Acceleration Describes Change in Velocity
Knowing an object’s acceleration tells us how its velocity is changing, but it does not automatically tell us the velocity itself. The distinction is like knowing the slope of a graph without knowing its vertical position: slopes describe change, while an additional value is needed to locate the graph vertically.
Let \(v(t)\) be an object’s velocity at time \(t\). Where \(v\) is differentiable, its acceleration is the derivative of velocity. Thus acceleration describes the instantaneous rate at which velocity changes. Its units are velocity units per time unit, such as metres per second per second.
That last qualification matters. A positive acceleration at one instant says that the instantaneous rate of change of velocity is positive then. It does not, by itself, prove that velocity is increasing throughout a nearby interval. To conclude that velocity is increasing on an interval, we need acceleration to be positive throughout that interval. Likewise, negative acceleration throughout an interval means velocity is decreasing there.
Similarly, \(a(t_0)=0\) says the instantaneous rate of change of velocity is zero at \(t_0\). It does not say that velocity is constant around \(t_0\). For instance, if \(v(t)=t^3\), then \(a(t)=3t^2\), so \(a(0)=0\), even though \(v\) is not constant on any interval around \(0\).
Reading Acceleration Graphs and Tables
An acceleration graph plots \(a(t)\) against time. The graph’s height at a particular time is the acceleration value at that time. A table provides the same kind of information at listed times. In either case, acceleration values describe rates of change of velocity, not velocity values.
When acceleration is positive throughout an interval, velocity increases throughout that interval. When acceleration is negative throughout an interval, velocity decreases. If acceleration is zero throughout an interval, velocity is constant there. These interval-wide conclusions require information about the acceleration throughout the interval, not merely at one time or at a few sampled times.
| Acceleration information | What it tells us about velocity |
|---|---|
| \(a(t_0)>0\) at one time | The instantaneous rate of change of velocity is positive at \(t_0\). |
| \(a(t)>0\) throughout an interval | Velocity is increasing throughout that interval. |
| \(a(t_0)=0\) at one time | The instantaneous rate of change of velocity is zero at \(t_0\). |
| \(a(t)=0\) throughout an interval | Velocity is constant throughout that interval. |
A table of acceleration values at selected times does not generally reveal what acceleration does between those times. For example, values at times \(0\), \(1\), and \(2\) can all be positive even if acceleration becomes negative between two of the listed times. To infer a trend throughout an interval, the data must support a conclusion about acceleration across the interval.
Worked Example: Interpreting an Acceleration Table
A table records acceleration at three times:
At \(t=1\), the acceleration is \(2\text{ m/s}^2\). Therefore the instantaneous rate of change of velocity is positive at that time.
At \(t=2\), the acceleration is \(0\). Therefore the instantaneous rate of change of velocity is zero at that time. The table does not establish that velocity is constant near \(t=2\), or even at any interval around it.
At \(t=3\), the acceleration is \(-3\text{ m/s}^2\). Therefore the instantaneous rate of change of velocity is negative at that time.
Check: Each conclusion matches the corresponding acceleration value: positive gives a positive instantaneous rate, zero gives a zero instantaneous rate, and negative gives a negative instantaneous rate. Because the table only gives three sampled values, it does not establish the acceleration’s sign at every time between them.
Constant Acceleration Gives a Velocity Change
When acceleration is constant on an interval, we can calculate the change in velocity over that interval directly. If the constant acceleration is \(c\) and the interval has duration \(\Delta t\), then the velocity change is \(c\Delta t\). The final velocity is the initial velocity plus this change.
On an acceleration-time graph, constant acceleration appears as a horizontal segment. Over its time interval, the velocity change is the signed area of the rectangle between the segment and the time axis: a segment above the axis gives a positive change, and one below the axis gives a negative change. This area interpretation describes the velocity change, not the velocity itself.
