What an Acceleration-Time Graph Tells You
An acceleration-time graph shows acceleration as time changes. Because acceleration is the derivative of velocity, the graph tells you the rate at which velocity changes. Its vertical position relative to the time axis determines whether velocity is increasing, decreasing, or momentarily changing at a rate of zero.
This is different from a velocity-time graph. A velocity-time graph gives velocity values directly; an acceleration-time graph gives the rate of change of velocity. The graph alone does not generally tell you the particle’s velocity at a particular time unless an initial velocity or other velocity information is also provided.
These statements concern the trend in velocity, not whether velocity itself is positive or negative. In particular, negative acceleration means that velocity is decreasing; it does not, by itself, mean that velocity is negative.
Read the Sign of Acceleration
First locate the horizontal line \(a=0\), which is the time axis on an acceleration-time graph. On an interval where the graph lies above that line, acceleration is positive and velocity increases. Where it lies below, acceleration is negative and velocity decreases. Where the graph coincides with the line throughout an interval, acceleration is zero and velocity stays constant.
| Position of the acceleration graph | Conclusion about velocity |
|---|---|
| Above the time axis | Velocity is increasing |
| Below the time axis | Velocity is decreasing |
| On the time axis throughout an interval | Velocity is constant |
The intervals matter. If the acceleration graph crosses the time axis, velocity may decrease on one side of the crossing and increase on the other. A single point where \(a(t)=0\) does not mean that velocity is constant over an interval; it means only that the instantaneous rate of change is zero at that time.
Why the Sign Determines the Trend
The conclusion follows from the definition of derivative. Suppose \(a(t)=v'(t)>0\) at every interior point of an interval. Choose any two times \(t_1<t_2\) in that interval. Since \(v\) is continuous on \([t_1,t_2]\) and differentiable between the endpoints, the Mean Value Theorem gives a time \(c\) between them such that
Here \(v'(c)>0\) and \(t_2-t_1>0\), so \(v(t_2)-v(t_1)>0\). Thus \(v(t_2)>v(t_1)\) whenever \(t_2>t_1\): velocity is increasing. If \(a(t)<0\) throughout the interval, the same equation gives \(v(t_2)-v(t_1)<0\), so velocity is decreasing. If \(a(t)=0\) throughout, the Mean Value Theorem gives \(v(t_2)-v(t_1)=0\), so velocity is constant.
Signed Area and Change in Velocity
The sign of acceleration tells you the trend. The signed area between the acceleration graph and the time axis tells you the amount of velocity change over an interval. Area above the axis counts positively; area below counts negatively. Thus positive signed area means a positive change in velocity, while negative signed area means a negative change.
This relationship follows from \(a(t)=v'(t)\) and the Fundamental Theorem of Calculus: the integral of the derivative of velocity over an interval is the final velocity minus the initial velocity. For a graph made of constant or linear pieces, the signed area can often be found with rectangles, triangles, or trapezoids.
Worked Examples
Worked Example: Constant Acceleration on Two Intervals
A particle has initial velocity \(v(0)=4\) meters per second. Its acceleration is \(2\) meters per second squared from \(t=0\) to \(t=3\) seconds, then \(-1\) meter per second squared from \(t=3\) to \(t=5\) seconds. Describe the velocity trend and find the velocity at \(t=3\) and \(t=5\).
On \(0<t<3\), acceleration is \(2>0\), so velocity increases. The graph makes a rectangle above the time axis with width \(3\) seconds and height \(2\) meters per second squared. Its signed area is
Therefore, the velocity at \(t=3\) is
On \(3<t<5\), acceleration is \(-1<0\), so velocity decreases. The signed area is
Thus
Check: The changes are \(10-4=6\) meters per second on the first interval and \(8-10=-2\) meters per second on the second. These match the two signed areas. The velocity increases first and decreases next, as the acceleration signs predict.
Worked Example: Acceleration Crossing the Time Axis
On \(0\le t\le4\) seconds, an acceleration graph is the line \(a(t)=t-2\), measured in meters per second squared. The initial velocity is \(v(0)=6\) meters per second. Find where velocity decreases and increases, and calculate \(v(2)\) and \(v(4)\).
Set the acceleration equal to zero to locate the crossing:
For \(0<t<2\), \(t-2<0\), so velocity decreases. The acceleration graph forms a triangle below the axis from \(t=0\) to \(t=2\). Its base is \(2\) seconds and its height is \(2\) meters per second squared, so its signed area is
Hence
For \(2<t<4\), \(t-2>0\), so velocity increases. The graph forms a triangle above the axis with base \(2\) seconds and height \(2\) meters per second squared. Its signed area is
Therefore,
Check: An antiderivative of \(a(t)=t-2\) is \(v(t)=\frac12t^2-2t+C\). Using \(v(0)=6\) gives \(C=6\), so \(v(t)=\frac12t^2-2t+6\). Differentiating gives \(v'(t)=t-2=a(t)\). Also, \(v(2)=2-4+6=4\) and \(v(4)=8-8+6=6\), matching the area calculations.
Worked Example: Zero Acceleration Followed by Positive and Negative Acceleration
A particle has velocity \(v(0)=5\) meters per second. Its acceleration graph is on the time axis from \(t=0\) to \(t=1\) second, at \(3\) meters per second squared from \(t=1\) to \(t=2\) seconds, and at \(-2\) meters per second squared from \(t=2\) to \(t=5\) seconds. Describe the velocity trend and find the velocity at the interval boundaries.
On \(0<t<1\), acceleration is zero throughout, so velocity is constant. Therefore, \(v(1)=5\) meters per second.
On \(1<t<2\), acceleration is positive, so velocity increases. The signed area is a rectangle:
Thus \(v(2)=5\text{ m/s}+3\text{ m/s}=8\text{ m/s}\).
On \(2<t<5\), acceleration is negative, so velocity decreases. The signed area is
It follows that
Check: The interval changes are \(v(1)-v(0)=5-5=0\), \(v(2)-v(1)=8-5=3\), and \(v(5)-v(2)=2-8=-6\) meters per second. Each agrees with the corresponding signed area. The velocity is constant, then increasing, then decreasing.
A Common Interpretation Error
A frequent mistake is to read the vertical value of the acceleration graph as if it were velocity. For example, if \(a(t)=-2\text{ m/s}^2\), it is incorrect to conclude that \(v(t)=-2\text{ m/s}\). These quantities have different meanings and different units: acceleration describes how quickly velocity changes.
Another error is to say “velocity is negative” when the acceleration graph lies below the time axis. The justified conclusion is “velocity is decreasing.” For instance, if velocity changes from \(7\text{ m/s}\) to \(5\text{ m/s}\), it has decreased even though both values are positive. To determine velocity values, use an initial velocity and the signed area; do not infer them from the acceleration sign alone.
When reading a graph, check the time interval, the sign of the acceleration there, and—if a numerical change is required—the signed area over that same interval. Keep the units consistent: acceleration multiplied by time has units of velocity.
Check Your Understanding
For each question, distinguish what the acceleration sign tells you from what would require additional velocity information.
- If an acceleration-time graph lies below the time axis throughout an interval, what happens to velocity on that interval?
- An acceleration graph is on the time axis for \(2\) seconds. What can you conclude about the change in velocity during those \(2\) seconds?
- A graph has constant acceleration \(4\text{ m/s}^2\) for \(3\) seconds. What is the change in velocity?
- Acceleration is negative before \(t=5\) and positive after \(t=5\). Describe the velocity trend on each side of \(t=5\).
- Why does a negative acceleration value not establish that velocity is negative?
