General Physics: Average Velocity, Acceleration, and Centripetal Acceleration Study Guide

Linear Kinematics: Displacement, Velocity, Distance, and Speed

• Average Velocity Definition: Total change in displacement divided by the total time taken. It indicates how fast an object changes its position and includes direction.

• Displacement:

  • Definition: The total change in position from an initial reference point to a final position.
  • Formula:         $\Delta d = d_f - d_i$
  • Where $\Delta d$ is displacement, $d_f$ is final position, and $d_i$ is initial position.

• Velocity:

  • Formula:         $v = \frac{\Delta d}{\Delta t}$
  • Where $v$ is average velocity, $\Delta d$ is displacement, and $\Delta t$ is total time elapsed.

• Distance:

  • Calculation from initial position in $\text{km}$ to the reference point ($0\,\text{km}$).
  • Calculation from the reference point to final position in $\text{km}$.
  • Total Distance: Sum of distance segments added together in $\text{km}$.

• Speed:

  • Formula:         $s = \frac{d}{t}$
  • Where $s$ is speed, $d$ is total distance, and $t$ is total time.

Directional Sign Conventions

• Positive ($+$) Direction Assignments:

  • North
  • East
  • Right
  • Up
  • Note: Applies depending on the designated positive axis.

• Negative ($-$) Direction Assignments:

  • South
  • West
  • Left
  • Down
  • Note: Represents the opposite direction of the chosen positive axis.

Kinematics Example Problem: Runner in Uniform Motion

• Problem Statement: A runner travels with uniform motion from a position of $3.0\,\text{km, west}$ to a position of $9.0\,\text{km, east}$ in $0.60\,\text{h}$.

• Given Parameters:

  • Initial position: $d_1 = -3.0\,\text{km, west}$ (or $-3.0\,\text{km}$)
  • Final position: $d_2 = 9.0\,\text{km, east}$ (or $9.0\,\text{km}$)
  • Time interval: $\Delta t = 0.60\,\text{h}$

• Part A: Displacement Calculation

  • Formula:         $\Delta d = d_f - d_i$
  • Substitution:         $\Delta d = 9.0\,\text{km} - (-3.0\,\text{km})$
  • Result:         $\Delta d = 12\,\text{km, east}$

• Part B: Velocity Calculation

  • Formula:         $v = \frac{\Delta d}{\Delta t}$
  • Substitution:         $v = \frac{12\,\text{km, east}}{0.60\,\text{h}}$
  • Result:         $v = 20\,\text{km/h, east}$

• Part C: Distance Calculation

  • Distance from $3.0\,\text{km, west}$ to reference point ($0\,\text{km}$): $3.0\,\text{km}$
  • Distance from reference point ($0\,\text{km}$) to $9.0\,\text{km, east}$: $9.0\,\text{km}$
  • Total distance addition:         $d = 3.0\,\text{km} + 9.0\,\text{km} = 12\,\text{km}$

• Part D: Speed Calculation

  • Formula:         $s = \frac{d}{t}$
  • Substitution:         $s = \frac{12\,\text{km}}{0.60\,\text{h}}$
  • Result:         $s = 20\,\text{km/h}$

Linear Acceleration

• Definition: Acceleration is the rate of change of velocity with respect to time. It quantifies how quickly an object's velocity changes over a given time interval.

• Formula:     $a = \frac{v_f - v_i}{t}$

• Variable Definitions:

  • $a$ = acceleration
  • $\Delta v$ = change in velocity ($v_f - v_i$), where $v_f$ is final velocity and $v_i$ is initial velocity
  • $\Delta t$ = change in time ($t$)

• SI Unit of Acceleration:     $\text{m/s}^2$

• Example Problem: Train Acceleration

  • Problem Statement: A train increases its velocity from $10\,\text{m/s}$ to $50\,\text{m/s}$ in $20\,\text{s}$. What is its acceleration?
  • Given Parameters:
    • Initial velocity: $v_i = 10\,\text{m/s}$
    • Final velocity: $v_f = 50\,\text{m/s}$
    • Time: $t = 20\,\text{s}$
  • Solution Step-by-Step:         $a = \frac{v_f - v_i}{t}$$a = \frac{50\,\text{m/s} - 10\,\text{m/s}}{20\,\text{s}}$$a = \frac{40\,\text{m/s}}{20\,\text{s}}$$a = 2\,\text{m/s}^2$

Centripetal Acceleration

• Definition: Centripetal acceleration refers to the inward-pointing acceleration of an object moving in a circular path, which alters the direction of the velocity vector without modifying its speed.

• Directional Property: Centripetal acceleration always points directly toward the center of the circular path.

• Variables and Units:

  • $a_c$ = centripetal acceleration, expressed in $\text{m/s}^2$
  • $v$ = tangential velocity or speed, expressed in $\text{m/s}$
  • $r$ = radius of the circular path, expressed in $\text{m}$

• Primary Formula:     $a_c = \frac{v^2}{r}$

• Example 1: Motorcycle Around a Circular Curve

  • Problem Statement: A motorcycle moves around a circular curve at a speed of $15\,\text{m/s}$. If the radius of the curve is $50\,\text{m}$, what is its centripetal acceleration?
  • Given Parameters:
    • Speed: $v = 15\,\text{m/s}$
    • Radius: $r = 50\,\text{m}$
  • Target Variable: $a_c$
  • Solution:         $a_c = \frac{v^2}{r}$$a_c = \frac{(15\,\text{m/s})^2}{50\,\text{m}}$$a_c = \frac{225\,\text{m}^2/\text{s}^2}{50\,\text{m}}$$a_c = 4.5\,\text{m/s}^2$
  • Answer: $a_c = 4.5\,\text{m/s}^2$ (or rounded to $5\,\text{m/s}^2$)

• Example 2: Ball Swinging in a Horizontal Circle

  • Problem Statement: What is the speed of a ball attached to a string that swings in a horizontal circle of radius $2.0\,\text{m}$ with centripetal acceleration of $15\,\text{m/s}^2$?
  • Given Parameters:
    • Radius: $r = 2.0\,\text{m}$
    • Centripetal acceleration: $a_c = 15\,\text{m/s}^2$
  • Target Variable: $v$
  • Derived Formula:         $v = \sqrt{a_c r}$
  • Solution:         $v = \sqrt{(15\,\text{m/s}^2)(2.0\,\text{m})}$$v = \sqrt{30\,\text{m}^2/\text{s}^2}$$v = 5.48\,\text{m/s}$
  • Answer: $v = 5.48\,\text{m/s}$ (or rounded to $5.5\,\text{m/s}$)

• Example 3: Car on a Circular Track

  • Problem Statement: A car moves around a circular track at a speed of $25\,\text{m/s}$. If its centripetal acceleration is $5\,\text{m/s}^2$, what is the radius of the circular track?
  • Given Parameters:
    • Speed: $v = 25\,\text{m/s}$
    • Centripetal acceleration: $a_c = 5\,\text{m/s}^2$
  • Target Variable: $r$
  • Derived Formula:         $r = \frac{v^2}{a_c}$
  • Solution:         $r = \frac{(25\,\text{m/s})^2}{5\,\text{m/s}^2}$$r = \frac{625\,\text{m}^2/\text{s}^2}{5\,\text{m/s}^2}$$r = 125\,\text{m}$
  • Answer: $r = 125\,\text{m}$