# The messenger walks 30 m north, 25 m east, 12 m south, and then rises in the building by elevator

**The messenger walks 30 m north, 25 m east, 12 m south, and then rises in the building by elevator to a height of 36 m. What is displacement?**

To solve the problem, we can schematically depict the path of the messenger.

It is known that it passes to the north – 30 meters, to the east – 25 meters, to the south – 12 meters, and then still rises by elevator to a height of 36 meters.

Let’s write in the form of coordinates of the vector OL the passed path of the messenger (25; 18; 36).

Let’s find all the way that the messenger went:

L = 30 + 25 + 12 + 36 = 103 meters.

To calculate the displacement, we must find the module of the displacement vector.

Let’s apply the formula for this:

S = √ ((x – xo) ^ 2 + (y – yo) ^ 2 + (z – zo) ^ 2),

where xo = 0, yo = 0, zo = 0.

Substitute the values and calculate:

S = √ (252 + 182 + 362) = 47.4 meters.

Answer: L = 103 meters, S = 47.4 meters.