity equation
- velocity of boat relative to bank = velocity of boat relative to water + velocity of water relative to bank
to find the boat's movement relative to the bank. Now and are measured relative to the bank, and we have found that
So these equations, with and , give the location of the boat at time . If you like, you can express this as a single vector equation using , or