Since the both goes straight across its 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!), m/min. So you know the velocity vector is and you must haveThe 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.
Solve that for .