Problema Solution
A boat capable of a speed of 6 miles per hour in still water attempts to go directly cross a river. As the boat crosses the river, it drifts 30 degrees from its intended path. How strong is the current? What is the effective speed of the boat?
Answer provided by our tutors
we suppose that the river is 1 mile across so that we have real numbers to manipulate
the boat crosses the river and drifts some distance 'x' along the opposite shore, making an angle of 30-degrees relative to the starting point
tan(30) = sqrt(3)/3. = x/1 = x
we let a vector of magnitude 6 represent the speed of the boat across the river with the current, then a vector of magnitutde 'x' along the far bank represents the speed of the current
these two vectors, 6 and x, comprise the hypotenuse and 'side opposite' of a right triangle and we with to solve for the magnitude of 'x' and the magnitude of the 'side adjacent'
we know that x/6 = sin(30)
therefore x = 6*sin(30) = 3
we let 'y' represent the 'side adjacent' and we also know that:
y/6 = cos(30)
y = 6*cos(30) = 5.196
the speed of the current is 3 mph, the effective speed of the boat is approximately 5.2 mph