Lesson 1.3 · Unit 1 · CED topic 1.3 (representations)
Representing motion with graphs
The same motion can be told in several languages: a motion diagram, a story in words, an equation, or a graph. AP Physics 1 expects you to move freely among them, and the exam often hands you one representation and asks for another. Graphs are the most compact of these languages, because a single picture shows how position, velocity and acceleration change together. This lesson teaches you to read the three motion graphs by asking two questions of any point or region: what is the slope here? and what is the area here?
Describe the position, velocity, and acceleration of an object using representations of that object’s motion.
- Motion can be represented by motion diagrams, figures, graphs, equations, and narrative descriptions.
- Graphs of position, velocity, and acceleration as functions of time can be used to find the relationships between those quantities.
- An object’s instantaneous velocity is the rate of change of the object’s position, which is equal to the slope of a line tangent to a point on a graph of the object’s position as a function of time.
- An object’s instantaneous acceleration is the rate of change of the object’s velocity, which is equal to the slope of a line tangent to a point on a graph of the object’s velocity as a function of time.
- The displacement of an object during a time interval is equal to the area under the curve of a graph of the object’s velocity as a function of time.
- The change in velocity of an object during a time interval is equal to the area under the curve of a graph of the acceleration of the object as a function of time.
AP Physics 1 does not expect students to quantitatively analyze nonuniform acceleration. However, students will be expected to be able to qualitatively analyze, sketch appropriate graphs of, and discuss situations in which acceleration is nonuniform.
Position against time: the slope is the velocity
On a graph of \(x\) against \(t\), the slope is rise over run, \(\Delta x/\Delta t\), which is exactly the definition of velocity. A steeper line means a larger magnitude of velocity. A line sloping upward to the right means positive velocity, a line sloping downward means negative velocity, and a horizontal line means the object is at rest. When the graph is a curve, the velocity changes from moment to moment, and the velocity at a particular instant is the slope of the tangent line drawn at that point, the straight line that just touches the curve there (Figure 1.2c in the last lesson showed secant lines closing in on the tangent).
Read curves by describing how the slope changes. A curve that gets steeper to the right means the velocity is growing. A curve that gets flatter means the velocity is shrinking. A hilltop, where the slope passes through zero from positive to negative, is the moment the object turns around.
Velocity against time: slope is acceleration, area is displacement
The same two questions apply one level up. On a graph of \(v\) against \(t\), the slope is \(\Delta v/\Delta t\), the acceleration. A horizontal line is constant velocity, a straight sloped line is constant acceleration, and a curved line is changing acceleration. The area between the graph and the time axis is the displacement, because velocity times time is displacement. Area above the axis is positive displacement (moving in the positive direction) and area below the axis is negative displacement.
The vertical position of the line also carries information. A point above the axis means positive velocity; below means negative; where the line crosses the axis, the object is momentarily at rest and its direction reverses. The line can be rising while it is below the axis: that is a negative velocity getting closer to zero, which means the object is slowing down while moving in the negative direction.
Acceleration against time: the area is the change in velocity
On a graph of \(a\) against \(t\), the area between the graph and the time axis is \(a\Delta t = \Delta v\), the change in velocity. The graph does not tell you the velocity itself; you also need the starting velocity. Constant acceleration appears as a horizontal line, and zero acceleration as a line lying on the axis.
| Graph | Slope gives | Area gives |
|---|---|---|
| position against time | velocity | (nothing useful) |
| velocity against time | acceleration | displacement |
| acceleration against time | (rate of change of acceleration, not used here) | change in velocity |
Figure 1.3b shows one trip in all three forms. Reading across the dashed lines, you can follow how a feature of one graph produces a feature of the next. A corner in the velocity graph (at 4 s and 8 s) becomes a jump in the acceleration graph, because the acceleration changed suddenly, and a change of slope in the position graph, because the velocity changed. A flat piece of the velocity graph becomes a zero on the acceleration graph and a straight sloped line on the position graph. Each graph is the slope of the one above it, and the area under each graph builds the one above.
Nonuniform acceleration, qualitatively
Real motion rarely has constant acceleration. A car that accelerates from a stoplight gets less acceleration as it gains speed, so its velocity graph curves over and flattens. AP Physics 1 does not ask you to calculate with such graphs, only to describe and sketch them. Use the same two questions. If the \(v\)–\(t\) curve gets flatter, the slope, and so the acceleration, is decreasing. If it gets steeper, the acceleration is increasing. The position graph in that case bends more and more gently upward, with the slope still increasing but at a slower rate.
Figure 1.3d shows an object’s position over 8 s. Describe the motion, find the velocity in each interval, and find the average velocity and average speed for the whole trip.
Step 1, slopes. From 0 to 2 s the position rises from 0 to 8 m: \(v = 8/2 = +4\ \text{m/s}\). From 2 s to 4 s the line is flat: \(v = 0\). From 4 s to 8 s the position falls from 8 m to \(-4\ \text{m}\): \(v = (-4 - 8)/4 = -3\ \text{m/s}\).
