What a Velocity-Time Graph Shows
A velocity-time graph displays a particle’s velocity \(v\) as time \(t\) changes. The horizontal coordinate is time, and the vertical coordinate is velocity. A point \((t,v)\) therefore tells you the particle’s velocity at time \(t\).
Velocity is signed: its sign tells which coordinate direction the particle is moving. Speed is the magnitude of velocity, so it is never negative. The slope of the velocity-time graph describes how velocity changes with time; in particle motion, that rate of change is acceleration.
Reading Velocity and Speed
Let \(v(t)\) denote the particle’s velocity at time \(t\). If \(v(t)>0\), the particle moves in the positive coordinate direction. If \(v(t)<0\), it moves in the negative coordinate direction. If \(v(t)=0\), its instantaneous velocity is zero at that moment.
The speed at time \(t\) is \(\lvert v(t)\rvert\). For example, a velocity of \(-5\) meters per second means motion in the negative direction at a speed of \(5\) meters per second. The negative sign describes direction, not a negative amount of speed.
To read these quantities from a graph, first locate the specified time on the horizontal axis, then read the graph’s vertical coordinate. A point above the horizontal axis has positive velocity; a point below it has negative velocity. The vertical distance from the horizontal axis gives the speed.
Why Slope Represents Acceleration
For two times \(t_1\) and \(t_2\), the average acceleration on the interval is the change in velocity divided by the change in time. On a velocity-time graph, that is exactly the slope of the secant line through the two corresponding points.
The vertical change is a change in velocity, and the horizontal change is a change in time. Thus the units of the slope are velocity units divided by time units. If velocity is measured in meters per second and time in seconds, the slope has units of meters per second per second, or meters per second squared.
At a single time, instantaneous acceleration is the derivative of velocity, when that derivative exists. Graphically, it is the slope of the tangent line to the velocity-time graph at that time. The secant slope gives average acceleration over an interval; a tangent slope gives instantaneous acceleration at a particular time.
This interpretation follows from the meaning of a derivative. Acceleration is the rate of change of velocity with respect to time, and a derivative is the instantaneous rate of change. Therefore, the derivative of velocity is the instantaneous acceleration. On the graph, the derivative is represented by tangent slope.
At the stated time, find the graph’s velocity \(v(t)\).
The sign gives coordinate direction, and \(\lvert v(t)\rvert\) gives speed.
Use a secant slope for average acceleration over an interval or a tangent slope for instantaneous acceleration.
Report acceleration in velocity-units per time-unit and identify whether it is average or instantaneous.
Worked Examples
Worked Example: Reading a Piecewise-Linear Velocity Graph
A velocity-time graph consists of straight segments joining \((0,2)\), \((2,6)\), \((5,6)\), and \((7,0)\), where time is in seconds and velocity is in meters per second. Find the direction, speed, and acceleration at \(t=1\), \(t=3\), and on the interval from \(t=5\) to \(t=7\).
At \(t=1\): The first segment runs from \((0,2)\) to \((2,6)\). Its slope is
Since the segment is straight, its velocity changes at a constant rate. At \(t=1\), one second after \(t=0\), the velocity is \(2+(2)(1)=4\text{ m/s}\). This is positive, so the particle moves in the positive direction; its speed is \(\lvert 4\rvert=4\text{ m/s}\). Its acceleration is the segment’s slope, \(2\text{ m/s}^2\).
At \(t=3\): The graph is horizontal from \(t=2\) to \(t=5\), at \(v=6\text{ m/s}\). Thus the direction is positive, the speed is \(6\text{ m/s}\), and the slope—and therefore acceleration—is \(0\text{ m/s}^2\).
From \(t=5\) to \(t=7\): The velocity changes from \(6\text{ m/s}\) to \(0\text{ m/s}\). The average acceleration is
Because this whole section is a straight segment, its slope is constant, so the acceleration is \(-3\text{ m/s}^2\) throughout the segment, including its interior. At the endpoint \(t=7\), a corner may prevent a two-sided derivative from existing, so the graph alone does not give a unique instantaneous acceleration there.
Check: The first segment increases by \(4\text{ m/s}\) over \(2\text{ s}\), giving \(2\text{ m/s}^2\). The final segment decreases by \(6\text{ m/s}\) over \(2\text{ s}\), giving \(-3\text{ m/s}^2\). These match the calculated slopes.
