Distance and Displacement Calculations in Mechanics

Ratio and Relationship Between Distance and Displacement

• Numerical Ratio of Displacement to Distance:

  • Displacement represents the shortest straight-line distance from the initial position to the final position, whereas distance represents the total path length traversed.

  • For any moving object, displacement magnitude is less than or equal to the total distance covered.

  • Therefore, the numerical ratio of displacement to distance for a moving object is always equal to or less than 1 ($\text{Ratio} \le 1$).

Rules and Calculations for Circular Motion

• Guideline for Value of $\pi$:

  • Use $\pi = \frac{22}{7}$ whenever the given radius $r$ is a multiple of $7$.

• Full Revolutions:

  • General Formulas:

    • $\text{Distance} = n \times 2\pi r$ (where $n$ is the total number of complete revolutions).

    • $\text{Displacement} = 0$ (since initial and final points coincide).

  • Single Revolution Example ($r = 7\,\text{cm}$):

    • A particle starts from point 'A' and completes $1$ full revolution in a circle of radius $r = 7\,\text{cm}$.

    • $\text{Distance} = 2\pi r = 2 \times \frac{22}{7} \times 7\,\text{cm} = 44\,\text{cm}$

    • $\text{Displacement} = 0\,\text{cm}$

  • Five Revolutions Example ($n = 5$, $r = 14\,m$):

    • A particle moves along a circular path of radius $r = 14\,m$ and completes $n = 5$ revolutions.

    • $\text{Distance} = n \times 2\pi r = 5 \times 2 \times \frac{22}{7} \times 14\,m = 5 \times 2 \times 22 \times 2\,m = 440\,m$

    • $\text{Displacement} = 0\,m$

• Fractional Circular Motion (Three-Quarters of a Circle):

  • General Formulas:

    • $\text{Distance} = \frac{3}{4} \times 2\pi r = \frac{3}{2}\pi r$

    • $\text{Displacement} = \sqrt{OA^2 + OB^2} = \sqrt{r^2 + r^2} = r\sqrt{2}$ (calculated as the straight-line hypotenuse across the two perpendicular radii $OA$ and $OB$).

  • Example 1 ($r = 7\,m$):

    • A particle completes $\frac{3}{4}\text{th}$ part of a circle of radius $r = 7\,m$.

    • $\text{Distance} = \frac{3}{4} \times 2 \times \frac{22}{7} \times 7\,m = \frac{3 \times 11}{1}\,m = 33\,m$

    • $\text{Displacement} = \sqrt{7^2 + 7^2}\,m = 7\sqrt{2}\,m \approx 7 \times 1.414\,m = 9.898\,m \approx 9.9\,m$

  • Example 2 ($r = 14\,\text{cm}$ in the MKS System):

    • A boy runs in a circular path of $\frac{3}{4}$ of the circle with a radius $r = 14\,\text{cm}$.

    • In the MKS (Meter-Kilogram-Second) system, the radius is converted to meters: $r = 0.14\,m$

    • $\text{Displacement} = r\sqrt{2} = 0.14\sqrt{2}\,m$

Linear Motion along Coordinate Axes

• 1D/2D Axis Path Problem:

  • Deekshitha starts at origin $(0,0)$ and moves along the y-axis to $(0,25\,m)$, and then moves to $(0,-25\,m)$.

  • Distance Calculation:

    • First movement from $(0,0)$ to $(0,25\,m)$ = $25\,m$

    • Second movement from $(0,25\,m)$ to $(0,-25\,m)$ = $| -25 - 25 |\,m = 50\,m$

    • $\text{Total Distance} = 25\,m + 50\,m = 75\,m$

  • Displacement Calculation:

    • Initial position = $(0,0)$

    • Final position = $(0,-25\,m)$

    • $\text{Magnitude of Displacement} = |-25\,m - 0\,m| = 25\,m$

  • Distance to Displacement Ratio:

    • $\text{Ratio} = \frac{\text{Distance}}{\text{Displacement}} = \frac{75\,m}{25\,m} = \frac{3}{1} = 3:1$

Geometric and Vector Displacement Calculations

• Hypotenuse Calculations using Pythagorean Theorem:

  • Right-Angled Triangle $\triangle AOB$ ($AO = 14\,\text{cm}$, $OB = 14\,\text{cm}$):

    • $AB^2 = AO^2 + OB^2$

    • $AB^2 = 14^2 + 14^2 = 196 + 196 = 392$

    • $AB = \sqrt{392}\,\text{cm}$

  • Segment $BE$ ($AB = 3\,m$, AE = 11\,m$)**:\n * BE^2 = AB^2 + AE^2\n * BE^2 = 3^2 + 11^2 = 9 + 121 = 130\n * BE = \sqrt{130}\,m\n * **Segment DF$($DE = 7\,m$,$FE = 5\,m$):

    • $DF^2 = DE^2 + FE^2$

    • $DF^2 = 7^2 + 5^2 = 49 + 25 = 74$

    • $DF = \sqrt{74}\,m$

The uploaded files cover key concepts and numerical problems in mechanics focused on Distance and Displacement:

1. Ratio of Displacement to Distance:

   • Defines displacement as the shortest straight-line distance from initial to final position, and distance as total path length.