Worked Example: Finding Velocity Changes from Constant Acceleration
An object has initial velocity \(v(0)=-1\text{ m/s}\). Its acceleration is \(2\text{ m/s}^2\) from \(t=0\) to \(t=3\text{ s}\), then \(-1\text{ m/s}^2\) from \(t=3\) to \(t=5\text{ s}\). Find its velocity at \(t=3\text{ s}\) and \(t=5\text{ s}\).
During the first interval, the duration is \(3-0=3\text{ s}\). The velocity change is
Therefore,
During the second interval, the duration is \(5-3=2\text{ s}\). The velocity change is
Therefore,
Check: Over the first interval, the velocity change is \(5-(-1)=6\text{ m/s}\), matching the calculated \(6\text{ m/s}\). Over the second, it is \(3-5=-2\text{ m/s}\), matching the calculated \(-2\text{ m/s}\).
Why Acceleration Alone Does Not Determine Velocity
Acceleration is the derivative of velocity. Adding a constant to a function does not change its derivative. Consequently, if a velocity function is consistent with given acceleration data, shifting that velocity function by a constant produces another velocity function with exactly the same acceleration.
Suppose \(v(t)\) has acceleration \(a(t)\), so \(v'(t)=a(t)\).
For a constant \(C\), define a second velocity \(w(t)=v(t)+C\).
Since the derivative of a constant is zero, \(w'(t)=v'(t)+0=a(t)\).
The two velocity functions have identical acceleration, even though their velocity values differ by \(C\).
This is a complete reason that acceleration cannot specify the vertical level of velocity: every velocity function compatible with the data has constant shifts that are compatible too. An initial velocity fixes this ambiguity by specifying the velocity at one time. For a fully specified acceleration function, an initial velocity then determines the velocity function.
Worked Example: Same Acceleration, Different Velocities
Consider two velocity functions, measured in metres per second:
Their accelerations are
Thus both have acceleration \(3\text{ m/s}^2\) at every time. But at \(t=0\),
The velocities differ by \(4-(-2)=6\text{ m/s}\). In fact, \(v_1(t)=v_2(t)+6\) for every \(t\), so the constant shift leaves the derivative unchanged.
Check: Differentiating each function gives the same acceleration, \(3\text{ m/s}^2\). Their distinct initial velocities confirm that acceleration alone does not tell us which velocity function describes the object.
A Common Mistake: Treating One Value as an Interval Trend
A frequent error is to use an acceleration value at one instant as if it described the whole surrounding interval. For example, from \(a(2)=0\), one might incorrectly conclude that velocity is constant near \(t=2\). A zero derivative at a single input does not imply constancy on an interval.
Worked Example: Zero Acceleration at One Time
Let \(v(t)=t^3\). Its acceleration is
At \(t=0\),
But at \(t=1\), \(v(1)=1\), and at \(t=2\), \(v(2)=8\). The velocity is not constant on the interval from \(0\) to \(2\), despite having zero acceleration at \(t=0\).
Correction: The valid conclusion from \(a(0)=0\) is only that the instantaneous rate of change of velocity at \(0\) is zero. To conclude that velocity is constant on an interval, acceleration must be zero throughout that interval.
What the Data Can and Cannot Establish
Acceleration information is valuable because it describes how velocity changes. A graph or table can show instantaneous rates, and acceleration known throughout an interval can establish whether velocity increases, decreases, or stays constant there. Under constant acceleration, it can also determine the exact velocity change across an interval.
But acceleration data alone do not give the velocity at a particular time. To determine velocity values, one needs a starting value, such as \(v(t_0)\), as well as enough acceleration information to track the changes from that time. A table containing only isolated samples may also leave the behaviour between samples unknown.
Check Your Understanding
Answer each question using the distinction between an instantaneous rate and a velocity value.
- If \(a(4)>0\), what does this tell you at \(t=4\), and what does it not establish about a whole interval?
- What can you conclude if \(a(t)<0\) throughout an interval?
- If acceleration is constantly \(-2\text{ m/s}^2\) for \(5\text{ s}\), what is the velocity change over that interval?
- Why can two different velocity functions have the same acceleration?
- What additional information is needed to determine velocity values from acceleration data?
