Why Compare the Three Graphs?
For motion along a line, position tells where an object is, velocity describes how its position changes, and acceleration describes how its velocity changes. Their graphs are connected by derivatives, so they cannot be matched independently.
A position graph can rise while getting flatter, for example. Its velocity is still nonnegative wherever the graph is differentiable, but velocity may equal zero at an instant. A velocity graph, meanwhile, can be above the horizontal axis while sloping downward: the object is moving in the positive direction, but its velocity is decreasing.
The key is to read each graph in the right way. The vertical value of a position graph is position, while its slope is velocity. The vertical value of a velocity graph is velocity, while its slope is acceleration.
What Each Graph Tells You
| Graph | Vertical value | Slope |
|---|---|---|
| Position \(x\) versus time \(t\) | Position \(x(t)\) | Velocity \(v(t)\) |
| Velocity \(v\) versus time \(t\) | Velocity \(v(t)\) | Acceleration \(a(t)\) |
| Acceleration \(a\) versus time \(t\) | Acceleration \(a(t)\) | Rate of change of acceleration, not needed for this matching task |
Keep the axes and units in mind. If position is measured in meters and time in seconds, then velocity has units of meters per second and acceleration has units of meters per second squared. A steep position graph indicates a large velocity magnitude; it does not necessarily indicate a large position.
At a particular time, a positive position-graph slope means positive velocity, a negative slope means negative velocity, and a horizontal tangent means velocity zero. Similarly, a rising velocity graph has positive acceleration wherever its slope is positive; a horizontal velocity graph has zero acceleration; and a falling velocity graph has negative acceleration.
Why Slope Gives the Derivative
The derivative is defined as the limit of average rates of change over shorter and shorter time intervals. For position, the average velocity from \(t\) to \(t+h\) is \(\frac{x(t+h)-x(t)}{h}\). When the limit exists, it is the instantaneous velocity at \(t\). The same reasoning applied to velocity gives acceleration.
On a graph, each quotient is the slope of a secant line. As the time interval shrinks, the secant slopes approach the tangent slope, when that limit exists. This is why matching the slope of the position graph to the velocity value, and the slope of the velocity graph to the acceleration value, is not just a visual shortcut: it follows from the derivative definition.
A Reliable Matching Process
When several graphs are presented, first identify what each axis represents. Then compare features at the same times, rather than judging each curve by its overall appearance alone.
On the position graph, note where the tangent slope is positive, negative, zero, or undefined. These features constrain the velocity graph.
Check whether the velocity graph is above, below, or on the time axis at those same times. Its vertical values must agree with the position slopes.
Check where velocity increases, decreases, or is locally constant. These slopes constrain the sign or value of acceleration.
At horizontal tangents, zeros, corners, and endpoints, check whether the derivative exists and whether the graphs agree about its value.
Worked Examples
Worked Example: Matching a Piecewise-Linear Position Graph
A particle’s position is measured in meters and time in seconds. Its position graph is made of straight segments joining \((0,1)\), \((2,5)\), \((4,5)\), and \((6,2)\), where each ordered pair is \((t,x)\). Find the velocity on the open time intervals and describe what this implies for acceleration.
On each straight segment, velocity is the slope \(\frac{\Delta x}{\Delta t}\).
From \(t=0\) to \(t=2\):
From \(t=2\) to \(t=4\):
From \(t=4\) to \(t=6\):
So the velocity graph is horizontal at \(2\) on \(0<t<2\), at \(0\) on \(2<t<4\), and at \(-1.5\) on \(4<t<6\). Since each velocity is constant on its open interval, its slope there is zero, so \(a=0\) on those intervals.
Check: over the first two seconds, constant velocity \(2\text{ m/s}\) changes position by \(2\cdot2=4\text{ m}\), from \(1\) to \(5\). Over the next two seconds, \(0\cdot2=0\text{ m}\), so position stays at \(5\). Over the last two seconds, \((-1.5)\cdot2=-3\text{ m}\), so position changes from \(5\) to \(2\), as given.
At \(t=2\) and \(t=4\), the position graph has corners: its slope changes abruptly. The ordinary derivative, and hence the instantaneous velocity as defined by that derivative, does not exist at either corner. The velocity graph has jumps there, so its derivative is not defined at those times either. Do not assign a finite acceleration from the slopes of the open intervals at the corners.
Worked Example: Matching Velocity and Acceleration Graphs
A velocity graph consists of straight segments joining \((0,-2)\), \((2,0)\), \((4,3)\), and \((6,3)\), with coordinates \((t,v)\). Find acceleration on each open interval and interpret the signs of velocity and acceleration.
Acceleration is the slope of the velocity graph. From \(t=0\) to \(t=2\):
From \(t=2\) to \(t=4\):
From \(t=4\) to \(t=6\):
Thus an acceleration graph must be horizontal at \(1\) on \(0<t<2\), at \(1.5\) on \(2<t<4\), and at \(0\) on \(4<t<6\). The velocity is negative before \(t=2\), zero at \(t=2\), and positive afterward. Its positive slope means acceleration is positive on the first two open intervals, even while velocity is negative on the first one.
