Vanessa can swim 3.0 m/s in still water. While trying to swim directly across a river from west to east, Vanessa is pulled by a current flowing southward at 2.0 m/s.
A) What is the magnitude of Vanessa’s resultant velocity?
B) If Vanessa wants to end up exactly across stream from where she began, at what angle to the shore must she swim upstream?
A) √(3^2+2^2) = √13
B) Draw a diagram. cosθ = 2/3
a. Vr = sqrt(3^2+2^2) = 3.61 m/s.
b. Tan A = Y/X = Vc/Vs = 2/3 = 0.66666.
A = 33.7o S. of E. due to the current.
Direction = 33.7o N. of E. to offset the current.
Vc = Velocity of current.
Vs = Velocity of swimmer.
To determine the magnitude of Vanessa's resultant velocity, we can use the Pythagorean theorem. The resultant velocity is the vector sum of her velocity in still water and the velocity of the current. Let's denote Vanessa's velocity in still water as V and the velocity of the current as C.
A) Magnitude of Vanessa's resultant velocity:
The magnitude of the resultant velocity (R) can be calculated using the Pythagorean theorem:
R = √(V^2 + C^2)
Given that Vanessa's velocity in still water (V) is 3.0 m/s and the velocity of the current (C) is 2.0 m/s, we can substitute these values into the equation:
R = √(3.0^2 + 2.0^2)
R = √(9.0 + 4.0)
R = √13
R ≈ 3.61 m/s
Therefore, the magnitude of Vanessa's resultant velocity is approximately 3.61 m/s.
B) To determine the angle Vanessa must swim upstream to end up exactly across the stream from where she began, we can use trigonometry. Let's denote the angle she needs to swim as θ.
Since Vanessa wants to end up exactly across the stream from her starting point, she needs to swim perpendicular to the current. Therefore, the angle θ is the complement of the angle formed between the resultant velocity and the velocity of the current. Let's denote this angle as α.
The tangent of α can be calculated as:
tan(α) = (V/C)
Given that Vanessa's velocity in still water (V) is 3.0 m/s and the velocity of the current (C) is 2.0 m/s, we can substitute these values into the equation:
tan(α) = (3.0/2.0)
tan(α) = 1.5
To find the angle α, we can take the inverse tangent (arctan) of both sides:
α = arctan(1.5)
α ≈ 56.31°
Since θ is the complement of α, we can calculate it by subtracting α from 90°:
θ = 90° - α
θ ≈ 90° - 56.31°
θ ≈ 33.69°
Therefore, Vanessa must swim at an angle of approximately 33.69° upstream to end up exactly across the stream from where she began.
To solve this problem, we can use vector addition to find the resultant velocity of Vanessa.
Let's consider the north direction as positive and use a coordinate system. We can break down Vanessa's velocity and the current velocity into their respective components as follows:
Vanessa's velocity (Vv) = 3.0 m/s (east)
Current velocity (Vc) = 2.0 m/s (south)
Now, let's determine the magnitude and direction of the resultant velocity:
A) Magnitude of Vanessa's Resultant Velocity:
The resultant velocity, Vr, can be found by adding Vanessa's velocity vector (Vv) and the current velocity vector (Vc):
Vr = Vv + Vc
To find the magnitude of Vr, we can use the Pythagorean theorem:
|Vr|^2 = |Vv|^2 + |Vc|^2
|Vr|^2 = (3.0 m/s)^2 + (2.0 m/s)^2
|Vr|^2 = 9.0 m^2/s^2 + 4.0 m^2/s^2
|Vr|^2 = 13.0 m^2/s^2
Taking the square root of both sides, we find:
|Vr| ≈ √(13 m^2/s^2)
|Vr| ≈ 3.61 m/s
Therefore, the magnitude of Vanessa's resultant velocity is approximately 3.61 m/s.
B) Angle to Swim Upstream:
To determine the angle at which Vanessa must swim upstream to end up exactly across the stream from where she began, we can use trigonometry.
Let's call the angle Vanessa needs to swim upstream θ. Since Vanessa wants to swim directly across the stream, the angle between the resultant velocity (Vr) and the north direction is also θ.
Using the components of Vanessa's resultant velocity, we can calculate θ as follows:
θ = tan^(-1)(opposite/adjacent)
θ = tan^(-1)(Vv/Vc)
Substituting the values we know:
θ = tan^(-1)(3.0 m/s / 2.0 m/s)
θ ≈ tan^(-1)(1.5)
Using a calculator or trigonometric tables, we find:
θ ≈ 56.31 degrees
Therefore, Vanessa must swim upstream at an angle of approximately 56.31 degrees to the shore in order to end up exactly across the stream from where she began.