Distance And Displacement

Distance and displacement are essential concepts in physics that describe motion, yet they represent different aspects of movement. Understanding the distinction is crucial: for instance, if you walk around a block and return to your starting point, your distance traveled may be significant, but your displacement is zero.

1.0Position Vector

•  It is used to specify the position of a certain particle.
• A vector that symbolizes the position of any given point with respect to any reference point like the origin.
• Direction: Always points from the Reference point (origin) of that vector towards a given point.

2.0Distance

• Actual length of the path traversed by the particle between initial and final position is called distance.

• Distance travelled by man through path ABCA = 3 + 4 + 5 = 12m

Key Point:

• Distance always depends on the path which is followed by the particle. Does not depend on initial and final position of the particle.

• Infinite distances are possible between two fixed points because infinite paths are possible between two fixed points.
• It is a Scalar Quantity.
• S.I. Unit- Metre (m)
• C.G.S. Unit- Centimeter (cm)                     
• Dimensional Formula: [M⁰L¹T⁰]
• Distance cannot be negative

3.0Displacement

• Shortest distance between initial & final position of the particle is called displacement.

Displacement of man when moving from A to B and B to C = 5m

Key Points:

• Displacement is a vector quantity, and its direction is always from initial position to final position.
• Displacement does not depend on the path followed by particles. It depends only on the initial & final position of the particle.
• (Displacement)₁ = (Displacement)₂

• Only a single value of displacement is possible between two fixed points.
•  Displacement may be positive, negative or zero.
• If motion is in straight line without change in direction then
• distance = |displacement| = magnitude of displacement.
• Magnitude of displacement may be equal or less than distance but never greater than distance.
•  Distance ≥ |displacement|

Displacement In Vector Form

Initial position vector ($\overrightarrow{OA}$)

$\overrightarrow{r_1}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}$

Initial position vector ($\overrightarrow{OB}$)

$\overrightarrow{r_2}=x_2\hat{i}+y_2\hat{j}+z_2\hat{k}$

Displacement Vector 

$\Delta \vec{r}={\vec{r}}_2-{\vec{r}}_1$

$\overrightarrow{\Delta r}=\left(x_2-x_1\right)\hat{i}+\left(y_2-y_1\right)\hat{j}+\left(z_2-z_1\right)\hat{k}$

Example: A particle starts from point A (1,-1,0) m and reaches at point B (5,3,-2) m, reference point is origin. Find

(a) Initial Position Vector (b) Final Position Vector

     (c) Displacement Vector  (d) Length of Shortest Path

Solution:

(a) Initial Position Vector $(\overrightarrow{\mathrm{OA}})=\hat{\mathrm{i}}-\hat{\mathrm{j}}$

(b) Final Position Vector $(\overrightarrow{OB})=5\hat{i}+3\hat{j}-2\hat{k}$

(c) Displacement Vector, $\Delta r={\vec{r}}_2-{\vec{r}}_1=\overrightarrow{OB}-\overrightarrow{OA}$

$=(5-1)\hat{i}+(3-(-1))\hat{j}+(-2-0)\hat{k}$

$=(4\hat{i}+4\hat{j}-2\hat{k})$

(d) Length of Shortest Path $|\overrightarrow{\Delta r}|=\sqrt{4^2+4^2+(-2{)}^2}=6\mathrm{m}$

4.0Distance And Displacement In Circular Path

1. One-Fourth Of The Complete Circle

$\text{ Distance }=\frac{\pi R}{2}$

$\text{ Displacement }=\sqrt{2}R$

2. Half of The Complete Circle

$\text{ Distance }=\pi R$

$\text{ Displacement=2R }$

5.0Useful Direction Conventions

Sense of Direction in terms of base vectors.

East   :    $\hat{i}$   West :     $\widehat{-i}$

    North :     $\hat{j}$ South :    $-\hat{j}$

      Up      :      $\hat{k}$    Down    : $-\hat{k}$

Example: A body moves 6m north 8m east and 10 m vertically upward, what is magnitude of its resultant displacement from initial position?

Solution:

$\vec{r}=x\hat{i}+y\hat{i}+z\hat{k}$

$r=\sqrt{x^2+y^2+z^2}=\sqrt{6^2+8^2+10^2}=10\sqrt{2}\mathrm{m}$

$\vec{r}=x\hat{i}+y\hat{i}+z\hat{k}$

6.0Displacement -Time Graph

(a) Stationary Particle

(b) particle moving with constant speed

(c) particle moving with increasing speed

(d) particle moving with decreasing speed

7.0Sample Questions on Distance And Displacement

If, in a given motion, displacement is directly proportional to the square of the time transpired, what can we infer about the nature of its acceleration—constant or variable? Explain.

Sol.

Given $x\propto t^2$

$x=c{t}^2$

Velocity, $v=\frac{dx}{dt}=c\times 2t$

Acceleration, $a=\frac{dv}{dt}=2c=\text{ Constant }$

Hence the object is moving with constant acceleration.