A man walked 5cm east and 10cm north calculate the distance covered and the magnitude of his displacement
d = d1 + d2 = 5m + 10m = 15 m.
Disp. = Vector sum = Sqrt(5^2+10^2) =
Physics
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To calculate the distance covered by the man, we can use the Pythagorean theorem.
According to the given information, the man walked 5cm east and 10cm north. Since these two movements are at right angles to each other, we can form a right triangle. The distance covered by the man is the hypotenuse of this right triangle.
Using the Pythagorean theorem:
distance^2 = (east)^2 + (north)^2
distance^2 = 5^2 + 10^2
distance^2 = 25 + 100
distance^2 = 125
Taking the square root of both sides, we find:
distance = √125
distance ≈ 11.180 cm
So, the distance covered by the man is approximately 11.180 cm.
Now, let's calculate the magnitude of his displacement. Displacement is a vector quantity that represents the shortest path from the initial position to the final position, regardless of the path taken. To find the magnitude of the displacement, we need to find the straight-line distance from the initial position to the final position.
The man walked 5cm east and 10cm north. Therefore, his displacement is a vector that can be represented as (5cm east, 10cm north).
Again, we can use the Pythagorean theorem to find the magnitude (or length) of this vector:
magnitude^2 = (east)^2 + (north)^2
magnitude^2 = 5^2 + 10^2
magnitude^2 = 25 + 100
magnitude^2 = 125
Taking the square root of both sides, we find:
magnitude = √125
magnitude ≈ 11.180 cm
So, the magnitude of the man's displacement is approximately 11.180 cm.