Position Information Constrains Velocity
A position record describes where an object is at particular times. Even when the motion between those times is not known, the positions can force conclusions about the object’s velocity at some instant. These conclusions are useful because they do not require a complete formula for the motion.
For motion along a line, let \(s(t)\) denote position at time \(t\). The velocity is \(s'(t)\), which records both the rate and direction of motion. If \(s'(t)>0\), position is increasing; if \(s'(t)<0\), it is decreasing. Speed is the nonnegative quantity \(|s'(t)|\).
The key connection is between the change in position over a time interval and the velocity at some time inside it. The Mean Value Theorem makes that connection precise. Rolle’s Theorem gives a particularly useful consequence when two positions are equal.
The Motion Form of the Mean Value Theorem
Continuity means that the position has no jump on the closed time interval; differentiability ensures that velocity exists at every interior time. These conditions matter: the theorem’s conclusion is not guaranteed if its assumptions fail.
The result concerns signed velocity. If the object ends to the left of where it began, the displacement and the average velocity are negative, so the theorem guarantees a negative instantaneous velocity somewhere. It does not say that the object moved in that direction throughout the interval.
Worked Example: Finding a Velocity Forced by Two Positions
Suppose an object has position \(s(t)=t^2\) metres for \(1\le t\le5\), with time measured in seconds. Find a time at which its instantaneous velocity equals its average velocity on this interval.
First calculate the endpoint positions: $$ s(1)=1^2=1\text{ m},\qquad s(5)=5^2=25\text{ m}. $$ The displacement is \(25-1=24\text{ m}\), and the elapsed time is \(5-1=4\text{ s}\). Therefore the average velocity is $$ \frac{s(5)-s(1)}{5-1}=\frac{24}{4}=6\text{ m/s}. $$ The position function is continuous on \([1,5]\) and differentiable on \((1,5)\), so the Mean Value Theorem applies. Its velocity is $$ s'(t)=2t\text{ m/s}. $$ Set this equal to the average velocity: $$ 2c=6,\qquad c=3\text{ s}. $$ Check: \(3\) lies in \((1,5)\), and \(s'(3)=2(3)=6\text{ m/s}\), as required. The positions also give \((25-1)/(5-1)=6\text{ m/s}\).
Repeated Positions Force a Moment of Rest
When an object has the same position at two distinct times, its net displacement between those times is zero. Rolle’s Theorem turns this endpoint fact into a conclusion about velocity inside the interval.
This is Rolle’s Theorem applied to the position function. It is also the Mean Value Theorem with \(s(b)-s(a)=0\): the average velocity is zero, so the instantaneous velocity is zero at some interior time. In one-dimensional motion, \(s'(c)=0\) means the object is momentarily at rest.
Worked Example: Equal Positions at Different Times
Let the position, in metres, be \(s(t)=t^2-6t+20\) on \(1\le t\le5\), with time measured in seconds. Find the time when velocity is zero.
Evaluate the endpoint positions: $$ s(1)=1^2-6(1)+20=1-6+20=15\text{ m}, $$ $$ s(5)=5^2-6(5)+20=25-30+20=15\text{ m}. $$ The positions are equal. The function is a polynomial, so it is continuous on \([1,5]\) and differentiable on \((1,5)\). Rolle’s Theorem therefore guarantees an interior time with zero velocity.
Differentiate: $$ s'(t)=2t-6\text{ m/s}. $$ Solve \(s'(c)=0\): $$ 2c-6=0,\qquad 2c=6,\qquad c=3\text{ s}. $$ Check that \(3\in(1,5)\) and \(s'(3)=2(3)-6=0\text{ m/s}\). Also, \(s(3)=9-18+20=11\text{ m}\); the object is at that position at the instant when its velocity is zero.
The conclusion is that velocity is zero at least once between the two equal-position times. It does not say that the object stayed still, or that its velocity was zero at either endpoint. An object can move away and return, passing through a moment when its direction changes.
Several Position Constraints Can Force Different Velocities
Apply the theorems separately to separate time intervals. This can reveal more than one necessary feature of a motion. In particular, an interval with equal endpoint positions forces zero velocity somewhere in that interval, while an interval with nonzero displacement forces the corresponding average velocity somewhere inside it.
Worked Example: A Return Followed by Further Motion
Let \(s(t)=t(t-2)\) metres for \(0\le t\le4\), where time is in seconds. Determine the velocity values guaranteed by the positions at times \(0\), \(2\), and \(4\).
