Question

A boat that travels 3.00 m/s relative to the water is crossing a river that is 1.00 km wide. The destination on the far side of the river is 0.500 km downstream from the starting point. (a) If the river current is 2.00 m/s, in what direction should the boat be pointed in order to reach the destination? (b) How much time will the trip take?

Answer 1

Answer:

a) 10.29° upstream

b) t=338.7s

Explanation:

If the river is 1km wide and the destination point is 0.5km away downstream, then the angle and distance the the boat has to travel is:

$\alpha =atan(\frac{0.5}{1})=26.56°$

$D=\sqrt{1^2+0.5^2}=1.118km$

The realitve velocity of the boat respect to the water is:

$V_{B/W}=[3*cos\beta ,3*sin\beta ]$  where β is the angle it has to be pointed at.

From the relative mvement equations:

$V_{B/W}=V_B-V_W$  where $V_B=[V*cos\alpha ,-V*sin\alpha ]$

From this equation we get one equation per the x-axis and another for the y-axis. If we square each of them and add them together, we will get 2 equations:

$(3*cos\beta )^2+(3*sin\beta )^2=(V*cos\alpha )^2+(-V*sin\alpha +2)^2$

$V^2-4*V*sin\alpha -5=0$  Solving for V:

V = 3.3m/s   and   V=-1.514m/s   Replacing this value into one of our previous x or y-axis equations:

$\beta =acos(\frac{V*cos\alpha }{3} ) = 10.29°$

The amount of time:

$t = D/V = \frac{1118m}{3.3m/s} =338.7s$