a ferryboat is travelling in a direction of 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 2.20 m/s due east relative to the boat. What is the magnitude and direction of the velocity of the passenger with respect to the water. Give directional Angle to due east.
boat north rel to water = 4.4 sin 28
boat east rel to water = 4.4 cos 28
pass north rel to water = 4.4 sin 28 same as boat = 2.07
pass east rel to water = 4.4 cos 28 + 2.2 = 3.88 + 2.2 = 6.08
mag = sqrt (2.07^2 + 6.08^2)
tan angle N of E = 2.07/6.08
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 velocities into their x and y components:
Velocity of the ferryboat relative to the water:
Vf_x = 4.40 m/s * cos(28°) (east)
Vf_y = 4.40 m/s * sin(28°) (north)
Velocity of the passenger relative to the boat:
Vp_x = 2.20 m/s (east)
Vp_y = 0 m/s (north)
Now, we can add the components to find the resultant velocity of the passenger relative to the water:
Vr_x = Vf_x + Vp_x
Vr_y = Vf_y + Vp_y
Vr_x = 4.40 m/s * cos(28°) + 2.20 m/s
Vr_y = 4.40 m/s * sin(28°) + 0 m/s
Calculating the components:
Vr_x = 3.988 m/s (east)
Vr_y = 2.042 m/s (north)
To find the magnitude and direction of the resultant velocity, we can use the Pythagorean theorem and inverse tangent functions:
Magnitude of the velocity:
Vr = sqrt(Vr_x^2 + Vr_y^2)
Direction of the velocity (angle with respect to due east):
θ = atan2(Vr_y, Vr_x)
Calculating the magnitude and direction:
Vr = sqrt((3.988 m/s)^2 + (2.042 m/s)^2)
θ = atan2(2.042 m/s, 3.988 m/s)
After calculating:
Vr ≈ 4.53 m/s (magnitude)
θ ≈ 27.13° (angle)
Therefore, the magnitude of the velocity of the passenger with respect to the water is approximately 4.53 m/s, and the direction is 27.13 degrees north of east.