A boat heads directly across a river with a velocity of 12 m/s. If the river flows at 6.0 m/s find the magnitude and direction (with respect to the shore) of the boat's resultant velocity.
How you find angle 63.4?
Please help I can't find angle how are you doing
To find the magnitude and direction of the boat's resultant velocity, we need to use vector addition. We can break down the velocity of the boat and the velocity of the river into their respective horizontal and vertical components.
Let's call the boat's velocity VB (boat) and the river's velocity VR (river). The magnitude and direction of VB can be determined by the given information:
Magnitude of VB = 12 m/s (given in the question)
Now, let's find the horizontal and vertical components of VB and VR.
Horizontal component of VB (VBx) = magnitude of VB × cos(angle between VB and the x-axis)
Since the boat is heading directly across the river, the angle between VB and the x-axis will be 0 degrees.
VBx = 12 m/s × cos(0 degrees) = 12 m/s
Vertical component of VB (VBy) = magnitude of VB × sin(angle between VB and the x-axis)
Since the boat is heading directly across the river, the angle between VB and the x-axis will be 0 degrees.
VBy = 12 m/s × sin(0 degrees) = 0 m/s
Now let's find the horizontal and vertical components of VR:
Horizontal component of VR (VRx) = 0 m/s (since the river flows perpendicular to the shore)
Vertical component of VR (VRy)= magnitude of VR = 6.0 m/s (given in the question)
Next, we can find the resultant velocity of the boat (VB + VR) by adding the horizontal and vertical components.
Horizontal component of VB + VR = VBx + VRx = 12 m/s + 0 m/s = 12 m/s
Vertical component of VB + VR = VBy + VRy = 0 m/s + 6.0 m/s = 6.0 m/s
To find the magnitude (V) and direction (θ) of the resultant velocity, we can use the Pythagorean theorem and trigonometry.
Magnitude of VB + VR (V) = sqrt((Horizontal component)^2 + (Vertical component)^2)
V = sqrt((12 m/s)^2 + (6.0 m/s)^2) = sqrt(144 m^2/s^2 + 36 m^2/s^2) = sqrt(180 m^2/s^2) = 13.4 m/s (approx.)
Direction (θ) of VB + VR can be found using the inverse tangent function:
θ = arctan((Vertical component)/(Horizontal component))
θ = arctan((6.0 m/s)/(12 m/s)) = arctan(0.5)
θ ≈ 26.6 degrees (approx.)
Therefore, the magnitude of the boat's resultant velocity is approximately 13.4 m/s, and its direction with respect to the shore is approximately 26.6 degrees.
12 east
6 north
magnitude = sqrt(12^2+6^2) = 13.4
tan A = 12/6 = 2
so A = 1.11 radians = 63.4 degrees