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

a) 10.29° upstream

b) t=338.7s

Step-by-step 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((0.5)/(1))=26.56°`

`D=√(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((V*cos\alpha )/(3) ) = 10.29°`

The amount of time:

`t = D/V = (1118m)/(3.3m/s) =338.7s`