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Can you please check my solution of this kinematics problem?

  1. Jun 28, 2016 #1
    1. The problem statement, all variables and given/known data
    A boat is crossing a river between via points A and B which are d=400m apart. The speed of the river is V2=2 m/s and is constant along the AB line. The angle of the boats trajectory is α=45 degrees. At what speed V2 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

    2. Relevant equations

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

    3. 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.
     
  2. jcsd
  3. Jun 28, 2016 #2

    Simon Bridge

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    Try checking by another method: ie. use the river bank as your x coordinate.
    Or maybe you can use your two diagrams to sum the velocities head-to-tail, making two triangles. You can then use the sine and cosine rules to solve for the sides using the information you have about the magnitude and direction of the final velocities ... perhaps.

    Note:
    Note: in the diagram, v1 is the river velocity, which makes angle alpha to the AB line which is your x axis.
    So you appear to have the angles backwards here: ##v_1\sin\alpha = v_2\sin\beta##
    You realise of course that: ##\sin\alpha = \cos\alpha = 1/\sqrt{2}##
    ... but do we take ##|AB|=d## or ##|AB| = d\sqrt{2}##? ie. is "d" measured along the bank or directly from A to B?
     
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