Worked Example: Finding Average Acceleration and Speed from Two Graph Points
A velocity-time graph passes through \((1,-4)\) and \((3,-1)\), with time in seconds and velocity in meters per second. Find the speed and direction at each listed time, then find the average acceleration from \(t=1\) to \(t=3\).
At \(t=1\), the velocity is \(-4\text{ m/s}\). The negative sign means motion in the negative coordinate direction, and the speed is \(\lvert-4\rvert=4\text{ m/s}\).
At \(t=3\), the velocity is \(-1\text{ m/s}\). The direction is still negative, and the speed is \(\lvert-1\rvert=1\text{ m/s}\).
The average acceleration over the interval is the secant slope:
Check: The change in velocity is \(-1-(-4)=3\text{ m/s}\), and the elapsed time is \(2\text{ s}\). Dividing gives \(3\text{ m/s}/2\text{ s}=1.5\text{ m/s}^2\), as found. The calculation gives average acceleration over the interval; without more information about the graph between the two points, it does not determine the instantaneous acceleration at every time there.
Worked Example: Reading Instantaneous Acceleration from a Tangent
At \(t=4\text{ s}\), a curved velocity-time graph has velocity \(1\text{ m/s}\). The tangent line at that point passes through \((2,3)\) and \((6,-1)\), with time in seconds and velocity in meters per second. Find the particle’s direction and speed at \(t=4\), and its instantaneous acceleration there.
The velocity at the specified time is \(1\text{ m/s}\), which is positive. The particle moves in the positive direction, and its speed is \(\lvert1\rvert=1\text{ m/s}\).
The instantaneous acceleration is the tangent slope. Using the two points on the tangent line,
Check: The tangent’s vertical change is \(-4\text{ m/s}\) while its horizontal change is \(4\text{ s}\). Their quotient is \(-1\text{ m/s}^2\), so the slope calculation agrees with the instantaneous acceleration indicated by the tangent.
Reading Graphs Carefully
A graph may be horizontal, slanted, or curved. A horizontal section has slope zero, so the velocity is constant there and the acceleration is zero on the interior of that section. A straight slanted section has constant slope, so its acceleration is constant on the section’s interior. A curved section generally has changing tangent slopes, so the instantaneous acceleration can vary with time.
At a corner, the left-hand and right-hand slopes may differ. In that case the graph does not have one tangent slope at the corner, and instantaneous acceleration may be undefined at that exact time. You can still calculate average acceleration across an interval using its endpoints.
Always distinguish the graph’s value from its slope. The vertical coordinate is velocity, not acceleration. The slope is acceleration, not velocity. They also have different units: velocity might be measured in meters per second, while acceleration is measured in meters per second squared.
Common Mistake: Reversing the Slope
Suppose a graph moves from \((1,-2)\) to \((5,6)\), with time in seconds and velocity in meters per second. A common incorrect calculation reverses the change and divides time change by velocity change:
This is not the graph’s slope: it divides seconds by meters per second, giving units of \(\text{s}^2/\text{m}\). For acceleration, divide the change in velocity by the change in time:
Correction: The slope must be \(\frac{\Delta v}{\Delta t}\), not \(\frac{\Delta t}{\Delta v}\). Also, do not report a negative velocity as a negative speed: if \(v=-2\text{ m/s}\), the speed is \(\lvert-2\rvert=2\text{ m/s}\).
Why These Interpretations Matter
Velocity-time graphs combine several kinds of information in one picture. A point gives the particle’s velocity at an instant; its sign and magnitude describe direction and speed. The graph’s slope describes how velocity changes, which is acceleration. Keeping these readings separate prevents a common source of errors in contextual questions.
When a prompt specifies an interval, check whether it asks for average acceleration or describes a straight graph segment whose slope is constant. When it specifies a single time, use the tangent slope if the graph is differentiable there. In either case, include units and make clear what the number represents.
Check Your Understanding
Use the sign, magnitude, and slope of a velocity-time graph to answer each question.
- A graph shows \(v(2)=-7\text{ m/s}\). What are the particle’s direction and speed at \(t=2\text{ s}\)?
- A graph passes through \((0,1)\) and \((4,9)\), with time in seconds and velocity in meters per second. What is the average acceleration from \(t=0\) to \(t=4\)?
- What does a horizontal section of a velocity-time graph tell you about velocity and acceleration on the interior of that section?
- A curved graph has a tangent slope of \(3\text{ m/s}^2\) at \(t=5\text{ s}\). What quantity does this slope represent?
- Why might the graph fail to determine instantaneous acceleration at a sharp corner?
