7.1 Describing Motion
Key Takeaways
- Distance is a scalar path length; displacement is a vector from start to finish and can be smaller than distance when the path curves or reverses.
- Average speed equals total distance divided by time; average velocity equals displacement divided by time and includes direction.
- Acceleration is the rate of change of velocity; speeding up, slowing down, or changing direction all count as acceleration.
- On a velocity-time graph for constant acceleration, the slope equals acceleration and the area under the line equals displacement.
- Praxis 5442 often pairs kinematics vocabulary with teaching scenarios that ask which quantity students measured or which graph matches a motion description.
7.1 Describing Motion
Quick Answer: Distance is scalar path length; displacement is the vector from start to finish. Average speed = distance ÷ time; average velocity = displacement ÷ time (needs direction). Acceleration is any change in velocity — speeding up, slowing down, or changing direction. On a constant-acceleration velocity-time graph, slope = acceleration and area under the line = displacement.
Domain II Physical Science makes up about 30% of Praxis Middle School Science (5442). Within II.B.1, ETS expects you to describe motion with the correct quantities — and to catch the common student confusions that show up in teaching-scenario items. This section builds the language of kinematics before Newton's laws appear in 7.2.
On the exam you will not have a calculator. Keep arithmetic simple, track units carefully, and always ask whether the quantity in the stem is a scalar (magnitude only) or a vector (magnitude and direction).
Distance vs. Displacement
| Quantity | Type | Definition | Units (SI) | Path dependence |
|---|---|---|---|---|
| Distance | Scalar | Total length of the path traveled | meters (m) | Depends on the route taken |
| Displacement | Vector | Straight-line change in position from start to finish | meters (m), plus direction | Depends only on initial and final positions |
Distance answers "How far did you travel along the path?" Displacement answers "How far and in which direction are you from where you started?"
Worked example — out and back
A student walks 40 m east to a locker, then 40 m west back to the classroom door.
- Distance = 40 m + 40 m = 80 m
- Displacement = 0 m (final position = initial position)
Worked example — L-shaped path
A cart rolls 3 m north, then 4 m east.
- Distance = 3 + 4 = 7 m
- Displacement magnitude = √(3² + 4²) = √25 = 5 m, directed northeast along the hypotenuse from start to finish
Exam trap: If a stem says "how far from the starting point," that is usually displacement magnitude, not distance. If it says "how far did the object travel," that is distance.
Speed vs. Velocity
| Quantity | Type | Formula | What direction means |
|---|---|---|---|
| Average speed | Scalar | total distance ÷ elapsed time | Speed has no direction |
| Average velocity | Vector | displacement ÷ elapsed time | Velocity includes direction; sign or compass heading matters |
| Instantaneous speed/velocity | — | Value at one moment (speedometer / slope of position-time) | Instantaneous velocity is the slope of position vs. time |
Worked example — same numbers, different meanings
A runner completes a 400 m lap on a circular track in 80 s and finishes at the starting line.
- Average speed = 400 m ÷ 80 s = 5 m/s
- Average velocity = 0 m ÷ 80 s = 0 m/s (displacement is zero)
If the same runner covers only the first 100 m straightaway in 20 s due east:
- Average speed = 100 ÷ 20 = 5 m/s
- Average velocity = 5 m/s east
Teaching-scenario cue: Students often report a "velocity" of 5 m/s with no direction. On Praxis items that ask what feedback you should give, the correct move is to require a direction (or note that they measured speed, not velocity).
Acceleration
Acceleration is the rate at which velocity changes:
[ a = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{t} ]
SI units are m/s². Acceleration is a vector: it has magnitude and direction.
An object accelerates whenever velocity changes — that includes:
- Speeding up in a straight line (velocity and acceleration in the same direction)
- Slowing down (velocity and acceleration in opposite directions — often called deceleration in classrooms, but physicists still call it acceleration)
- Changing direction at constant speed (for example, uniform circular motion)
Worked example — speeding up
A bicycle's velocity increases from 2 m/s to 8 m/s in 3 s in a straight line forward.
- Δv = 8 − 2 = 6 m/s
- a = 6 ÷ 3 = 2 m/s² in the forward direction
Worked example — slowing down
A skateboarder moving +6 m/s (right) comes to rest in 2 s.
- Δv = 0 − 6 = −6 m/s
- a = −6 ÷ 2 = −3 m/s² (acceleration directed left, opposite the velocity)
Constant velocity means zero acceleration
If velocity is constant (same magnitude and direction), acceleration is zero even if the object is moving quickly. A car cruising at a steady 25 m/s on a straight highway has a = 0.
