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.
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?
2 answers
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.
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.
1 answer