A boat is rowed 6.4 km up a river and back again and this takes in total 2 hours. The stream velocity is 2.4 km/h. What velocity would the boat have been moving in if the water was standstill?

I started by drawing a picture of the two directions (using paranthesis instead of arrows to indicate direction):

(---  v(current)
----------------------

----------------------
---)  v(boat)

|---------------------|
6.4 km

(---  v(current)
----------------------

----------------------
(---  v(boat)

|---------------------|
6.4 km


$s_{tot} = 6.4 \cdot 2 = 12.8$ km

$t_{tot} = 2$ h

$v_{avg} = \frac{12.8}{2} = 6.4$

$v_{up} = v_{boat} - v_{current}$

$v_{down} = v_{boat} + v_{current}$

Getting an expression for the average based on velocity up and down.

$v_{avg} = \frac{v_{up} + v_{down}}{2}$

Adding the two equations for up and down velocity:

$v_{up} + v_{down} = v_{boat} - v_{current} + (v_{boat} + v_{current}) = 2v_{boat}$

This would indicate that the velocity without the current is the same as the average velocity with the current. But the answer is 7.2 km/h, so there are probably multiple errors here and the stream velocity has not really been used.

I suspect I have to weigh the velocity up and down with their respective times since it takes longer to go up the river (and the vector addition does not seem quite correct), but not sure how to proceed. The question also appears somewhat equivocal since it is not clear if 6.4 km is one way or two way distance.

What are some more productive approaches to this question?

• if the water was at a standstill, all of the water would be in the ocean, in which case, there wouldn't be any water in the stream, thus the velocity of the boat would be zero, unless the boat was in the ocean, but how would the boat get to the ocean with no water in the stream? – John Joy Apr 26 '15 at 17:51

Consider the two legs of the trip. $$v_{\text{down}}t_{\text{down}} = d_{\text{down}}= 6.4\text{ km}= d_{\text{up}} = v_{\text{up}}t_{\text{up}}$$ As you pointed out:
$v_{\text{up}} = v_{\text{boat}} - v_{\text{river}}, \ v_{\text{down}} = v_{\text{boat}} + v_{\text{river}}$ and we are given that $t_{\text{up}} + t_{\text{down}} = 2\text{ hours} = T$.
$$d = v_{\text{down}}t_{\text{down}} = v_{\text{up}}t_{\text{up}} \implies$$ $$T = t_{\text{up}} + t_{\text{down}} = d\cdot\bigg(\frac1{v_{\text{down}}} + \frac1{v_{\text{up}}}\bigg) = \frac{2d\cdot v_{\text{boat}}}{v_{\text{boat}}^2 - v_{\text{river}}^2}$$
$$Tv_{\text{boat}}^2 - 2d\cdot v_{\text{boat}} - T\cdot v_{\text{river}}^2 = 0$$ $$\implies v_{\text{boat}} = \frac{2d + \sqrt{4d^2 + 4T^2v^2_{\text{river}}}}{2T} = \frac{d}{T} + \sqrt{\bigg(\frac{d}{T}\bigg)^2 + v_{\text{river}}^2}$$