motion with constant acceleration

Chapter 1: Introduction to Motion with Constant Acceleration

• Motion is defined as the change in position of an object with respect to time.

  • Position: Function of time, defining the location of an object along a line.

  • Velocity (v(t)): Derivative of the position function, showing the rate of change of position.

  • Acceleration (a(t)): Derivative of the velocity function, second derivative of position.

Constant Acceleration

• Special case where acceleration does not depend on time; represented as a(t) = a (a constant).

• Integral relationships:

  • Velocity as integral of acceleration: v(t) = v0 + at.

    • Here, v0 is initial velocity at time t=0.

  • Position as integral of velocity: x(t) = x0 + v0t + 1/2 at^2.

    • x0 is initial position at t=0.

Key Equations Derived

1. Position Function:

   • x(t) = x0 + v0t + 1/2 at^2 (Equation 1).

2. Velocity Function:

   • v(t) = v0 + at (Equation 2).

3. Velocity-Position Relationship:

   • v^2 = v0^2 + 2a(x - x0) (Equation 3).

Chapter 2: The Initial Velocity

• Average velocity (v̅) is defined as the displacement over time:

  • v̅ = (x - x0) / (t - 0).

• Relating v̅ to initial and final velocities gives:

  • v̅ = (v0 + v) / 2 (Equation 4).

• Using Equation 1 for average velocity:

  • x = x0 + v̅t (Equation 5).

• Summary of Kinematic Equations with constant acceleration in one dimension:

  1. x = x0 + v0t + 1/2 at^2.

  2. v = v0 + at.

  3. v^2 = v0^2 + 2a(x - x0).

  4. v̅ = (v + v0) / 2.

  5. x = x0 + v̅t.

Chapter 3: Position Over Time

Example 1

• Object with:

  • Initial position (x0) = 2 m.

  • Initial velocity (v0) = 3.5 m/s.

  • Constant acceleration (a) = 12.4 m/s^2.

• Find:

  1. Position at t=4.6 s:

     • x(4.6) = 149.3 m.

  2. Velocity at t=3.7 s:

     • v(3.7) = 49.4 m/s.

  3. Displacement from 1.7s to 3.8s:

     • Displacement = 78.9 m.

Example 2

• Object with:

  • Initial position (x0) = 1.8 m.

  • Initial velocity (v0) = 33.4 m/s (positive).

  • Constant acceleration (a) = -6.2 m/s^2 (negative).

• Find:

  1. Time to stop: t = 5.387 s.

  2. Distance to stop: 89.94 m displacement.

Chapter 4: The Initial Velocity

Stopping Time and Distance

• For an object initially moving with:

  • Initial velocity = 33.4 m/s,

  • a = -6.2 m/s^2:

  • Time to stop:

  • t = 33.4 / 6.2 = 5.387 s.

  • Distance to stop:

  • x_final = 91.74 m.

    • Displacement = 89.94 m.

• Two methods to calculate displacement yield consistent results:

  • Using kinematic equation v^2 = v0^2 + 2a(x - x0).

Chapter 5: Free Fall Motion

• Objects dropped experience gravitational acceleration:

  • g = -9.80 m/s^2 (negative indicating downward).

• Time to fall from height h:

  t = √(2h/g) (derived using y = y0 + 0 - 0.5gt^2).

• Velocity at impact:v = -√(2gh) when dropped from rest.

• Example:

  • Object thrown upwards with:

    • Initial velocity = 32.5 m/s, height = 120 m.

      • Time to highest point: t = 3.316 s; final height = 173.89 m.

      • Time to travel up and down to 5 m above ground: t = 9.19 s.

      • Velocity at that moment = -57.56 m/s.

Chapter 6: Conclusion

• Kinematic equations represent motion with constant acceleration along horizontal and vertical directions.

• Key concepts include calculating time, velocity, and displacement for objects under constant acceleration, including gravity.