With the Pythagoras' Theorem we can find that the distance traveled is:
39 m
The distance is the length of the path between two points and it is a scalar magnitude so if the path changes direction the Pythagoras' theorem should be used
d = [tex]\sqrt{(x-x_o)^2 + (y-y_o)^2 + (z-z_o)^2 }[/tex]
Where d is the distance, (x,y,z) is the interes point, (x₀,y₀,z₀) is de reference point.
In this case, let's set a reference system in the lower part of the school, take the z-axis as vertical and set the point of arrival at as the reference (0, 0, 0).
The distance that the students descend is d₁ = 30 k ^ m, when they arrive from the bottom of the school they travel d₂ = 15 j ^ m and d₃ = 20 i ^ m
let's calculate
d =[tex]\sqrt{(20-0)^2 + (15-0)^2 + ( 30-0)^2 }[/tex]
d = 39.05 m
Notice that the distance by being a scalar does not have unit vectors
In conclusion using the Pythagoras' Theorem we can find that the distance traveled is 39 m
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