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November 24, 2014

November 24, 2014

Posted by **Anonymous** on Tuesday, February 19, 2008 at 10:29pm.

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?

- Math: Calculus - Vectors -
**drwls**, Tuesday, February 19, 2008 at 11:14pmHow 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.

- Math: Calculus - Vectors -
**Quidditch**, Tuesday, February 19, 2008 at 11:14pmIf 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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