1. ## Velocity of boat

A straight river is 20 m 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.

a. Find the velocity of the boat.
b. Find the location of the boat at time t.
c. Sketch the path of the boat.

I figured the velocity of the boat, since traveling in the positive direction on the x axis, would also be radical 20 m/min. The boat seems to cross the river and end up DIRECTLY across, so I tried to set up a right triangle where the hypotenuse represented the boats velocity (radical 20) and the vertical leg represented the river velocity (-40 m/min? calculated using the integral of the equation from 1 to 20?). I'm not sure what I'm doing and an totally stuck. Thank you.

2. Originally Posted by jennifer1004
A straight river is 20 m 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.

a. Find the velocity of the boat.
b. Find the location of the boat at time t.
c. Sketch the path of the boat.

I figured the velocity of the boat, since traveling in the positive direction on the x axis, would also be radical 20 m/min.
Do you understand the difference between "velocity" and "speed"? The velocity is a vector while speed is a number. So the velocity CANNOT be "radical 20 m/min", that's not a vector. The speed is the magnitude of velocity vector. If the velocity at any point is $v_x\vec{i}+ v_y\vec{j}$, then the speed is $\sqrt{v_x^2+ v_y^2}$.

The boat seems to cross the river and end up DIRECTLY across, so I tried to set up a right triangle where the hypotenuse represented the boats velocity (radical 20) and the vertical leg represented the river velocity (-40 m/min? calculated using the integral of the equation from 1 to 20?). I'm not sure what I'm doing and an totally stuck. Thank you.
Since the both goes straight across its $\vec{j}$ component of velocity must oppose the velocity of the river (which is NOT -40 m/min. The river speed is dy/dt- integrating (dy/dt)dx doesn't give anything meaningful!), $-\frac{3x(20-x)}{100}\vec{j}$ m/min. So you know the velocity vector is $v_x\vec{i}+\frac{3x(20-x)}{100}\vec{j}$ and you must have $\sqrt{v_x^2+ \left(\frac{3x(20-x)}{100}\right)^2}= 20$
Solve that for $v_x$.

3. Would $v_x=20$ if x=0? $v_x=4$ (rather than -4) if x=20? or just $20-[(60x-3x^2)/100]$?