Math: Calculus  Vectors
posted by Anonymous .
A boat travels at a speed of 5 m/s in still water. The boat moves directly across a river that is 70m wide. The water in the river flows at a speed of 2 m/s. How long does it take the boat to cross the river? In what direction is the boat headed when it starts the crossing.
Textbook Answer: 15.3 m/s, 66.4 degrees to shore
My answer: 14 seconds and 22 degrees from the shore.
This is what I did: I found the speed of the resultant by doing the Pythagorean (5.4 m/s),then found the distance of the resultant by "5/70 = 2/x (28m)" and then using the Pythagorean (75.4m), and then I found the time by doing "t = d/s". Then, I found the direction of the boat by using trig, "tanx = 2/5"
MY QUESTION: What did I do wrong? And can you please help me fix my mistake?

How can the textbook answer be 15.3 m/s if they are asking for the crossing time?
To go directly across, the boat must aim upstream so that the velocity component upstream relative to the water is 2 m/s. That makes its velocity component across the water sqrt[5^2  2^2) = sqrt 21 = 4.58 m/s. The crossing time is
70 m/4.58 m/s = 15.3 s. The pointing angle of the boat is cos^1 4.58/5 = 23.7 degrees relative to the crossstream direction, or 66.3 degrees relative to the shore. 
If the boat moves directly across then there is enough of the boats speed vector to counteract the river flow. That is 2m/s.
So, the vector parallel to the shore (the river flow) speed is 2m/s. The TOTAL speed of the boat is 5m/s. Now solve this for the speed perpendicular to the shore.
SQRT(5^2  2^2).
Divide that into the distance (70m).
Use any of the trig values to find the angle.