Recall:

  1. Displacement: Dx = xf - xi
  2. Average velocity: vavg = Dx / Dt
  3. Instantaneous velocity: the average velocity in the limit the time interval goes to zero, also the slope on the position versus time graph.
  4. Average acceleration: aavg = Dv / Dt = (vf-vi)/(tf-ti). Note that vf and vi are instantaneous velocities.
Example: P2.16

A car is travelling initially at +7.0m/s. It accelerates at a rate of +0.80m/s2 for 2.0s. What is the final velocity?

  1. Read the problem.
  2. Make a sketch.
  3. Identify the data: vi = 7.0m/s, a = 0.80m/s2, Dt = 2.0s.
  4. We will use the equation a = (vf - vi) / Dt
  5. Rewrite the equation to solve for vf = vi + aDt = 7.0m/s + (0.80m/s2)(2.0s) = 8.6m/s
  6. Check the answer.

Motion Diagram

1-D Motion with Constant Acceleration

Free Fall

Chapter 3: Vectors and 2-D Motion

Vectors and Scalars

Some Properties of Vectors

Components of a Vector