A boats speed is 1,70 m/s in still water.
It must cross a river with a width of 260 m.
The boats starting point is the origin on the xy-axsis (on the shore).
It has to dock 110 m to the right(in the positive x-direction) opposite of the starting point on the other shore(i.e. the point parallel to the starting point on the other side + 110 m).
The boat must sail in a 45° angle relative to the shore(x-axis) to arrive at that point.

What is the speed of the water current(water flows to the negative x-direction)?

Picture of the problem

What I have done:

It semms to be a pretty simple vector problem.
Just subtract the vector of the boat in moving water from the vector of the boat in still water(direct route) to get the vector of water flow.

I did this and got a nonzero y component of the water flow, which can't be true. How can it even be zero if only the sin(0°+180°*n)= 0 and the y components of the vectors aren't equal?

Thank you for your help


2 Answers 2


Let $v_b$ be the speed of boat in still water, $v_r$ is the speed of river. The speed of boat in $x$ direction (in still water at 45 degree angle): $v_x=v_b/\sqrt{2}$, the speed of boat in $y$ direction is $v_y=v_b/\sqrt{2}$. The river speed has negative $x$ direction as you correctly concluded. Thus, we need to subtract it from $v_x$. We have: $(v_b/\sqrt{2}-v_r)\cdot t=110$, $v_b/\sqrt{2} \cdot t=260$ (where $t$ is the time to cross the river). This will give us the equation to find $v_r$: $$\frac{v_b/\sqrt{2}}{v_b/\sqrt{2}-v_r}=\frac{260}{110}$$ Can you complete from here?

  • $\begingroup$ Where did you get sqrt(2) from? $\endgroup$
    – ΟΥΤΙΣ
    Apr 20, 2018 at 18:14
  • $\begingroup$ @ΟΥΤΙΣ: $\cos (\pi/4)=\frac{1}{\sqrt{2}}$ $\endgroup$
    – Vasili
    Apr 21, 2018 at 2:04
  • $\begingroup$ Thank you I see it now, because $\cos (\pi/4)=\frac{\sqrt{2}}{2}$ $\endgroup$
    – ΟΥΤΙΣ
    Apr 21, 2018 at 7:05
  • $\begingroup$ I've got (- 1,9) for an answer. Thank you very much for your help. I couldn't see beyond vectors addition, so I couldn't think of a different approach. $\endgroup$
    – ΟΥΤΙΣ
    Apr 21, 2018 at 7:08
  • $\begingroup$ @ΟΥΤΙΣ: Check your math, I am getting about -0.7. You can use vector approach too just remember that the magnitude of vectors will depend on time. $\endgroup$
    – Vasili
    Apr 21, 2018 at 14:25

Sailing on a still 'river', the boat would arrive to the opposide side at the point $260\,m$ across the river and $260\,m$ along the river, due to the angle of $45^\circ$.

What is the length of this route? What time would it take the boat to sail it?

You know the boat actually ends its travel $110\,m$ along the river. That means water shifted the boat by $260-110=150$ meters during the jorney (or $260+110=370$ meters, depending on the direction of the $45^\circ$ angle). Divide it by the travel time and you'll get the river speed.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .