a ferry boat is traveling in a direction 28 degrees north of east with a speed of 4.40 m/s relative to the water. A passenger is walking with a velocity of 2.02 m/s due east relative to the boat. What is (A) the magnitude and (B) the direction of the velocity of the passenger with respect to the water?
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(A) The magnitude of the velocity of the passenger with respect to the water can be found using the Pythagorean theorem. We have the horizontal component due to the boat's velocity (4.40 m/s) and the horizontal component due to the passenger's velocity (2.02 m/s). So, using the Pythagorean theorem:
Magnitude = √(4.40^2 + 2.02^2)
Magnitude = √(19.36 + 4.08)
Magnitude ≈ √23.44
Magnitude ≈ 4.84 m/s
So, the magnitude of the velocity of the passenger with respect to the water is approximately 4.84 m/s.
(B) To find the direction of the velocity of the passenger with respect to the water, we can use trigonometry. The direction is given as 28 degrees north of east. Considering east as the positive x-axis and north as the positive y-axis, we can break down the velocity of the passenger into its x (east) and y (north) components.
The x-component can be found using the formula:
X-component = Magnitude * cos(θ)
X-component = 4.84 m/s * cos(28°)
X-component ≈ 4.84 m/s * 0.881
The y-component can be found using the formula:
Y-component = Magnitude * sin(θ)
Y-component = 4.84 m/s * sin(28°)
Y-component ≈ 4.84 m/s * 0.471
So the velocity is approximately (4.84 m/s * 0.881) east and (4.84 m/s * 0.471) north.
To find the direction, we can use the arctangent of the y-component over the x-component:
Direction = arctan(Y-component / X-component)
Direction = arctan((4.84 m/s * 0.471) / (4.84 m/s * 0.881))
Direction = arctan(0.471 / 0.881)
Direction ≈ 0.498
So, the direction of the velocity of the passenger with respect to the water is approximately 0.498 radians.
To determine the magnitude and direction of the velocity of the passenger with respect to the water, we can use vector addition.
Let's break down the given information into components:
Ferry boat velocity relative to the water:
Magnitude (Vb) = 4.40 m/s
Direction (28 degrees north of east)
Passenger's velocity relative to the boat:
Magnitude (Vp) = 2.02 m/s
Direction (due east)
Using trigonometry, we can find the x and y components of both velocities.
Ferry boat velocity components (Vbx and Vby):
Vbx = Vb * cos(28°)
Vby = Vb * sin(28°)
Vbx = 4.40 m/s * cos(28°) ≈ 3.946 m/s
Vby = 4.40 m/s * sin(28°) ≈ 2.022 m/s
Passenger's velocity components (Vpx and Vpy):
Vpx = Vp * cos(0°)
Vpy = Vp * sin(0°)
Vpx = 2.02 m/s * cos(0°) ≈ 2.02 m/s
Vpy = 2.02 m/s * sin(0°) ≈ 0 m/s
Now, we can find the total velocity of the passenger with respect to the water by adding the x and y components.
Vx = Vbx + Vpx
Vy = Vby + Vpy
Vx = 3.946 m/s + 2.02 m/s ≈ 5.966 m/s
Vy = 2.022 m/s + 0 m/s ≈ 2.022 m/s
To find the magnitude of the velocity (A), use the Pythagorean theorem:
A = √(Vx^2 + Vy^2)
A ≈ √(5.966 m/s)^2 + (2.022 m/s)^2
A ≈ √35.648 m^2/s^2 + 4.088 m^2/s^2
A ≈ √39.736 m^2/s^2
A ≈ 6.306 m/s
Therefore, the magnitude of the velocity of the passenger with respect to the water is approximately 6.306 m/s.
To find the direction (B) of the velocity, use the inverse tangent function:
B = atan(Vy / Vx)
B = atan(2.022 m/s / 5.966 m/s)
B ≈ atan(0.338)
B ≈ 18.95°
Therefore, the direction of the velocity of the passenger with respect to the water is approximately 18.95 degrees.
To find the magnitude and direction of the velocity of the passenger with respect to the water, we can use vector addition.
Let's denote the velocity of the ferry boat with respect to the water as Vboat and the velocity of the passenger with respect to the boat as Vpassenger.
Given:
- The magnitude of Vboat is 4.40 m/s.
- The angle between the direction of Vboat and due east is 28 degrees.
- The magnitude of Vpassenger is 2.02 m/s in the due east direction.
To find the magnitude and direction of Vpassenger with respect to the water, we need to add the vectors Vboat and Vpassenger.
Step 1: Resolve Vboat into its northward and eastward components.
- The northward component of Vboat (Vnorth) can be found using trigonometry:
Vnorth = Vboat * sin(angle)
Vnorth = 4.40 m/s * sin(28 degrees)
Vnorth ≈ 2.18 m/s (rounded to two decimal places)
- The eastward component of Vboat (Veast) can also be found using trigonometry:
Veast = Vboat * cos(angle)
Veast = 4.40 m/s * cos(28 degrees)
Veast ≈ 3.92 m/s (rounded to two decimal places)
Step 2: Add the eastward components and northward components separately:
- The eastward component of the total velocity (Vtotal_east) is the sum of Veast and Vpassenger.
Vtotal_east = Veast + Vpassenger
Vtotal_east = 3.92 m/s + 2.02 m/s
Vtotal_east ≈ 5.94 m/s (rounded to two decimal places)
- The northward component of the total velocity (Vtotal_north) remains the same as Vnorth.
Vtotal_north = Vnorth
Vtotal_north ≈ 2.18 m/s (rounded to two decimal places)
Step 3: Use the components to find the magnitude and direction of the total velocity with respect to the water.
- The magnitude (Vtotal) can be found using Pythagoras' theorem:
Vtotal = sqrt(Vtotal_east^2 + Vtotal_north^2)
Vtotal = sqrt((5.94 m/s)^2 + (2.18 m/s)^2)
Vtotal ≈ 6.41 m/s (rounded to two decimal places)
- The direction (θ) can be found using tan^-1(Vtotal_north / Vtotal_east):
θ = tan^-1(Vtotal_north / Vtotal_east)
θ = tan^-1(2.18 m/s / 5.94 m/s)
θ ≈ 20.13 degrees (rounded to two decimal places)
Therefore, the (A) magnitude and (B) direction of the velocity of the passenger with respect to the water are approximately 6.41 m/s and 20.13 degrees, respectively.