Distance is the difference in the position of an object between two points in time.

Imagine that at a given initial time, \( t_1 \), the particle is in a position \( p_1 = p(t_1) \). At a later time, \( t_2 \), the particle is in position \( p_2 = p(t_2) \). If we subtract the two positions, we will get the distance travelled by that particle between the initial time and the later time:

$$ d_1 = p_2 - p_1 $$

Consider another position of the particle given by a third time, \( p_3 = p(t_3) \). If we subtract this position from the previous position of \( p_2 \), we get the distance travelled by the particle between \( t_2 \) and \( t_3 \):

$$ d_2 = p_3 - p_2 $$

If we add the two distances travelled by the particle together from \( t_1 \) to \( t_2 \) and \( t_2 \) to \( t_3 \), we will get the total distance travelled by the particle between \( t_1 \) and \( t_3 \).

$$ d_3 = d_1 + d_2 $$

Displacement is defined by the distance from the initial position of the particle to the final position. In many cases, this is not the same as the total distance travelled by the particle.

Consider the case where the particle were to travel from the origin to position \( p_2 \), and then travel back to the origin. Although the total distance travelled was double the distance from the origin to position \( p_2 \) - as the particle had to travel to \( p_2 \) and back from the origin - the total displacement was actually zero, since the particle ended up where it started at the end.

For example, the object on the right starts at \( p_1 = 0 \) and begins moving away from the origin. When it reaches \( p_2 = 3 \), the distance and displacement is 3. However, when it starts moving back towards the origin, the distance continue to increase, while the displacement decreases, until the object reaches the origin again. At that point in time, the distance travelled would be 6 units, while the displacement would be 0 units, since the object is back at the starting position.