Two Different Claims About Zero Acceleration
Acceleration describes how velocity changes. The phrase “acceleration is zero” can refer either to one instant or to every instant in an interval. Those are different claims, and they support different conclusions.
At one instant, zero acceleration says that the velocity function has a horizontal tangent there, if the derivative exists. It does not say the velocity stays the same at nearby times. If acceleration is zero throughout an interval, a stronger conclusion is possible: velocity is constant, provided the required continuity and differentiability conditions hold.
The endpoint conditions matter. A derivative that is zero at every interior point is not, by itself, enough to guarantee that velocity is constant all the way to an endpoint where the velocity function is discontinuous.
Acceleration at One Instant
If an object’s velocity at time \(t\) is \(v(t)\), its acceleration is the derivative of velocity, when that derivative exists:
Consequently, \(a(t_0)=0\) means \(v'(t_0)=0\). In terms of the velocity graph, the tangent is horizontal at \(t_0\). This describes the instantaneous rate of change at \(t_0\); it does not determine the velocity at other times. In particular, it does not imply that the object is at rest: rest at an instant requires \(v(t_0)=0\), not \(v'(t_0)=0\).
Worked Example: Zero Acceleration at One Instant
Suppose a particle’s position, measured in metres, is \(x(t)=t^3-3t^2+4t\), where \(t\) is measured in seconds. Find its velocity and acceleration, and interpret what happens at \(t=1\).
Differentiate position to obtain velocity:
Differentiate velocity to obtain acceleration:
At \(t=1\) second, the acceleration is
while the velocity is
The particle is therefore not at rest at this instant. To check whether its velocity is constant, evaluate it at other times:
The velocity is not constant, even though acceleration is zero at \(t=1\). The derivatives confirm the result: \(a(0)=-6\ \text{m/s}^2\), \(a(1)=0\), and \(a(2)=6\ \text{m/s}^2\).
Zero Acceleration Throughout an Interval
Now consider the stronger condition that acceleration is zero at every interior time of an interval. To conclude that velocity is constant, we need enough regularity to apply the Mean Value Theorem between any two times in the interval.
The hypotheses have specific roles. Differentiability allows us to talk about acceleration at interior times. Continuity ensures the Mean Value Theorem applies to each closed subinterval, including when one of its endpoints is an endpoint of \(I\).
Why the Theorem Works
Take any two times \(s\) and \(t\) in \(I\) with \(s<t\). The interval \([s,t]\) lies inside \(I\). Since \(v\) is continuous on \(I\), it is continuous on \([s,t]\). Since \(v\) is differentiable at every interior point of \(I\), it is differentiable on \((s,t)\). Thus the Mean Value Theorem applies on \([s,t]\).
That theorem says that for some \(c\) between \(s\) and \(t\), the change in \(v\) between the two endpoints equals its derivative at \(c\) times the elapsed time. Since the derivative is zero, the change must be zero.
Therefore \(v(t)=v(s)\). Because \(s\) and \(t\) were arbitrary points of \(I\), every pair of values of \(v\) on \(I\) is equal. That is exactly what it means for \(v\) to be constant on \(I\).
Worked Example: Zero Acceleration on a Closed Time Interval
A particle has position \(x(t)=10+4t\) metres for \(0\leq t\leq5\) seconds. Find its velocity and acceleration, and verify the velocity behavior on this interval.
Differentiate the position function:
Differentiate velocity:
The position function is continuous on \([0,5]\), and it is differentiable in the interval’s interior. Its velocity is therefore constant there; in fact, the explicit derivative gives \(v(t)=4\ \text{m/s}\) at every time. For instance,
As a direct check, the acceleration obtained by differentiating \(v(t)=4\) is zero throughout the interval. Thus the position model and its derivatives agree with the constant-velocity conclusion.
Zero at One Time Versus Zero Throughout
A useful way to keep the two claims separate is to ask how much information about \(a\) has been provided. One equation, \(a(t_0)=0\), specifies the derivative of velocity at one time. The condition \(a(t)=0\) for all interior \(t\) in an interval specifies the derivative throughout that interval. Only the second condition, with the theorem’s hypotheses, supports the conclusion that velocity is constant there.
Worked Example: A Horizontal Tangent Does Not Make Velocity Constant
Let a velocity function, measured in metres per second, be \(v(t)=6+(t-2)^2\), for \(0\leq t\leq4\) seconds. Determine where acceleration is zero and check whether velocity is constant.
Differentiate velocity to find acceleration:
Set this expression equal to zero:
At that time, the velocity is
Compare with the velocities at two other times:
Thus acceleration is zero at \(t=2\), but velocity is not constant on \([0,4]\). The calculation verifies the distinction: \(a(2)=2(2-2)=0\), while \(v(1)=7\ \text{m/s}\) and \(v(2)=6\ \text{m/s}\) are unequal.
Why Continuity at Endpoints Cannot Be Omitted
The theorem’s continuity condition is not a technical detail that can be ignored. If a velocity function is defined at an endpoint but jumps there, knowing that its derivative is zero at all interior times does not force the endpoint value to match the rest of the function.
Worked Example: A Missing Endpoint Condition
Consider the function on \([0,1]\) defined by
At every interior time \(t\in(0,1)\), the function is constant in a neighbourhood of \(t\), so
Nevertheless, \(v(0)=1\) and \(v(1)=0\), so the function is not constant on \([0,1]\). It is not continuous at \(0\): values \(v(t)=0\) for \(t>0\) do not approach \(v(0)=1\) as \(t\) approaches \(0\) from the right.
This example does not contradict the theorem. The theorem requires continuity on the whole interval, and that requirement fails at \(0\). It also shows why a derivative condition only at interior points cannot, on its own, determine a possibly discontinuous endpoint value.
Common Pitfall: Applying the Mean Value Theorem Without Checking Continuity
A tempting but invalid argument for the function in the preceding example is: “The derivative is zero between \(0\) and \(1\), so the Mean Value Theorem gives \(v(1)-v(0)=0\).” The arithmetic difference is actually
The contradiction is a signal that the theorem has been applied without its hypotheses. The Mean Value Theorem requires continuity on the closed interval \([0,1]\), and this function is not continuous there. It is not valid to infer that the endpoint values agree from the interior derivative condition alone.
Interpreting the Result Carefully
Constant velocity means that the velocity has the same value at every time in the interval. The value need not be zero. Thus zero acceleration throughout an interval describes an object whose velocity does not change; it does not, by itself, describe an object at rest.
Conversely, a zero of acceleration at a single time does not identify a whole interval of constant velocity. It only says the instantaneous rate of change of velocity is zero at that time. To make a conclusion about velocity across an interval, use information about acceleration throughout that interval and check the continuity and differentiability conditions needed for the argument.
Check Your Understanding
For each question, distinguish a statement about one instant from a statement about an interval.
- If \(a(3)=0\), what does this tell you about the velocity function at \(t=3\)? What does it not tell you?
- Can an object have zero acceleration and nonzero velocity? Explain using the relationship between acceleration and velocity.
- State the continuity and differentiability conditions used to conclude that \(v\) is constant when \(v'=0\) throughout an interval.
- Why is continuity at an included endpoint relevant to applying the Mean Value Theorem?
- For \(v(t)=5+(t-1)^2\), find \(a(t)\), identify where acceleration is zero, and decide whether velocity is constant on \([0,2]\).