First calculate: $$ s(0)=0(0-2)=0\text{ m},\qquad s(2)=2(2-2)=0\text{ m}, $$ $$ s(4)=4(4-2)=8\text{ m}. $$ Since \(s(0)=s(2)\), Rolle’s Theorem on \([0,2]\) guarantees a time with zero velocity. Since \(s\) is a polynomial, the continuity and differentiability conditions hold. Differentiate: $$ s'(t)=2t-2\text{ m/s}. $$ Solving \(2c-2=0\) gives \(c=1\text{ s}\). Check: \(1\in(0,2)\) and \(s'(1)=2-2=0\text{ m/s}\).
Now apply the Mean Value Theorem on \([2,4]\). The displacement is \(8-0=8\text{ m}\), and the elapsed time is \(4-2=2\text{ s}\), so the average velocity is $$ \frac{s(4)-s(2)}{4-2}=\frac{8}{2}=4\text{ m/s}. $$ The theorem guarantees a time \(d\in(2,4)\) with \(s'(d)=4\). Solving \(2d-2=4\) gives \(2d=6\), hence \(d=3\text{ s}\). Check: \(3\in(2,4)\) and \(s'(3)=6-2=4\text{ m/s}\). Thus the motion has zero velocity at \(t=1\) and velocity \(4\text{ m/s}\) at \(t=3\).
Displacement Bounds and Speed Limits
The Mean Value Theorem also gives a necessary bound on how far an object can travel between two times if its velocity is bounded. Suppose \(s\) satisfies the theorem’s hypotheses on \([a,b]\) and \(|s'(t)|\le M\) for every \(t\in(a,b)\), where \(M\ge0\). At the time \(c\) supplied by the theorem, $$ |s(b)-s(a)|=|s'(c)|(b-a)\le M(b-a). $$ So a speed limit \(M\) implies that the magnitude of the displacement cannot exceed \(M\) times the elapsed time.
Worked Example: Testing Whether a Speed Limit Is Possible
Suppose a continuously differentiable position function describes motion over \(0\le t\le4\) seconds, and the positions are \(s(0)=2\) metres and \(s(4)=18\) metres. Can the speed remain at most \(3\text{ m/s}\) throughout the interval?
The displacement is \(18-2=16\text{ m}\), and the elapsed time is \(4\text{ s}\). Thus the magnitude of average velocity is $$ \left|\frac{s(4)-s(0)}{4-0}\right| =\left|\frac{16}{4}\right| =4\text{ m/s}. $$ The Mean Value Theorem guarantees some \(c\in(0,4)\) with \(s'(c)=4\text{ m/s}\). Therefore the speed at that time is \(|s'(c)|=4\text{ m/s}\), which exceeds \(3\text{ m/s}\). The proposed speed limit is impossible.
The displacement bound gives the same contradiction: a speed no greater than \(3\text{ m/s}\) would require \(|s(4)-s(0)|\le3(4)=12\text{ m}\), but the stated displacement is \(16\text{ m}\). Check: \(16\le12\) is false, so the positions and speed limit cannot all hold.
What the Conclusions Do—and Do Not—Say
These results force a velocity at some interior time. They do not generally identify when that time occurs unless the position function is also known well enough to solve \(s'(c)\) explicitly. Nor do they say that velocity is constant throughout the interval.
A common mistake is to treat average velocity as the velocity at every time. For \(s(t)=t^2\) on \([1,5]\), the average velocity is \(6\text{ m/s}\), but the endpoint velocities are \(s'(1)=2\text{ m/s}\) and \(s'(5)=10\text{ m/s}\). The correct conclusion is that \(s'(c)=6\text{ m/s}\) for at least one \(c\in(1,5)\); indeed, \(c=3\). Equality at one instant does not imply equality throughout the interval.
Another mistake is to infer that equal endpoint positions mean the object remained at rest. In the example \(s(t)=t^2-6t+20\), the position changes from \(15\text{ m}\) to \(11\text{ m}\) by \(t=3\), even though it returns to \(15\text{ m}\) at \(t=5\). Rolle’s Theorem guarantees a moment of zero velocity, not an interval of rest.
Check Your Understanding
Use the theorem hypotheses and distinguish conclusions about one instant from claims about an entire interval.
- If a position function is continuous on \([2,7]\), differentiable on \((2,7)\), and changes from \(10\) metres to \(25\) metres, what velocity must occur at some interior time?
- If the positions at times \(a<b\) are equal and the Mean Value Theorem hypotheses hold, what does Rolle’s Theorem guarantee about velocity?
- Why does a speed bound of \(5\text{ m/s}\) over \(3\) seconds rule out a displacement of \(16\) metres?
- Does an average velocity of zero prove that velocity was zero at every time? Explain.
- For \(s(t)=t^2\) on \([1,5]\), verify the time when instantaneous velocity equals average velocity.
