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Hi!
I'd love for some help on this vector word problem I'm confused on:
"A river has a constant current of 3 km/h due south. At what angle to a boat dock should a motorboat,
capable of maintaining a constant speed of 20 km/h, be headed in order to reach a point directly opposite the dock?
If the river is ½ a km wide, how long will it take to cross?"
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Make a sketch.
Let A be the point on a bank of the river, where motorboat starts.
Let B be the point on the other bank of the river, right opposite to the point A
(I assume that the banks of the river are parallel lines: quite natural assumption).
Let C be the target point on the opposite bank of the river: the captain HEADS the boat to the point C.
So, you have the right angled triangle ABC with the legs AB (across the river)
and CB (along the opposite bank of the river).
Let t be the time (in hours) to get the opposite bank.
+-----------------------------------------------------------------------+
| When crossing the river, the boat participates in TWO MOVES: |
| it is headed to the point C; |
| at the same time the river flow drifts the boat along the river. |
+-----------------------------------------------------------------------+
In other words, the boat is headed along the hypotenuse AC;
at the same time the river flow drifts the boat along leg CB.
Let "a" be the angle CAB.
During the time t, the boat crosses river, having the speed 20 km/h relative the water. It means that
20*cos(a)*t = 0.5 mile (the move across the river) (1)
During the same time, the river flow drifts the boat at the distance 20*sin(a)*t,
and this distance (the leg CB) is 3*t, due to the river flow speed.
It gives you the second equation
20*sin(a)*t = 3t (the move along the river) (2)
Thus you have these two "governing equations"
20*cos(a)*t = 0.5 (1)
20*sin(a)*t = 3t (2)
Now we should solve this system of equations.
For it, cancel "t" in both sides of the equation (2). You will get
20*sin(a) = 3, or sin(a) =
It determines the angle "a" : a = = 0.150568 radians = 8.63 degrees (approximately).
and the first question of the problem is just ANSWERED.
Then cos(a) = = = = = 0.9886 (approximately).
Now you can determine the time required to cross over the river from equation (1)
t = = = 0.02529 of an hour = 1.517 minute = 1 minute 31 seconds.
The problem is just SOLVED.
You got from me not only the solution, but the standard mantra to pronounce when solving such problems.
It is one of good standard classical problems for students learning Physics (Mechanics).
It is a University Physics level, or a renowned College Physics level, or an advanced high school level.
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Do not forget to post your deepest "THANKS" to me for my teaching.
. . . . . . . . .
One wise teacher (old professor) said once:
"Good problems for students are the same as tale stories for children:
from them, the students learn about the world."
This problem is exactly of that category.