4. The boat cross the river with constant velocity with respect to the water, v=0.3 m/s. The boat’s velocity is directed perpendicular to the shores of the river. River width b=63 m. River flow is not constant and changes across the river as a parabolic function u = uo - ((4uo)/(b^2))(x-b/2)^2 where x – distance from the shore, uo =5 m/s. Find the drift s of the boat along the river when it reaches the opposite shore.

Question

4.	The boat cross the river with constant velocity with respect to the water, v=0.3 m/s. The boat’s velocity is directed perpendicular to the shores of the river. River width b=63 m. River flow is not constant and changes across the river as a parabolic function  u = uo - ((4uo)/(b^2))(x-b/2)^2 where x – distance from the shore, uo =5 m/s. Find the drift s of the boat along the river when it reaches the opposite shore.

anonymous · Answer

please someone help me!! 
the attached file is the pic!

queelius · Answer

Ok, it's moving at a constant velocity v = 3/10 m/s toward the shore across the river, and at a varying velocity (river flow gradient) u perpendicular to the shore.

So, how long does it take to cross the river? This question only concerns v = 3/10, so b = vt, t = b/v, it takes time b/v to cross the river. Let's look at u(x)... let's put this in terms of u(t) instead to help us integrate later on. x = vt, so replace the x in u with vt. Now we're almost done!

Let's think about this discretely first. We know that for some small time slice, dt, it will move u(vt) * dt. So, as dt -> 0, we get the integral. Let's integrate u(t) from 0 to b/v. Done.