Vectors in physics

Question

A boat sails 3km east then 5km south east then a further distance in an unknown direction. If his final position is 8km directly east od the starting point determine the magnitude and direction of the third leg of the journey

Answer 1

We can solve this problem by using vector addition to find the third leg of the journey.

Let's denote the distance and direction of the third leg as d and θ, respectively. We can represent the boat's journey as a series of displacements:

1. A displacement of 3 km to the east can be represented as a vector of magnitude 3 km in the positive x-direction (i.e., along the x-axis).
2. A displacement of 5 km to the southeast can be represented as a vector that makes a 45-degree angle with the positive x-axis and has a magnitude of 5 km. We can break this vector into components along the x- and y-axes as follows:
3. The x-component of this vector is 5 km * cos(45) = 3.536 km to the east.
4. The y-component of this vector is 5 km * sin(45) = 3.536 km to the south.
5. The third displacement can be represented as a vector that makes an angle θ with the positive x-axis and has a magnitude of d km.

The final position of the boat is 8 km directly east of the starting point. This means that the total displacement of the boat is 8 km to the east. We can represent this displacement as a vector of magnitude 8 km in the positive x-direction.

Using vector addition, we can write:

displacement = 3 km east + 3.536 km east - 3.536 km south + d km θ

The x-components of these vectors add up to 8 km, and the y-components add up to 0 km (since the final position is on the x-axis). Therefore:

8 km = 3 km + 3.536 km cos(θ) + d km cos(θ)

0 km = -3.536 km sin(45) + d km sin(θ)

Solving for d and θ, we get:

d = 5.8 km θ = 31.9 degrees south of east

Therefore, the magnitude of the third leg of the journey is 5.8 km, and it makes an angle of 31.9 degrees south of east with the positive x-axis.