Motion Graphs You Must Read Fluently
Praxis items frequently show position-time or velocity-time graphs and ask what is happening physically.
Position-time (x–t) quick reads
| Graph feature | Meaning |
|---|---|
| Horizontal line | Object at rest (position not changing) |
| Straight slanted line | Constant velocity (slope = velocity) |
| Curve getting steeper | Speeding up |
| Curve getting flatter | Slowing down |
| Negative slope | Motion in the negative direction |
Velocity-time (v–t) for constant acceleration
For constant acceleration, the v–t graph is a straight line.
| Graph feature | Meaning |
|---|---|
| Slope of the line | Acceleration (a = Δv/Δt) |
| Area under the line | Displacement during that interval (signed: below v = 0 counts negative) |
| Horizontal line above v = 0 | Constant positive velocity; a = 0 |
| Line sloping upward | Positive acceleration (speeding up if v > 0) |
| Line sloping downward toward zero | Negative acceleration (slowing down if v > 0) |
| Line crossing through zero | Object reverses direction |
Worked example — reading a constant-acceleration v–t graph
A cart's velocity increases uniformly from 0 m/s to 12 m/s over 4 s, then stays at 12 m/s for another 2 s.
Interval 0–4 s (constant acceleration):
- Slope = a = (12 − 0) ÷ 4 = 3 m/s²
- Displacement ≈ area of triangle = ½ × 4 × 12 = 24 m
Interval 4–6 s (constant velocity):
- Slope = a = 0
- Displacement = rectangle area = 2 × 12 = 24 m
Total displacement in 6 s = 48 m. Average velocity for the whole trip = 48 m ÷ 6 s = 8 m/s.
Connecting equations for constant acceleration
When acceleration is constant, these relationships are consistent with the v–t picture:
| Relationship | Meaning |
|---|---|
| ( v = v_0 + at ) | Final velocity after time t |
| ( \Delta x = v_0 t + \tfrac{1}{2}at^2 ) | Displacement (area under the v–t line) |
| ( v^2 = v_0^2 + 2a\Delta x ) | Useful when time is not given |
You rarely need heavy algebra on 5442, but you should recognize that a steeper v–t slope means larger |a|, and that zero slope means constant velocity.
Classroom Lab Connections (SEP-ready)
Many 5442 items (~40%) integrate Science and Engineering Practices. Motion labs are classic contexts:
- Ticker timers / motion sensors: Students confuse distance and displacement when the cart reverses.
- Constant-velocity carts on a level track: Position-time is linear; velocity-time is horizontal; acceleration is zero.
- Inclined planes: Velocity increases; v–t slope is positive and approximately constant if friction is small.
- Graph interpretation tasks: Give a story ("walks east, stops, returns west") and ask which graph matches — or give a graph and ask which student description is accurate.
Common student (and exam) misconceptions
| Misconception | Accurate idea |
|---|---|
| "If it is moving, it is accelerating" | Constant velocity → a = 0 |
| "Negative velocity means slowing down" | Negative velocity means motion in the negative direction; slowing down depends on whether a opposes v |
| "Distance and displacement are always equal" | Equal only for straight-line motion without reversing |
| "Speed and velocity are interchangeable" | Velocity requires direction; average velocity uses displacement |
| "Steeper position-time means higher acceleration" | Steeper x–t means higher speed/velocity; acceleration is slope of v–t |
Praxis Teaching-Scenario Pattern
A typical stem might show a student claiming a ball thrown straight up has zero acceleration at the top of its flight because "it stops." The correct instructional response: velocity is momentarily zero, but acceleration due to gravity is still about 9.8 m/s² downward. That single idea links this section to free-fall discussion in 7.2 and appears often in middle-grades physical science assessments.
Another frequent pattern: students report "velocity = 3 m/s" after timing a back-and-forth walk that returns to the start. Feedback should emphasize using displacement for average velocity (which is zero) versus distance for average speed.
Master these distinctions and graph readings before moving to forces — Newton's laws explain why velocity changes; kinematics describes how it changes.
A student walks 50 m east, then 20 m west. What is the student's displacement from the starting point?
A car travels around a circular track of circumference 2.0 km and returns to the starting point in 100 s. Which statement is correct?
On a velocity-time graph, a straight line slopes from +8 m/s down to +2 m/s over 3 s. What is the acceleration?
Which classroom observation best indicates that an object has nonzero acceleration?