A motor boat due east at 10m/s on a current moving towards S 30 degree W @ 5.0m/s. Find the resultant value of the boat and the direction.
The velocity of the boat with respect to land equals the VECTOR sum of the water velocity and the boat's speed with relative to the water.
If i and j are unit vectors East and north,
V(with respect to land) =
10 i + -5 sin30 i - 5 cos30 j
= 7.5 i - 4.33 j
Speed = sqrt[(7.5)^2 + (4.33)^2]
= 8.66 m/s
The direction is arctan(4.33/7.5)= 30 degrees South of East
To find the resultant value and direction of the boat, we will resolve the velocities into their horizontal and vertical components.
Given:
Velocity of boat (Vb) = 10 m/s due east
Velocity of current (Vc) = 5.0 m/s towards S 30° W
Step 1: Resolving the velocities
Resolve the velocity of the boat into its horizontal and vertical components. Since the boat is moving due east, the horizontal component (Vbx) is equal to the velocity of the boat, and the vertical component (Vby) is 0.
Resolve the velocity of the current into its horizontal and vertical components. To do this, we can use trigonometry. The current is moving towards S 30° W, which means it is at an angle of 150° clockwise from the positive x-axis.
Vcx = Vc * cos(150°)
= 5.0 m/s * cos(150°)
≈ -4.33 m/s (negative because it is in the opposite direction of the positive x-axis)
Vcy = Vc * sin(150°)
= 5.0 m/s * sin(150°)
≈ -2.50 m/s (negative because it is in the opposite direction of the positive y-axis)
Step 2: Finding the resultant value
The resultant horizontal component (Vrx) is the sum of the horizontal components of boat and current:
Vrx = Vbx + Vcx
= 10 m/s + (-4.33 m/s)
≈ 5.67 m/s towards the east
The resultant vertical component (Vry) is the sum of the vertical components of boat and current:
Vry = Vby + Vcy
= 0 m/s + (-2.50 m/s)
≈ -2.50 m/s downwards
Step 3: Finding the resultant value and direction
The resultant velocity (Vr) can be found using the Pythagorean theorem:
|Vr| = sqrt((Vrx)^2 + (Vry)^2)
= sqrt((5.67 m/s)^2 + (-2.50 m/s)^2)
≈ 6.27 m/s
The direction of the resultant velocity (θ) can be found using inverse trigonometry:
θ = atan(Vry / Vrx)
= atan((-2.50 m/s) / (5.67 m/s))
≈ -24.94°
Since θ is negative, we add 180° to get the resultant direction:
θ_resultant = -24.94° + 180°
= 155.06°
Therefore, the resultant value of the boat is approximately 6.27 m/s, and its direction is towards S 34.94° E.