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## Homework Statement

A boat is crossing a river between via points A and B which are d=400m apart. The speed of the river is V

_{2}=2 m/s and is constant along the AB line. The angle of the boats trajectory is α=45 degrees. At what speed V

_{2}and with what angle β should the boat be directed so that it crosses the ABA path in t=4min? The angle β is the same in all directions of boats motion.

Picture: http://pokit.org/get/?c13e83eda46604a35787df84c222522d.jpg

## Homework Equations

[/B]

All kinematic. Equation for the roots of a quadratic polynomial. Basic trigonometry.

## The Attempt at a Solution

So first of all I set my coordinate plane so that the AB line lies on the x-axis. Then we can draw vector diagrams and deduce that if the boat is moving strictly on the AB line then it holds $$v_2sinα=v_1sinβ$$. The speed of the boat in the direction of AB is given by $$v_{AB}=v_2cosβ+v_1cosα$$ and the speed of the boat in the direction of BA is given by $$v_{BA}=v_1cosα-v_2cosβ$$. The given speed equations can be deduced from studying the following two diagrams:

http://pokit.org/get/?2751e09a819e7d327fd72e25fef6c7eb.jpg

The total time is given by $$t=t_1+t_2 =>\frac d {v_2cosβ+v_1cosα}+ \frac d {v_2cosβ-v_1cosα} = t$$ which after some algebra yields $$β=arccot(\frac {d+\sqrt {d^2+v_1^2t^2cos^2α}} {v_1tsinα})$$. Now the problem is that I obtain the value of $$β=20.16°$$ and then the boat speed from the equation $$v_2sinβ=v_1sinα$$ as $$v_2=4.10 m/s$$. However the answers given in the textbook are as follows $$β≈35° ~and~ v_2=2.49 m/s$$

**I do not see how were the results from the texbook obtained, so if you could check my results so I know if I've done it correctly.**Thanks.