A ship sailed 15 km north, then 20km east followed by 10km south. Find its displacement from its starting point
The first and last legs of the journey are equivalent to a displacement of 5 km north.
Add, as vectors, the net displacement vectors of 5 km north and 20 km east. Use the Pythagorean theorem.
sqrt(5^2 + 20^2) = sqrt(425) = ___ km
with the sqare root of 425,the answer will be 20.6km
To find the displacement of the ship from its starting point, we can use the concept of vectors. Displacement can be thought of as the straight-line distance between the starting point and the final position.
1. Draw a diagram: Start by drawing a diagram to visualize the ship's movement. Label the north direction as positive and the south direction as negative.
|
|
|
|------------------ S (Final position)
|
|
| N (Starting point)
2. Break down the ship's movement into components: The ship sailed 15 km north, then 20 km east, and finally 10 km south. We can break down this movement into two perpendicular components: the north-south (vertical) component and the east-west (horizontal) component.
|
| 15 km 10 km
|------------------ S
|
| 20 km
|
| N
3. Calculate the vertical displacement: The north-south component of the ship's movement is 15 km north and 10 km south, which cancel each other out. Therefore, the vertical displacement is 15 km - 10 km = 5 km to the north.
4. Calculate the horizontal displacement: The east-west component of the ship's movement is 20 km to the east. There is no westward movement, so the horizontal displacement is simply 20 km to the east.
5. Calculate the displacement: Now, we have the vertical displacement of 5 km north and the horizontal displacement of 20 km east. To find the displacement, we can use the Pythagorean theorem since the vertical and horizontal displacements form a right triangle.
Displacement = √(vertical displacement^2 + horizontal displacement^2)
= √(5 km^2 + 20 km^2)
= √(25 km^2 + 400 km^2)
= √(425 km^2)
≈ 20.62 km
Therefore, the displacement of the ship from its starting point is approximately 20.62 km.