3  Motion in a Straight Line

3.1 Syllabus inquiry question

  • How is the motion of an object moving in a straight line described and predicted?
Feynman Insight

From The Feynman Lectures on Physics, Vol I, Chapter 8:

Motion becomes predictable when the change of velocity is treated as a measurable quantity rather than a mystery. The model does not explain why an object moves; it shows how to calculate its motion.

3.2 Learning Objectives

  • Distinguish scalars from vectors in one-dimensional motion.
  • Define displacement, velocity, and acceleration with SI units.
  • Interpret average and instantaneous quantities.
  • Apply constant-acceleration ideas to straight-line motion.

3.3 Content

3.3.1 Scalars and vectors in one dimension

A scalar has magnitude only. A vector has magnitude and direction. In straight-line motion, the direction is represented by a positive or negative sign along a chosen axis.

3.3.2 Describing motion

Displacement describes the change in position. Velocity describes the rate of change of displacement. Acceleration describes the rate of change of velocity.

\[v_{avg} = \frac{\Delta s}{\Delta t}, \quad a = \frac{\Delta v}{\Delta t}\]

3.3.3 Interactive: Motion Diagram

Explore how objects move under constant acceleration. The dots show position at equal time intervals, and the arrows show velocity at each instant.

Observe:

  • The dots get closer together as the object slows down
  • The velocity arrows get shorter over time
  • The negative acceleration reduces velocity until the object momentarily stops

3.3.4 Instantaneous values

Instantaneous velocity is the slope of a displacement-time graph at a point. It represents motion at a single instant rather than over an interval.

Try This

Hover over the position-time graph below to see the instantaneous velocity at any point. The tangent line shows the slope (velocity) at that instant.

Key observations:

  • The curve is a parabola (constant acceleration)
  • The gradient (slope) at any point equals the instantaneous velocity
  • Where the curve is steepest, velocity is greatest
  • At the maximum point, the gradient (and velocity) is zero

3.4 Worked Examples

3.4.1 Example 1: Average velocity

A cyclist moves from 120 m to 20 m in 25 s along a straight road.

  1. Displacement: \(\Delta s = 20 - 120 = -100\) m.
  2. Average velocity: \(v_{avg} = -100/25 = -4.0\) m/s.
  3. The negative sign indicates motion in the chosen negative direction.

3.4.2 Example 2: Acceleration from velocity change

A car slows from 18 m/s to 6 m/s in 4.0 s.

  1. Change in velocity: \(\Delta v = 6 - 18 = -12\) m/s.
  2. Acceleration: \(a = -12/4.0 = -3.0\) m/s\(^2\).
  3. The negative sign indicates deceleration.

3.4.3 Example 3: Constant acceleration displacement

A trolley starts at 2.0 m/s and accelerates at 0.50 m/s\(^2\) for 6.0 s.

  1. Use \(s = ut + \frac{1}{2}at^2\).
  2. Substitute: \(s = 2.0(6.0) + 0.5(0.50)(6.0)^2\).
  3. \(s = 12 + 9 = 21\) m.

3.4.4 Interactive Example: Velocity-Time Analysis

The velocity-time graph shows how velocity changes over time. The area under the curve equals displacement.

Hover over the graph to see:

  • Time (t)
  • Velocity (v) at that instant
  • Displacement (s) from the start

Observe:

  • The shaded blue area represents positive displacement
  • The shaded red area (when v < 0) represents displacement in the negative direction
  • The gradient of the line equals the acceleration

3.5 Common Misconceptions

Common Misconceptions

  • Misconception: Speed and velocity are interchangeable. Correction: Velocity includes direction; speed does not.

  • Misconception: A negative velocity means the object is slowing down. Correction: It only indicates direction relative to the axis.

  • Misconception: Zero acceleration means zero velocity. Correction: Zero acceleration means velocity is constant.

3.6 Practice Questions

3.6.1 Easy (2 marks)

A runner moves from 40 m to 10 m in 10 s. Calculate average velocity.

  • Correct displacement and sign (1)
  • Correct division and units (1)

Answer: \(v_{avg} = (10-40)/10 = -3.0\) m/s

3.6.2 Medium (4 marks)

A train travels at 12 m/s, then accelerates uniformly to 20 m/s in 16 s. Find the acceleration and describe its direction.

  • Correct change in velocity (1)
  • Correct acceleration calculation (2)
  • Direction statement (1)

Answer: \(a = (20-12)/16 = 0.50\) m/s\(^2\) in the direction of motion

3.6.3 Hard (5 marks)

An object moves with constant acceleration. It travels 8.0 m in the first 2.0 s and 18.0 m in the next 2.0 s. Determine the initial velocity and acceleration.

  • Correct setup using two displacement equations (2)
  • Correct solution for acceleration (2)
  • Correct initial velocity with units (1)

Hint: Use \(s = ut + \frac{1}{2}at^2\) for each interval.

Answer: \(u = 2.5\) m/s, \(a = 2.5\) m/s\(^2\)

3.7 Multiple Choice Questions

Test your understanding with these interactive questions:

3.8 Summary

Key Takeaways
  • Straight-line motion uses signed quantities to encode direction.
  • Displacement, velocity, and acceleration define the kinematic model.
  • Instantaneous values come from gradients on graphs.
  • Constant acceleration allows predictive equations of motion.

3.9 Self-Assessment

Check your understanding:

After studying this section, you should be able to: