Posted by Anonymous on Tuesday, February 19, 2008 at 10:29pm.
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 cross-stream 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.
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