Acceleration and Velocity Concepts

Acceleration as a Vector

• Acceleration is defined as the change in velocity over time.
• Represented by the formula:
  $a = \frac{\Delta V}{t}$
  where ( \Delta V = Vf - Vi ) and ( t ) is the time taken.
• Average Velocity (AV): In the example where an object slows from 9 m/s to 2 m/s:
  • Use the formula:
    $AV = V<em>f - V</em>i = 2 \text{ m/s} - 9 \text{ m/s} = -7 \text{ m/s}$

Velocity/Time Graphs

• Different scenarios can be noted on velocity/time graphs:
  a) Object going faster (increase in velocity towards the right)
  b) Object maintains constant speed (horizontal line on graph)
  c) Object slows down and stops (decrease in velocity and hits 0)
  d) Object moves faster backwards (velocity decreases but in negative direction)
  e) Constant speed backwards (constant negative velocity)
  f) Object slowing down before coming to a stop.

Example Calculations

Example #1 (Acceleration)

• Given:
  • ( V_i = 0 \text{ km/hr} )
  • ( V_f = 90 \text{ km/hr} )
  • Time (t) = 10 seconds
• Calculate the average acceleration:
  • Find change in velocity:
    $\Delta V = V<em>f - V</em>i = 90 - 0 = 90 \text{ km/hr}$

Example #2 (Finding Acceleration)

• Time redefined in hours to seconds for conversion:
  $t = 0.003 \text{ hr} * 3600 \text{ s/hr} = 10.8 \text{ s}$
• Given:
  • ( V_i = 20 \text{ m/s} )
  • Calculate acceleration ( a ):
    $a = \frac{V<em>f - V</em>i}{t}$

Further Examples

• Distance Calculation: If an object moves:
  • Formula:
    $d = V_it + \frac{1}{2} a t^2$
• Example values yield: $d = (7.0 \times 10^3 \text{ m/s}) \times (50 \text{ s}) + \frac{1}{2} (20.0) \times (50)^2$
  • Final distance outcome can be approximated as:
    $d = 375000 \text{ m} = 375 \text{ km}$

Summary of Velocity Concepts

• Understanding and applying formulas for velocity, acceleration, and distance are crucial for interpreting motion. Ensure to be familiar with both scalar speeds and vector velocities, especially when directions change (e.g., going backwards).