   • Establishes that the numerical ratio of displacement to distance is always equal to or less than 1 ($\text{Ratio} \le 1$).

2. Circular Motion Rules and Calculations:

   • Full Revolutions: Total distance is $n \times 2\pi r$, while displacement is always $0$ because the initial and final positions coincide.

   • Fractional Revolutions (Three-Quarters Circle): Distance is $\frac{3}{4} \times 2\pi r = \frac{3}{2}\pi r$, and displacement is calculated using the perpendicular radii as $r\sqrt{2}$.

   • Guidelines: Use $\pi = \frac{22}{7}$ when the radius $r$ is a multiple of 7, and convert measurements to meters when requested in the MKS system.

3. 1D/2D Motion along Coordinate Axes:

   • Demonstrates calculating total path distance versus net straight-line displacement for movements along axes (such as moving from origin $(0,0)$ to $(0,25\,m)$ and then to $(0,-25\,m)$), resulting in a distance-to-displacement ratio of $3:1$.

4. Geometric Displacement via Pythagorean Theorem:

   • Uses the formula $a^2 + b^2 = c^2$ to find net displacement across perpendicular segments (e.g., calculating hypotenuses like $AB = \sqrt{392}\,\text{cm}$, $BE = \sqrt{130}\,m$, and $DF = \sqrt{74}\,m$).

Here is an explanation of each question covered in your notes and uploaded sources:

1. Numerical Ratio of Displacement to Distance:

   • Concept: Displacement is the shortest straight-line path between the initial and final positions, while distance is the total path length traveled.
   • Explanation: Since a straight line is the shortest distance between two points, the magnitude of displacement can never exceed total distance. Therefore, the numerical ratio of displacement to distance is always equal to or less than 1 ($\text{Ratio} ile 1$).

2. Circular Motion Problems:

   • 5 Revolutions ($r = 14\,m$):
     • Distance: For $n = 5$ complete revolutions, total distance is $n \times 2\pi r = 5 \times 2 \times \frac{22}{7} \times 14\,m = 440\,m$.
     • Displacement: Since the particle returns to its starting point after full revolutions, the initial and final positions coincide, so displacement is $0\,m$.
   • 3/4 Circle ($r = 7\,m$):
     • Distance: The path traveled is $\frac{3}{4}$ of the total circumference: $\frac{3}{4} \times 2\pi r = \frac{3}{4} \times 2 \times \frac{22}{7} \times 7\,m = 33\,m$.
     • Displacement: The straight line between the start and end points forms the hypotenuse of a right triangle with two perpendicular radii of length $7\,m$. Using the Pythagorean theorem: $\text{Displacement} = \sqrt{7^2 + 7^2} = 7\sqrt{2}\,m \approx 9.9\,m$.
   • 1 Revolution ($r = 7\,\text{cm}$):
     • Distance: $2\pi r = 2 \times \frac{22}{7} \times 7\,\text{cm} = 44\,\text{cm}$.
     • Displacement: $0\,\text{cm}$ because it ends at the starting point.
   • 3/4 Circle in MKS System ($r = 14\,\text{cm}$):
     • In the MKS system, distance units are meters, so $r = 0.14\,m$. Displacement is calculated as $r\sqrt{2} = 0.14\sqrt{2}\,m$.

3. 1D/2D Axis Path Problem (Deekshitha's Motion):

   • Question: Deekshitha starts at $(0,0)$, moves to $(0,25\,m)$, and then moves to $(0,-25\,m)$.
   • Distance: First movement = $25\,m$. Second movement from $(0,25\,m)$ to $(0,-25\,m)$ = $50\,m$. Total distance = $25\,m + 50\,m = 75\,m$.
   • Displacement: Net position change from origin $(0,0)$ to $(0,-25\,m)$ has a magnitude of $25\,m$.
   • Ratio: The ratio of distance to displacement is $\frac{75\,m}{25\,m} = 3:1$.

4. Geometric Calculations using Pythagorean Theorem:

   • $AB$ with legs $14\,\text{cm}$ and $14\,\text{cm}$: $AB = \sqrt{14^2 + 14^2} = \sqrt{392}\,\text{cm}$.
   • $BE$ with legs $3\,m$ and $11\,m$: $BE = \sqrt{3^2 + 11^2} = \sqrt{130}\,m$.
   • $DF$ with legs $7\,m$ and $5\,m$: $DF = \sqrt{7^2 + 5^2} = \sqrt{74}\,m$.