Acceleration Study Notes

LESSON 1: ACCELERATION

FOCUS QUESTION

• What are two ways velocity can change?

Nonuniform Motion Diagrams

• An object in uniform motion moves along a straight line with an unchanging velocity.
• Nonuniform motion is more common, where velocity is changing.
• Examples of nonuniform motion include:
  • Balls rolling down hills.
  • Cars braking to a stop.
  • Falling objects.
• Later modules will address nonuniform motion not confined to a straight line, such as:
  • Circular motion.
  • The motion of thrown objects, like baseballs.

Describing Nonuniform Motion

• Uniform motion feels smooth; one does not feel motion when moving uniformly.
• Non-uniform motion (e.g., turning or going over a roller coaster hill) induces feelings of being pushed or pulled.
• Figure 1 includes motion diagrams:
  • Diagram 1: Indicates motionless (e.g., runner waiting).
  • Diagram 2: Distances between positions are constant (uniform motion at constant velocity).
  • Diagram 3: Increasing distance indicates the jogger is speeding up.
  • Diagram 4: Decreasing distance indicates the jogger is slowing down.

Particle Model Diagrams

• Definition: A particle model motion diagram displays an object with changing velocity.
  • Indicates changes in velocity via the spacing of dots and lengths of velocity vectors.
• Characteristics:
  • If an object speeds up:
  • Each successive velocity vector is longer.
  • Spacing between dots increases.
  • If the object slows down:
  • Each velocity vector is shorter.
  • Spacing between dots decreases.

Displaying Acceleration on a Motion Diagram

• Acceleration is defined as the rate at which an object's velocity changes.
• Including acceleration vectors in motion diagrams provides a full picture of movement.
• Acceleration vector is calculated:
  • Find change in velocity: $riangle v = v<em>f - v</em>i$
  • Average over time interval: $a = \frac{ riangle v}{ riangle t}$ where $riangle t$ is the time interval.

Finding Acceleration Vectors

• Use the velocity vectors to draw the acceleration vector:
  • Start with the final velocity vector $v<em>f$, then draw initial velocity vector $v</em>i$.
  • Draw acceleration vector $a$ from tail of $v<em>i$ to tip of $v</em>f$.
• For constant acceleration, length and direction of the vector are determined by:
  • $a = \frac{ riangle v}{ riangle t}$

Direction of Acceleration

• Figure 4 illustrates four scenarios:
  1. Car speeding up in positive direction – Velocity and acceleration vectors point in the same direction.
  2. Car slowing down in positive direction – Velocity and acceleration vectors are opposite.
  3. Car speeding up in negative direction – Both vectors align in the negative direction.
  4. Car slowing down in negative direction – Vectors are in opposing directions.
• Important points:
  • Positive acceleration means acceleration vector points positively.
  • Negative acceleration indicates velocity vector points negatively.
• Observation: Sign of acceleration alone does not indicate speeding up or slowing down. Both velocity and acceleration directions are needed for analysis.

Velocity-Time Graphs

• Definition: A velocity-time graph plots velocity against time (velocity on vertical axis, time on horizontal axis).
• Example: Car accelerating from rest.
  • Graph shows a straight line indicating constant acceleration.
  • Acceleration calculated by slope: $ext{slope} = \frac{ ext{rise}}{ ext{run}}$.
  • For example, a slope of $5.00 ext{ m/s}^{2}$ indicates a change in velocity of 5.0 m/s in 1.0 seconds.
• Five runners’ motions shown in Figure 6:
  • Different graphs indicate varying accelerations and velocities.

Average and Instantaneous Acceleration

• Average Acceleration: Change in velocity during a specific time interval divided by that interval.
• Units: Measured in $ext{m/s}^2$.
• Instantaneous acceleration calculated using tangent line on a velocity-time graph.
• For constant acceleration, average and instantaneous accelerations are equal.

Example Problems

• Calculating average acceleration:
• If a car accelerates from 4.0 m/s to 36 m/s in 4.0 s:
  • Average acceleration = $\frac{36 ext{ m/s} - 4 ext{ m/s}}{4.0 ext{ s}} = 8 ext{ m/s}^2$.
• Average acceleration of a bus stopping from 25 m/s to 0 m/s in 3.0 s is:
  • $a = \frac{0 ext{ m/s} - 25 ext{ m/s}}{3 ext{ s}} = -8.33 ext{ m/s}^2$.

Additional Practice Problems

1. Sketch motion diagram for given velocity-time graph.
2. Determine train's acceleration intervals and average acceleration calculations.
3. Challenge: Create a velocity-time graph for an elevator's movement.

Conclusion

• Acceleration is a matter of both speed change and direction change. Understanding its implications is crucial for analyzing motion dynamics.