2.1 Distance and displacement

  Motion in One Dimension

Distance: is the actual path length traversed by a body or a particle.

Note:

• It is a scalar quantity
• Unit of distance is $m$
• Distance traversed by a body cannot be zero or negative.

Displacement: is the shortest distance between the initial and the final position.

In other words displacement is the length of the line joining the initial and the final

Note:

• It is a vector quantity
• Unit of displacement is $m$
• Displacement is independent of a path traversed by a body
• Displacement depends only on the initial and final position
• Displacement can be positive, negative or zero
• The value of displacement can never be greater than the distance travelled.

Difference between distance and displacement

Consider a particle has traveled along the path $ACB$ as shown in the figure.

$ACB$ is the distance covered by the body. Therefore, the red line shows the distance traveled by a body.

$AB$ is the displacement of the body. Displacement only depends on the final and initial position. It is irrespective of the actual path covered by a body. Therefore, the blue line represents the displacement.

Relation between distance and displacement

If the particle moves in a straight line without changing its direction, then the distance is equal to the magnitude of displacement.

$${\text{Distance}} = \left| {{\text{Displacement}}} \right|$$

In any other case magnitude of displacement will be less than the distance traversed by a body.
$${\text{Distance}} > \left| {{\text{Displacement}}} \right|$$
So generalized equation can be written as,
$${\text{Distance}} \geqslant \left| {{\text{Displacement}}} \right|$$

Question: A particle $A$ travel $3\ km$ towards east and then travels $4\ km$ towards north. Find the displacement and distance travelled by a body.

Solution: The motion can be described in diagram as,

$O:$ Initial position
$Q:$ Final position

$OP$ and $PQ$ is the actual path on which the body has traveled. So, the distance traveled is given by,
$$D = OP + PQ$$$$D = \left( {3 + 4} \right)\,km$$$$D = 7\,km$$
For displacement,

$O:$ Initial position
$Q:$ Final position

So, the length of $OQ$ is displacement.

In $\Delta OPQ$ we can write,
$$O{Q^2} = O{P^2} + P{Q^2}$$$$O{Q^2} = {3^2} + {4^2}$$$$O{Q^2} = 9 + 16$$$$O{Q^2} = 25$$$$OQ = 5\,km$$
So, the displacement is $5\ km$.

Alternate method

$\overrightarrow {OP} = 3\widehat i$
$\overrightarrow {PQ} = 4\widehat j$

From triangle law of addition we can write,
$$\overrightarrow {OQ} = \overrightarrow {OP} + \overrightarrow {PQ} $$$$\overrightarrow {OQ} = 3\widehat i + 4\widehat j$$$$\left| {\overrightarrow {OQ} } \right| = \sqrt {{3^2} + {4^2}} $$$$\left| {\overrightarrow {OQ} } \right| = 5\,km$$