From Average Velocity to an Instantaneous Velocity
A position function describes where an object is at each time. Its change in position over a time interval gives the object’s average velocity on that interval. The Mean Value Theorem connects this overall change to the object’s motion at a particular instant: under suitable conditions, the instantaneous velocity equals the average velocity at some time strictly inside the interval.
This is an existence result. It guarantees that at least one such time exists, but it does not generally tell us that the time is unique. The theorem also has conditions: the position function must be continuous on the closed time interval and differentiable at every time strictly inside it.
Average velocity uses displacement, the final position minus the initial position. It is not generally the same as average speed, which uses total distance traveled. This distinction matters especially when an object changes direction.
The Mean Value Theorem for Position
The theorem does not require the velocity to be constant, nor does it say that the matching time is the midpoint. It asserts that the value of the derivative at some interior point matches the slope of the line joining the endpoint positions. The continuity and differentiability assumptions are what allow the theorem to guarantee such a point.
Why the Matching Velocity Exists
The result can be seen by comparing the position function with the straight-line motion that has the same endpoint positions. Let \(L(t)\) be the linear function whose graph joins \((a,s(a))\) to \((b,s(b))\). Its slope is the average velocity. Subtracting \(L\) from \(s\) gives a function that is zero at both endpoints, so Rolle’s Theorem applies.
Define \(L(t)=s(a)+\dfrac{s(b)-s(a)}{b-a}(t-a)\). Then \(L(a)=s(a)\) and \(L(b)=s(b)\).
Set \(h(t)=s(t)-L(t)\). The assumed continuity and differentiability of \(s\), together with the linearity of \(L\), mean \(h\) is continuous on \([a,b]\) and differentiable on \((a,b)\).
Because \(h(a)=0\) and \(h(b)=0\), Rolle’s Theorem gives some \(c\in(a,b)\) with \(h'(c)=0\).
Since \(h'(c)=s'(c)-L'(c)=0\), and \(L'(c)=\dfrac{s(b)-s(a)}{b-a}\), the claimed equality follows.
This argument explains the geometry as well as the algebra. The graph of the position function and its endpoint line have the same height at each end of the interval. At some interior time, their slopes agree. Since the endpoint line’s slope is the average velocity, the position function’s slope at that time is the same average velocity.
Finding a Time with the Matching Velocity
To apply the theorem in a specific problem, first calculate the average velocity from the endpoint positions. Then differentiate the position function, set its derivative equal to that average, and solve for a time in the open interval. The theorem guarantees at least one solution under its hypotheses; the calculation identifies one.
Worked Example: A Quadratic Position Function
Suppose \(s(t)=t^2-4t+7\) on \([1,4]\). Find a time when instantaneous velocity equals the average velocity on this interval.
First evaluate the endpoint positions:
The elapsed time is \(4-1=3\), and the displacement is \(7-4=3\). Therefore the average velocity is
The instantaneous velocity is the derivative:
Set it equal to the average velocity and solve:
Since \(1<\frac{5}{2}<4\), this is an interior time. Check the velocity directly:
This equals the average velocity \(1\), as required. The theorem guarantees an interior match because this polynomial is continuous on \([1,4]\) and differentiable on \((1,4)\).
Worked Example: A Cubic Position Function
Let \(s(t)=t^3\) on \([0,2]\). Find a time when instantaneous velocity equals average velocity.
The endpoint positions are
Thus the displacement is \(8-0=8\), the elapsed time is \(2-0=2\), and the average velocity is
Differentiate and equate the instantaneous velocity to \(4\):
The solutions are \(t=\frac{2}{\sqrt{3}}\) and \(t=-\frac{2}{\sqrt{3}}\). Only the positive solution lies in \((0,2)\): indeed \(0<\frac{2}{\sqrt{3}}<2\), since \(0<2<2\sqrt{3}\). Check its velocity:
So the matching time is \(t=\frac{2}{\sqrt{3}}\), and its instantaneous velocity agrees with the average velocity \(4\). The function \(t^3\) is continuous and differentiable everywhere, so the theorem’s conditions hold on this interval.
Worked Example: The Same Average Velocity at an Interior Time
Consider \(s(t)=12t-2t^2\) on \([1,3]\). Find an interior time when the instantaneous velocity is the average velocity.
Evaluate the positions carefully:
The displacement is \(18-10=8\), and the elapsed time is \(3-1=2\). Hence
Differentiate:
Set this equal to \(4\) and solve:
Since \(1<2<3\), the time is in the required open interval. Substitution verifies the result:
The instantaneous velocity at \(t=2\) equals the average velocity on \([1,3]\). Again, the polynomial position function satisfies continuity on the closed interval and differentiability on its interior.
Average Velocity Is Not Average Speed
A common mistake is to use total distance traveled in the average-velocity calculation. The denominator is elapsed time in both average velocity and average speed, but their numerators differ: average velocity uses signed displacement, while average speed uses total distance. When motion reverses direction, these can produce different answers.
Common Pitfall: Using Distance Instead of Displacement
Consider \(s(t)=(t-1)^2\) on \([0,2]\). A wrong calculation adds the endpoint positions and divides by the elapsed time:
This is not the average velocity. The correct calculation subtracts the initial position from the final position:
The object moves from position \(1\) to position \(0\) and then back to position \(1\). Its total distance is \(1+1=2\), so its average speed is \(\frac{2}{2}=1\), not \(0\). Its instantaneous velocity is
Thus the instantaneous velocity at \(t=1\) matches the average velocity \(0\). The incorrect value \(1\) came from confusing endpoint positions with displacement; it happens to be the average speed here, not the average velocity.
What the Result Does—and Does Not—Say
The Mean Value Theorem provides a precise connection between motion over an interval and motion at an instant. When the hypotheses hold, at least one interior time has the average velocity as its instantaneous velocity. In practical problems, solving \(s'(t)=\dfrac{s(b)-s(a)}{b-a}\) locates candidate times, and checking that a candidate lies in \((a,b)\) completes the application.
The theorem does not assert that every instantaneous velocity equals the average, or that the matching instant is unique. If the equation has more than one solution in the interval, each such solution is a matching time. If a proposed position function is not continuous on the closed interval or not differentiable throughout the open interval, the theorem cannot be invoked to guarantee a match. A calculation might still find one, but that would need to be checked directly.
Finally, the result concerns signed velocity. A negative average velocity indicates net displacement in the negative direction; a positive one indicates net displacement in the positive direction. Neither sign by itself describes total distance traveled. Keeping endpoint order, displacement, and elapsed time straight makes the theorem’s conclusion both meaningful and verifiable.
Check Your Understanding
For each question, use displacement for average velocity and check the interval conditions before applying the theorem.
- State the two regularity conditions on \(s\) required to guarantee a matching instantaneous velocity on \([a,b]\).
- If \(s(0)=3\) and \(s(4)=11\), what is the average velocity on \([0,4]\)?
- For \(s(t)=t^2+1\) on \([1,3]\), find the average velocity and solve for an interior time when \(s'(t)\) equals it.
- Why is the average velocity of \(s(t)=(t-1)^2\) on \([0,2]\) different from its average speed?
- Does the Mean Value Theorem guarantee exactly one matching time, or at least one? Explain briefly.
