a sailboarder is zipping northeast across the river wit ha velocity of 10 m/s relative to the water the river current is moving the water north at a velocity of 3 m/s downstream. if the angle between the relative velocity of the sailboard and the river current is 30 degrees, what is the resultant or true velocity of the saildboard?

need help

To find the resultant or true velocity of the sailboard, you need to use vector addition. The sailboard's velocity consists of two components: its velocity relative to the water and the velocity of the river current.

First, let's break down the velocities into their x and y components:

Velocity of the sailboard relative to water (Vsw):
Magnitude Vsw = 10 m/s
Angle θ1 = 0 degrees (since the sailboard is zipping northeast, or directly along the x-axis)

Velocity of the river current (Vrc):
Magnitude Vrc = 3 m/s
Angle θ2 = 90 degrees (since the current is moving north, or along the y-axis)

Now, we can find the x and y components of each velocity using trigonometry.

For Vsw:
Vswx = Vsw * cos(θ1) = 10 m/s * cos(0 degrees) = 10 m/s
Vswy = Vsw * sin(θ1) = 10 m/s * sin(0 degrees) = 0 m/s

For Vrc:
Vrcx = Vrc * cos(θ2) = 3 m/s * cos(90 degrees) = 0 m/s
Vrcy = Vrc * sin(θ2) = 3 m/s * sin(90 degrees) = 3 m/s

Now, we can add the x and y components separately to find the resultant velocities:

Resultant velocity in the x-direction (Vx):
Vx = Vswx + Vrcx = 10 m/s + 0 m/s = 10 m/s

Resultant velocity in the y-direction (Vy):
Vy = Vswy + Vrcy = 0 m/s + 3 m/s = 3 m/s

Finally, we can find the magnitude and angle of the resultant velocity:

Magnitude Vr = √(Vx^2 + Vy^2) = √(10 m/s)^2 + (3 m/s)^2) = √(100 + 9) = √109

Angle θr = arctan(Vy / Vx) = arctan(3 / 10)

Therefore, the resultant or true velocity of the sailboard is approximately √109 m/s at an angle of arctan(3 / 10) degrees.