A women 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 is 10 mi/h and the river is flowing east at the rate of 5 mi/h, in what direction should she head the boat in order to arrive at the desired landing point?
the boat needs to follow a course that results in a 5 mph upstream component
5 = 10 cos(Θ) ... the cosine of 60º is 1/2
the boat should aim upstream at an angle of 60º to the shore
Did you make your diagram?
On mine:
sinØ = 5/10
Ø = 30°
so aim for N 30° W, assuming she want to head directly north.
To determine the direction the woman should head the boat, we need to consider the effect of both the boat's speed and the river's flow.
1. Calculate the actual speed of the boat relative to the ground:
The boat's speed is 10 mi/h, and the river is flowing at a rate of 5 mi/h in the east direction. Since the river's flow is in the opposite direction of the desired landing point, it will have an impact on the boat's motion. The resulting velocity can be calculated using vector addition.
The actual speed of the boat relative to the ground can be determined using the Pythagorean theorem. The boat's speed is the hypotenuse of a right triangle, with the components being its actual speed relative to the ground (unknown) and the river's velocity (5 mi/h).
Let's call the actual speed of the boat "Vb" and the river's velocity as "Vr". Then the equation becomes:
Vb^2 = (Vr^2) + (Boat's speed)^2
Substituting the values given:
Vb^2 = (5 mi/h)^2 + (10 mi/h)^2
Vb^2 = 25 mi^2/h^2 + 100 mi^2/h^2
Vb^2 = 125 mi^2/h^2
Taking the square root of both sides:
Vb = √125 mi/h
Vb ≈ 11.18 mi/h
2. Determine the angle of the boat's heading:
The woman needs to head the boat in a direction that ensures it will travel directly across the river and land on the opposite shore. Since the river is flowing to the east and the boat wants to reach the opposite shore (west), the angle of the boat's heading needs to compensate for the river's flow.
To find this angle, we can use trigonometry. Let's call this angle "θ".
tan(θ) = (River's velocity)/(Boat's speed)
tan(θ) = 5 mi/h / 10 mi/h
tan(θ) = 0.5
θ = tan^(-1)(0.5)
θ ≈ 26.57 degrees
Therefore, the woman should head the boat at an angle of approximately 26.57 degrees west of due north in order to arrive at the desired landing point.
To determine the direction the woman should head the boat, we need to consider the effect of the river's flow on the boat's path.
Let's break down the velocities involved:
- Velocity of the boat with respect to the ground (VBG) = 10 mi/h (forward)
- Velocity of the river flow (VRF) = 5 mi/h (eastward)
To calculate the resultant velocity, we can use vector addition. Since the velocities are at right angles to each other, we can treat them as perpendicular vectors in a right triangle.
The resultant velocity (VR) will be the diagonal connecting the boat's velocity and the river's velocity in the triangle:
|
|
VR |
|\
| \ VBG (10 mi/h forward)
|__\
VRF (5 mi/h eastward)
Using the Pythagorean theorem, we can calculate VR:
VR^2 = VBG^2 + VRF^2
VR^2 = 10^2 + 5^2
VR^2 = 100 + 25
VR^2 = 125
VR = √125
VR ≈ 11.18 mi/h
The magnitude of the resultant velocity (VR ≈ 11.18 mi/h) tells us that the boat will move at an average speed of approximately 11.18 mi/h relative to the ground.
Now, to determine the direction she should head the boat, we can use trigonometry. Let's define θ as the angle between the boat's velocity (VBG) and the resultant velocity (VR):
sin(θ) = VRF / VR
sin(θ) = 5 / 11.18
Using inverse sine (sin⁻¹) on both sides:
θ ≈ sin⁻¹(5 / 11.18)
θ ≈ 26.5°
Therefore, the woman should head the boat at an angle of approximately 26.5° east of straight across the river.