find distance and displacement of a man walk 1200m due north and 500m East
Distance = 1200 + 500 = 1700 m.
Displacement = sqrt(500^2 + 1200^2) =
To find the distance and displacement of a man who walks 1200m due north and 500m east, we need to use basic trigonometry.
Step 1: Draw a diagram representing the man's movement.
N
|
|
|
X------------->E
Step 2: Calculate the distance traveled.
The distance is the total path taken. In this case, the man traveled 1200m due north and 500m east. To find the distance, we can use the Pythagorean theorem.
Distance = √(1200m)^2 + (500m)^2
Distance = √(1440000m^2 + 250000m^2)
Distance = √1690000m^2
Distance ≈ 1301.65m
Therefore, the distance traveled is approximately 1301.65m.
Step 3: Calculate the displacement.
Displacement is the straight-line distance from the initial position to the final position. To find the displacement, we can draw a right triangle with the north and east distances as the sides.
Using trigonometry, we can find the angle between the north and east directions.
tan(θ) = opposite/adjacent
tan(θ) = 1200m/500m
θ ≈ 67.38 degrees
The displacement can be calculated as:
Displacement = √(1200m)^2 + (500m)^2
Displacement = √(1440000m^2 + 250000m^2)
Displacement = √1690000m^2
Displacement ≈ 1301.65m
Therefore, the displacement is approximately 1301.65m in the direction of 67.38 degrees east of north.
To find the distance and displacement of a person who walks 1200m due north and then 500m east, we can use basic concepts of vector addition.
First, let's consider the distance traveled by the person. Distance is the total length covered irrespective of the direction. In this case, the person has walked 1200m north and then 500m east. These two displacements can be represented by vectors. The north displacement is a vector of length 1200m pointing directly north, and the east displacement is a vector of length 500m pointing directly east.
To find the distance, we need to find the magnitude of the resultant vector formed by adding these two displacement vectors. We can do this using the Pythagorean theorem:
Distance = √(1200^2 + 500^2)
Distance ≈ 1283.67 meters
Therefore, the distance traveled by the person is approximately 1283.67 meters.
Next, let's consider the displacement of the person. Displacement is the shortest straight-line path from the initial to the final position, taking into account the direction. In this case, the person starts at some initial position and ends up at a final position. The distance between these two points is the resultant displacement vector.
To find the displacement, we can sum up the north and east displacements using vector addition. Since the north and east directions are perpendicular, we can create a right-angled triangle with the north displacement as the vertical leg (1200m) and the east displacement as the horizontal leg (500m). The hypotenuse of this triangle represents the displacement of the person.
Using the Pythagorean theorem, we can find the magnitude of the displacement:
Displacement = √(1200^2 + 500^2)
Displacement ≈ 1303.84 meters
Therefore, the displacement of the person is approximately 1303.84 meters, and the distance traveled is approximately 1283.67 meters.