Step 2, story. The object moves forward at 4 m/s, stops for 2 s, then moves backward at 3 m/s, passing through the origin and ending 4 m on the negative side.
Step 3, whole trip. The displacement is \(-4 - 0 = -4\ \text{m}\), so \(v_{\text{avg}} = -4/8 = -0.50\ \text{m/s}\). The distance is \(8 + 0 + 12 = 20\ \text{m}\), so the average speed is \(20/8 = 2.5\ \text{m/s}\). They differ because the object turned around.
Using the velocity graph in Figure 1.3b, find the displacement from 0 to 10 s and the average velocity.
Step 1, areas. The graph is a triangle (0 to 4 s, rising to 8 m/s), a rectangle (4 s to 8 s at 8 m/s) and a triangle (8 s to 10 s, falling to zero), shown in Figure 1.3c.
Step 2. \(\tfrac12(4)(8) = 16\ \text{m}\); \((4)(8) = 32\ \text{m}\); \(\tfrac12(2)(8) = 8\ \text{m}\). The displacement is \(16 + 32 + 8 = 56\ \text{m}\).
Step 3. \(v_{\text{avg}} = 56/10 = 5.6\ \text{m/s}\), which is less than the 8 m/s maximum, as an average over a trip with slow beginning and end should be.
A cart’s velocity falls steadily from \(+6\ \text{m/s}\) at \(t = 0\) to \(-6\ \text{m/s}\) at \(t = 6\ \text{s}\) (Figure 1.3e). Find its acceleration, the time it is at rest, its displacement and the distance it travels.
Acceleration. The slope is \((-6 - 6)/6 = -2\ \text{m/s}^2\).
Rest. The velocity is zero where the line crosses the axis, at \(t = 3\ \text{s}\), because \(6 - 2t = 0\).
Displacement. The triangle above the axis has area \(\tfrac12(3)(6) = +9\ \text{m}\); the one below has area \(-\tfrac12(3)(6) = -9\ \text{m}\). The total is \(0\): the cart returns to its start.
Distance. Distance counts both areas as positive: \(9 + 9 = 18\ \text{m}\).
- Say which graph you are reading. “The slope is the velocity” is correct on an \(x\)–\(t\) graph and wrong on a \(v\)–\(t\) graph, where the slope is the acceleration.
- Signed area. Area below the time axis is negative displacement. For distance, add the absolute values of the areas.
- Axis crossing is not “stopping for good.” A velocity graph crossing zero means the object is momentarily at rest and reversing.
- Check the axes. Read the labels and the scale before using a graph. A graph of speed (always positive) is different from a graph of velocity.
- Sketching. When asked to sketch a second graph from the first, mark where slopes change sign and where they are zero, and keep the same time scale on every graph.
- A position-time graph is a straight line from \((0\ \text{s}, 2\ \text{m})\) to \((4\ \text{s}, 14\ \text{m})\). Find the velocity.
Show answer
The slope is \((14 - 2)/(4 - 0) = +3.0\ \text{m/s}\). The velocity is constant at +3.0 m/s.
- A velocity graph is constant at \(+5\ \text{m/s}\) from 0 to 6 s and then falls straight to zero at 10 s. Find the total displacement.
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Rectangle: \(5 \times 6 = 30\ \text{m}\). Triangle: \(\tfrac12(4)(5) = 10\ \text{m}\). Total \(= 40\ \text{m}\).
- An acceleration graph shows \(+3.0\ \text{m/s}^2\) for 4.0 s. The object started at \(+2.0\ \text{m/s}\). What is its velocity at the end of the 4.0 s?
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The area under the graph is \((3.0)(4.0) = 12\ \text{m/s}\), the change in velocity. The final velocity is \(2.0 + 12 = +14\ \text{m/s}\).
- A velocity-time graph is a straight line sloping upward from \(-4\ \text{m/s}\) at \(t = 0\) to \(+4\ \text{m/s}\) at \(t = 4\ \text{s}\). Is the object speeding up or slowing down during the first 2 s? Explain.
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It is slowing down. Its velocity is negative but moving toward zero, so the speed decreases from 4 m/s to 0. The slope (acceleration) is \(+2\ \text{m/s}^2\), opposite in sign to the negative velocity, which is the condition for slowing down.
- Describe in words the motion shown by an \(x\)–\(t\) graph that starts at the origin, curves upward with an increasing slope for 3 s, then continues as a straight line.
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For the first 3 s the slope (velocity) keeps increasing, so the object speeds up with positive acceleration. After 3 s the slope is constant, so the object moves at a constant velocity equal to the slope it had reached at 3 s. The acceleration is zero.
- Two cars A and B have identical \(v\)–\(t\) graphs except that A’s line is steeper. Which has the greater acceleration? Which has the greater displacement after 5 s if both start from rest?
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A has the greater acceleration because the slope of a velocity graph is acceleration. Starting from rest, A also has the larger velocity at every later time, so the area under its graph, and therefore its displacement after 5 s, is larger.