How Are Velocity and Angle Calculated When Crossing a River?

[Adeopapposaurus]

Homework Statement

The speed of the boat on the water without stream is v_1 = 5 m/s. The water flows in the river parallel to the shore at a speed of v_2 = 3 m/s.
How should the boat be directed to cross the river in a direction perpendicular to the shores (What should be the angle between the boat and the shore)?
How long it takes for the boat across the river which is 80 meters wide?

Relevant Equations

arccos = (vr/vu)
a^2+b^2+c^2

v upstream = 5 - 3 =2
v of the water = 3
arccos = (2/3) = 48.19
v across the river = c^2 - a^2 = b^2
√3^2 - 2^2 = 6m/s
t = 80/6 = 13.3
can someone please point out my mistakes? am I even supposed to solve for angle like this or "perpendicular to the shores" just means that the angle is supposed to be 90 degrees?

 

[PeroK]

Homework Statement:: The speed of the boat on the water without stream is v_1 = 5 m/s. The water flows in the river parallel to the shore at a speed of v_2 = 3 m/s.
How should the boat be directed to cross the river in a direction perpendicular to the shores (What should be the angle between the boat and the shore)?
How long it takes for the boat across the river which is 80 meters wide?
Relevant Equations:: arccos = (vr/vu)
a^2+b^2+c^2

v upstream = 5 - 3 =2
v of the water = 3
arccos = (2/3) = 48.19
v across the river = c^2 - a^2 = b^2
√3^2 - 2^2 = 6m/s
t = 80/6 = 13.3
can someone please point out my mistakes? am I even supposed to solve for angle like this or "perpendicular to the shores" just means that the angle is supposed to be 90 degrees?

You need to explain to yourself and us what you are doing. In particular, you need a clear vector diagram of the motion of the boat relative to the shore.

A speed of $6m/s$ for motion across the river must be too high. It can't be greater than the $5m/s$ it would be without any current.

 

[Lnewqban]

Homework Statement:: The speed of the boat on the water without stream is v_1 = 5 m/s. The water flows in the river parallel to the shore at a speed of v_2 = 3 m/s.
...
v upstream = 5 - 3 =2
v of the water = 3

can someone please point out my mistakes? am I even supposed to solve for angle like this or "perpendicular to the shores" just means that the angle is supposed to be 90 degrees?

Do you know how to do a vector addition?
If not yet, please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html#vec1

Note that given $V_1$ is referenced to the water, while $V_2$ is referenced to the ground.
I believe that the problem asks about the trajectory corrections that the captain of that boat should do to travel between two docks, which are joined by a line that is perpendicular to the shores of the river.

For days when $V_2=0$, he does not need to correct that trajectory.
For days when water flows at $V_2=3 m/s$, what would you do to reach your destination and how long would the trip would take you, if you were that captain?

 

[haruspex]

v upstream = 5 - 3 =2
v of the water = 3
arccos = (2/3) = 48.19
v across the river = c^2 - a^2 = b^2
√3^2 - 2^2 = 6m/s
t = 80/6 = 13.3
can someone please point out my mistakes?

can someone please point out my mistakes?
I count five.

v upstream = 5 - 3 =2
Velocity of what upstream? If the boat points directly upstream then its velocity relative to the bank will be 2m/s, but it will never make progress across the river.

arccos = (2/3)
What angle is this supposed to be? You seem to be imagining a right angled triangle with hypotenuse 3 and a side 2, but the given 3 and calculated 2 are in the same direction.

√3^2 - 2^2 = 6m/s
where does the √3 come from?

What are the values of √3^2 and 2^2 ? What do you get from that subtraction?

Your equation said c^2 - a^2 = b^2, not c^2 - a^2 = b. So what did you forget to do?

Draw a diagram with a vector representing the current and a vector representing the boat's velocity relative to the current. This should point at some angle upstream. The sum of these two will be in the actual direction the boat takes. What direction does that need to be?