Chapter 2 class 13 11 advanced

Summary

The lesson explains how to connect acceleration, velocity, and position using motion-versus-time graphs, then applies those ideas to several exercises.

1. Two types of motion

• Uniform motion: Velocity is constant, so acceleration is zero.
  • Positive velocity means position increases over time.
  • Negative velocity means position decreases over time.
• Uniformly varied motion: Acceleration is constant, so velocity changes linearly with time and position follows a quadratic curve.

2. Reading and sketching motion graphs

A useful sequence for predicting graph shapes is:

Acceleration → velocity → position

• If acceleration is zero, velocity is constant and position is linear in time.
• If acceleration is constant, velocity is linear and position is quadratic.
• Positive velocity makes position increase; negative velocity makes it decrease.
• Positive acceleration makes velocity increase; negative acceleration makes it decrease.
• With constant positive acceleration, the position–time graph is concave upward. With constant negative acceleration, it is concave downward.
• A position graph can rise and then fall if velocity changes from positive to negative.

For a velocity–time graph:

• The slope gives acceleration: (a=\Delta v/\Delta t).
• The area under the graph gives displacement. If the object does not reverse direction, this also equals the distance traveled.
• When labeling a graph, distinguish a duration from an instant in time. For example, a 40-second phase beginning at (t=20) ends at (t=60), not (t=40).

3. Worked exercises

1. Runner: Accelerates from rest to (9\text{ m/s}) in 4 seconds, then maintains that speed for 8 seconds.

   • Acceleration: (9/4 = 2.25\text{ m/s}^2).
   • Displacement: area under the velocity–time graph, (18+72=90\text{ m}).

2. Car braking: Travels at (10\text{ m/s}) for 30 seconds, then slows uniformly to rest over the next 12 seconds.

   • Distance: rectangular area (300\text{ m}), plus triangular area (60\text{ m}).
   • Total: (360\text{ m}).

3. Cyclist: Accelerates from rest to (8\text{ m/s}) in 20 seconds, continues at that speed for 40 seconds, then stops uniformly in 15 seconds.

   • First-phase acceleration: (0.4\text{ m/s}^2).
   • Final-phase acceleration: approximately (-0.53\text{ m/s}^2).
   • Total displacement from the graph’s area: (460\text{ m}).
   • Each phase must be treated separately when using equations of motion.

4. Motorcycle: Accelerates from rest to (30\text{ m/s}) over 15 seconds, travels at that speed for (T) seconds, then slows to rest over 25 seconds. The total distance is (2.4\text{ km}).

   • The graph’s total area is (30T+600) metres.
   • Setting this equal to (2400) gives (T=60) seconds.
   • Total travel time: (T+40=\mathbf{100\text{ seconds}}).

5. Cart reversing direction: The cart moves in the positive (x)-direction through the origin, slows to a stop, then moves in the negative direction with increasing speed.

   • Position first increases, reaches a maximum, then decreases.
   • The lesson identifies choice A as the matching position–time graph.

6. Interpreting a velocity–time graph: Velocity is positive during the first interval and negative during the second.

   • Position therefore increases first, then decreases.
   • The lesson identifies choice A as the matching position–time graph.

Speakers and sources

• Instructor/narrator: An AP Physics teacher presenting the lesson. The video context identifies the channel as Amer Kabrit.
• Exercise sources: The lesson includes textbook-style motion problems and refers to one as an AP exercise. No other speakers are identifiable.