Question

A body moves $6m$ north, $8m$ east, and $10m$ vertically upwards, what is its resultant displacement from the initial position (only magnitude).

Answer 1

Step 1: Given data

A body moves $6m$ north, $8m$ east, and $10m$ vertically upwards.

Step 2: Displacement

1. The shortest distance between the final position and the initial position of the motion of the object.
2. It can be positive, negative, or zero, depending on the context.
3. It is a vector quantity.
4. It is denoted by s.

Step 3: Calculating the resultant displacement

Consider, the east direction to be the x-direction, the north direction to be the y-direction, and the vertically upward direction to be the z-direction.
Displacement of the particle is given by $\overrightarrow{s}=8\hat{i}+6\hat{j}+10\hat{k}$

The magnitude of displacement: $\left|s\right|=\sqrt{6^2+8^2+{10}^2}=10\sqrt{2}m$

Hence, the resultant displacement from the initial position is $10\sqrt{2}m$.