Check: the first velocity change is \(1\text{ m/s}^2\cdot2\text{ s}=2\text{ m/s}\), taking velocity from \(-2\) to \(0\). The second change is \(1.5\cdot2=3\text{ m/s}\), taking it from \(0\) to \(3\). The last change is \(0\cdot2=0\text{ m/s}\), leaving velocity at \(3\).
At \(t=2\) and \(t=4\), the velocity graph changes slope, so its derivative is not defined at those corners. The stated acceleration values apply on the open intervals, not at the corners.
Worked Example: A Horizontal Tangent Does Not Force a Direction Change
Consider a position graph described by \(x(t)=t^3\), with time in seconds and position in meters. Use its derivatives to identify the matching velocity and acceleration behavior, especially at \(t=0\).
Differentiate position to obtain velocity, then differentiate velocity to obtain acceleration:
At \(t=-1\), the position is \(-1\text{ m}\), velocity is \(3\text{ m/s}\), and acceleration is \(-6\text{ m/s}^2\). At \(t=0\), the position, velocity, and acceleration are all \(0\). At \(t=1\), the position is \(1\text{ m}\), velocity is \(3\text{ m/s}\), and acceleration is \(6\text{ m/s}^2\).
For every \(t\), \(v(t)=3t^2\geq0\), so the position graph is increasing; at \(t=0\), however, its slope and velocity are exactly zero. The velocity graph touches the time axis at \(0\), while decreasing before \(0\) and increasing after \(0\). Accordingly, the acceleration is negative before \(0\), zero at \(0\), and positive after \(0\).
Check the derivative relationships directly: the position slope is \(3t^2\), exactly the vertical value of the velocity graph; the velocity slope is \(6t\), exactly the vertical value of the acceleration graph. At the sample times these give \(v(-1)=3\), \(v(0)=0\), \(v(1)=3\), and \(a(-1)=-6\), \(a(0)=0\), \(a(1)=6\).
Sign and Shape Checks
Several quick checks can eliminate an incorrect graph match:
- If a position graph has positive slope at a time, its velocity value there must be positive. A negative slope requires negative velocity.
- If position is increasing on an interval and is differentiable there, its derivative is nonnegative at interior points where the derivative exists; it may equal zero at some points. Decreasing position similarly implies a nonpositive derivative wherever it exists.
- If position has a local maximum or minimum at an interior point and is differentiable there, its tangent slope is zero, so velocity is zero at that point. A zero velocity alone does not guarantee a local maximum or minimum.
- If velocity is rising, acceleration is positive where the slope exists; if velocity is falling, acceleration is negative. A horizontal velocity segment has zero acceleration in its interior.
- A velocity graph crossing the time axis shows velocity changing sign, which is consistent with a change in direction. Merely touching the axis does not show a sign change.
Common Mistake: Confusing Increasing with Positive Slope Everywhere
A common incorrect claim is: “The position graph is increasing, so velocity is strictly positive at every point.” That conclusion is too strong. For the position graph \(x(t)=t^3\), the graph is increasing, but at \(t=0\) its slope is \(x'(0)=3(0)^2=0\). Thus velocity is nonnegative, not strictly positive at every point.
The correction is to distinguish an interval trend from a pointwise derivative. Increasing position implies a nonnegative derivative where it exists, but some tangent slopes may be zero. Conversely, a strictly positive slope at a particular time guarantees positive velocity at that time. Apply the corresponding distinction to decreasing graphs: their derivatives are nonpositive, and may be zero at some points.
Another error is to read a velocity graph’s height as acceleration. For example, a velocity graph at \(v=4\text{ m/s}\) gives the velocity value, not acceleration. To find acceleration, examine the graph’s slope. A horizontal segment at \(v=4\text{ m/s}\) has acceleration zero, not \(4\text{ m/s}^2\).
Why the Connections Matter
Matching the graphs gives several ways to check a motion description without relying on one curve alone. A proposed position graph determines the velocity through its slopes; the velocity graph in turn determines acceleration through its slopes. If one proposed graph contradicts either relationship, the set cannot represent the same motion.
Pay particular attention to the exact time of a feature. A position graph’s horizontal tangent must correspond to a zero velocity value at that time, if the derivative exists. A velocity graph’s local extremum with a differentiable tangent has zero acceleration at that time. Corners require care: a sharp corner in position may mean velocity is undefined, while a corner in velocity may mean acceleration is undefined. These are not ordinary zero values.
Check Your Understanding
Use derivative relationships to answer each question. Assume the relevant derivatives exist unless the question says otherwise.
- A position graph has a negative tangent slope at \(t=3\). What must be true of velocity at \(t=3\)?
- A position graph is increasing on an interval. Must velocity be strictly positive at every point of that interval? Explain.
- A velocity graph is horizontal at \(v=-2\text{ m/s}\). What is the acceleration on that interval?
- A velocity graph has a negative slope at a time when its value is positive. State the signs of velocity and acceleration.
- What does a corner in a position graph imply about the derivative-based velocity at that time?
