Difficulty: Introductory | Prerequisites: Basic algebra, familiarity with Cartesian graphs.
Kinematics is the branch of mechanics that describes how objects move, without worrying about why they move. In PHYS 114, this is the foundation everything else builds on: forces, energy, and momentum all assume you can already describe motion precisely. Position-time graphs are the single most important visual tool in 1D kinematics, because they encode position, velocity, and acceleration all in one picture. If you missed the first few weeks or are shaky on graph reading, start here before tackling Newton's laws.
A position-time graph tells you everything about 1D motion. The slope at any point gives you the velocity at that instant, and the way the slope changes tells you about acceleration. Average velocity is just total displacement divided by total time, read straight off the graph.
Position (x)
The location of an object along a chosen axis at a given time, measured in metres (m). Think of it as "where on the number line the object sits right now."
Displacement (Δx)
The change in position: Δx = x_final - x_initial. It is a signed quantity (positive or negative depending on direction). In simple terms, this means how far and in which direction you have moved from where you started.
Average velocity (v_avg)
The total displacement divided by the total time interval: v_avg = Δx / Δt. It tells you the overall rate of position change across an interval, not what happens at any single moment. Think of it as the slope of the straight line connecting two points on a position-time graph.
Instantaneous velocity
The velocity at one specific moment in time, equal to the slope of the tangent line to the position-time graph at that point. In simple terms, this is what your speedometer reads (with a sign for direction).
Instantaneous acceleration
The rate at which velocity is changing at a single instant. On a position-time graph, this relates to the curvature: a straight-line segment has zero instantaneous acceleration, while a curved segment has nonzero acceleration. Think of it as how quickly the slope of the position-time graph is itself changing.
Position-time graph (x vs t graph)
A plot with time on the horizontal axis and position on the vertical axis. It is the central tool for analysing 1D motion. Every feature of the graph, its slope, its curvature, its intercepts, maps directly onto a physical quantity.
The vertical axis is position (in metres); the horizontal axis is time (in seconds).
A straight, upward-sloping line means constant positive velocity.
A straight, downward-sloping line means constant negative velocity (moving in the negative direction).
A horizontal line means the object is stationary (velocity = 0).
A curved section means the velocity is changing, which means acceleration is present.
Pick two points on the graph at times t₁ and t₂.
Read off the corresponding positions x₁ and x₂.
Compute: v_avg = (x₂ - x₁) / (t₂ - t₁).
This is the slope of the secant line (the straight line connecting the two points).
Example from the exam: between t = 2 s (x = 40 m) and t = 6 s (x = 20 m), v_avg = (20 - 40) / (6 - 2) = -20 / 4 = -5.0 m/s. The negative sign means the object moved in the negative x-direction on average.
Instantaneous acceleration is the rate of change of the slope of the position-time graph.
At any point where the graph is a straight line (even if sloped), the slope is constant, so the instantaneous acceleration is zero.
Acceleration is nonzero only where the graph curves.
Example from the exam: at t = 2 s, 4 s, and 5 s the graph consisted of straight-line segments. The slope was not changing at any of these instants, so the instantaneous acceleration was zero at all three.
v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{t_2 - t_1}v_{\text{inst}} = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}a_{\text{inst}} = \frac{dv}{dt} = \frac{d^2 x}{dt^2}On a position-time graph, v_avg is the slope of the secant line between two points, v_inst is the slope of the tangent line at one point, and a_inst corresponds to the curvature (concavity) of the graph.
Position-time graphs are exactly how GPS tracking systems log your location over time. The slope of that trace is your speed. Traffic engineers use these graphs to study congestion: a flat section means you were stuck, a steep section means you were moving freely.
Students often confuse the height of the graph with speed. A high position value does not mean the object is moving fast. The object could be sitting still at a high position (flat line at y = 60 m).
Students often think a downward slope means the object is decelerating. It does not. A downward slope means the object is moving in the negative direction. It might be speeding up in that direction.
Students often think average velocity equals the average of two velocities. It does not, unless acceleration is constant. Average velocity is displacement over time, full stop.
Students often read instantaneous acceleration as the slope of the position-time graph. It is not. The slope gives velocity. Acceleration is the rate at which that slope changes.
⚠️ Expect to be given a position-time graph and asked for average velocity over a specific interval. Know the formula and be careful with signs.
⚠️ A common exam question asks when instantaneous acceleration is zero. Look for straight-line segments on the position-time graph, because straight lines have constant slope (no change in velocity).
⚠️ Negative velocity does not mean slowing down. The exam will try to trip you up on the difference between "moving in the negative direction" and "decelerating."
True or false: the slope of a position-time graph gives you the acceleration.
False. The slope gives velocity. The rate of change of the slope gives acceleration.
Fill in the blank: average velocity equals ______ divided by ______.
Displacement (Δx) divided by the time interval (Δt).
True or false: if a position-time graph is a straight line with a negative slope, the instantaneous acceleration is negative.
False. A straight line (any slope) means zero acceleration.
True or false: an object at position x = 100 m must be moving faster than one at x = 5 m.
False. Position tells you where the object is, not how fast it is moving.
Q: A particle's position-time graph shows it at x = 40 m at t = 2 s and at x = 20 m at t = 6 s. What is the average velocity over this interval?
A: v_avg = (20 - 40) / (6 - 2) = -20 / 4 = -5.0 m/s. The negative sign indicates motion in the negative x-direction.
Q: On a piecewise-linear position-time graph, at points where the graph changes slope ("kinks"), is the instantaneous acceleration defined?
A: Strictly, no. The velocity changes abruptly at a kink, so the acceleration is not well-defined at that exact instant. In practice, exams typically ask about points on the straight segments, where acceleration is clearly zero.
Q: A position-time graph shows a straight line from (0 s, 0 m) to (5 s, 30 m). What are the average and instantaneous velocities?
A: Both are 6.0 m/s. For a straight-line graph (constant velocity), the average and instantaneous velocities are the same everywhere.
Q: Consider instants at t = 2 s, t = 4 s, and t = 5 s on a position-time graph made entirely of straight-line segments. At which instant is the instantaneous acceleration greatest?
A: The acceleration is zero at all three instants, because the graph is linear at each of those points. The slope is constant at each instant, so dv/dt = 0.
This material connects directly to 2D kinematics and projectile motion, where you apply the same ideas (displacement, velocity, acceleration) separately to x and y components. It also underpins Newton's second law: once you can describe motion precisely, forces become the explanation for why motion changes. If you understand slopes and curvature on a position-time graph, you already have the intuition for velocity-time and acceleration-time graphs, which appear later in the course.
Position-time graph, x-t graph, displacement vs distance, average velocity vs average speed, instantaneous velocity, tangent line slope, secant line slope, constant velocity, zero acceleration, piecewise linear motion, 1D kinematics, PHYS 114, general physics, University of Washington