A woman launches a boat from one shore of a straight river and wants to land at the point directly on the opposite shore. If the speed of the boat (relative to the water) is 12mi/h and the river is flowing east at the rate of 6mi/h, in what direction should she head the boat in order to arrive at the desired landing point?
To arrive at the desired landing point, the woman should head the boat in a direction that compensates for the effect of the river's flow.
Let's assume that the direction the boat should head in to offset the river's flow is θ degrees south of east.
The velocity of the boat relative to the ground can be represented by a vector addition of the velocity of the boat relative to the water and the velocity of the water current.
The velocity of the boat relative to the ground is equal to the vector sum of the velocity of the boat relative to the water and the velocity of the river current.
VBG = VBW + VRC
The magnitude of VBW can be found using the boat's speed relative to the water, which is given as 12 mi/h.
|VBW| = 12 mi/h
The magnitude of VRC can be found using the river's flow rate, which is given as 6 mi/h.
|VRC| = 6 mi/h
Since the boat needs to arrive directly across the river, the y-component of VBG should be zero.
VBGy = 0
Based on these considerations, the following equation can be set up:
VBW*sinθ + VRC*sin90 = 0
VBW*sinθ = -VRC
Simplifying:
12*sinθ = -6
sinθ = -6/12
sinθ = -0.5
Since the sin of an angle is negative in the third and fourth quadrants, θ must be in the third quadrant.
Using the inverse sine function to find the angle:
θ = invsin(-0.5)
θ ≈ -30.96 degrees
Therefore, the woman should head the boat in a direction approximately 30.96 degrees south of east to arrive at the desired landing point.
To determine the direction the woman should head the boat, we need to consider the velocity of the boat relative to the shore.
Let's break down the velocities involved:
1. Velocity of the boat relative to the water: 12 mi/h (in the direction the woman wants to go)
2. Velocity of the river: 6 mi/h (eastward)
Now, to determine the resultant velocity, we can use vector addition. The resultant velocity will point from the original position of the boat to its final position.
Since the vectors are at right angles to each other (one along the river and the other perpendicular to it), we can use the Pythagorean theorem to find the magnitude of the resultant velocity:
Resultant Velocity = sqrt((Velocity of the boat)^2 + (Velocity of the river)^2)
Resultant Velocity = sqrt((12 mi/h)^2 + (6 mi/h)^2)
= sqrt(144 + 36)
= sqrt(180)
≈ 13.42 mi/h
Now, we need to determine the direction in which the woman should head the boat. To do this, we can use trigonometry.
The tangent of the angle between the desired direction and the direction of the river is given by the ratio of the perpendicular velocity (12 mi/h) to the velocity along the river (6 mi/h):
Tangent (θ) = (Velocity of the boat relative to the water) / (Velocity of the river)
Tangent (θ) = 12 mi/h / 6 mi/h
= 2
Taking the inverse tangent of both sides, we find:
θ = arctan(2)
≈ 63.4 degrees
Therefore, the woman should head the boat at an angle of approximately 63.4 degrees (north of east) in order to arrive at the desired landing point directly opposite.