Velocity-Time Graphs and How the Three Motion Graphs Connect

If the slope of a position-time graph gives you velocity, what do you think the slope of a velocity-time graph gives you? Descartes' whole coordinate-plane idea never cared what was on the axes — position vs. time was just one choice. Following that same logic one level up, with a new pair of axes, reveals one of the most useful tools in kinematics.

On a velocity-time graph, the slope equals acceleration — the same relationship as before, just one derivative up the chain. But velocity-time graphs hold a second piece of information too: the area under the curve equals displacement. Picture a car speeding up from 0 to 20 m/s over 5 seconds, plotted as a straight diagonal line. The slope of that line (4 m/s²) is the acceleration. The triangular area underneath that same line works out to the total distance the car covered while speeding up. Slope and area — two different geometric features of the same graph, two different physical quantities.

This sets up a bigger idea: any single motion can be described by three connected graphs — position vs. time, velocity vs. time, and acceleration vs. time — and each one can be built from either of the others. Take an object starting at rest with constant positive acceleration: the acceleration-time graph is a flat horizontal line above zero (constant acceleration). The velocity-time graph is a straight line sloping upward from zero, since velocity keeps increasing at a steady rate. The position-time graph is the trickiest of the three — it's a curve that gets steeper and steeper over time, not a straight line, because the object's velocity itself is constantly increasing. Being able to move fluidly between all three representations — sketch any one given either of the other two — is one of the most tested skills in this unit.

Try It: Graph Runner

Play Area = Displacement to predict displacement and distance from a velocity-time graph, then Read the Story to match motions to graphs. Kinematics Castle: Graph Runner

Videos

• Motion Graphs - AP Physics 1: Unit 1 Review Supplement — Flipping Physics (YouTube)

Practice Problems

1. A v-t graph is a flat line at 10 m/s for 4 seconds. Find the displacement during this time.
2. A v-t graph starts at 0 and rises to 15 m/s over 3 seconds. Find the acceleration.
3. Using the graph from problem 2, find the displacement over those 3 seconds.
4. An a-t graph is a flat line at +2 m/s² for 5 seconds, starting with an initial velocity of 0. Sketch the matching v-t graph and label the velocity at t = 5 s.
5. A v-t graph shows a straight line decreasing from 20 m/s to 0 over 4 seconds. Find the acceleration and sketch the matching a-t graph.
6. A v-t graph has a rectangle section (constant 8 m/s for 3 s) followed by a triangle section (decreasing from 8 m/s to 0 over 2 s). a) Find the total displacement (sum of both areas). b) Explain in words what the object was doing during each section. c) Sketch the matching a-t graph for the full 5 seconds.
7. A classmate sketches an x-t graph as a straight line with constant positive slope, but sketches the matching v-t graph as also having a constant positive slope (not flat). Explain what's wrong with this classmate's v-t sketch.
8. A student wants to determine an object's acceleration at the exact instant t = 5 s, given a curved velocity-time graph (not a straight line). Which of the following methods would give the most accurate estimate of the instantaneous acceleration at t = 5 s? (A) Find the slope of the line connecting the points at t = 0 s and t = 10 s. (B) Find the slope of the line connecting the points at t = 4.9 s and t = 5.1 s. (C) Find the total area under the curve from t = 0 s to t = 10 s and divide by 10 s. (D) Find the average velocity between t = 0 s and t = 5 s.