Question

the velocity of a boat is 2o km/h in a direction 50 degree north of east. the wind velocity is 5 km/h from the west. the resultant velocity of the boat can be represented by the side of a triangle in which two velocities are the other side. determine the resultant velocity of the boat

Answer 1

Answer on Question #38912, Physics, Mechanics | Kinematics | Dynamics

The velocity of a boat is $20\mathrm{km / h}$ in a direction 50 degree north of east. The wind velocity is $5\mathrm{km / h}$ from the west. The resultant velocity of the boat can be represented by the side of a triangle in which two velocities are the other side. Determine the resultant velocity of the boat.

Solution:

We introduce the following notations:

$\mathsf{v_b} = 20\mathrm{km / h}$ (velocity of a boat),

$\mathsf{v_w} = 5\mathrm{km / h}$ (wind velocity),

$\gamma = 180{}^{\circ} - 50{}^{\circ} = 130{}^{\circ}$ (angle between the direction from the west and north-east direction).

I give the formula for the Law of Cosines and use it to find the missing side length of a triangle.

$$c ^ {2} = a ^ {2} + b ^ {2} - 2 a b \cos \gamma$$

In our notations the resultant velocity of the boat $\mathbf{v}_{\mathrm{r}}$ is:

$$v _ {r} ^ {2} = v _ {b} ^ {2} + v _ {w} ^ {2} - 2 v _ {b} v _ {w} \cos \gamma$$ $$v _ {r} ^ {2} = 2 0 ^ {2} + 5 ^ {2} - 2 \cdot 2 0 \cdot 5 \cdot \cos (1 3 0 {}^ {\circ}) = 4 0 0 + 2 5 + 1 2 8. 5 6 = 5 5 3. 5 6$$ $$v _ {r} = \sqrt {5 5 3 . 5 6} = 2 3. 5 \mathrm {k m / h}$$

Answer: $23.5\mathrm{km / h}$