From Average Velocity to Instantaneous Velocity
Average velocity describes the change in position over a time interval. If an object has position \(s(t)\) at time \(t\), its average velocity from \(t=a\) to \(t=b\), where \(a\ne b\), is the change in position divided by the elapsed time:
This gives one rate for the entire interval. It does not generally tell us the object's velocity at a particular instant within that interval. To describe motion at exactly time \(t\), we examine average velocities over intervals that begin at \(t\) and become shorter.
The quotient inside this limit is the average velocity from time \(t\) to time \(t+h\). Thus, instantaneous velocity is not an average over a nonzero time interval; it is the limit of those average velocities as the interval shrinks toward the instant.
If position is measured in meters and time in seconds, then \(s'(t)\) has units of meters per second. More generally, derivative units are position units divided by time units. The value is a signed rate: its sign and size belong to the velocity, not necessarily to the object's speed.
Why Velocity Is a Tangent Slope
On a position-time graph, time is the horizontal coordinate and position is the vertical coordinate. Pick two points on the curve, at times \(t\) and \(t+h\). The line through them is a secant line. Its slope is
That slope is exactly the average velocity over the interval. As the second point moves toward the first, the secant line approaches the tangent line when the curve has a well-defined tangent there. The limiting secant slope is therefore the tangent slope.
This connection explains why a tangent drawn on a graph can be used to estimate instantaneous velocity. It also explains why simply choosing two points on the curve and calculating their secant slope gives an average velocity, not automatically an instantaneous one.
Deriving the Tangent-Slope Connection
The relationship follows directly from the definition of derivative. The proof also identifies exactly what is being estimated when nearby position values are used.
For a nonzero time change \(h\), the two graph points are \((t,s(t))\) and \((t+h,s(t+h))\). Their slope is \(\frac{s(t+h)-s(t)}{h}\), which is the average velocity over that interval.
Let \(h\) approach zero. If the derivative exists, the secant slopes approach the finite limit \(\lim_{h\to0}\frac{s(t+h)-s(t)}{h}=s'(t)\).
At a differentiable point, the tangent line is the limiting position of the secant lines. Its slope is the limit of their slopes, so it equals \(s'(t)\), the instantaneous velocity.
The condition that the derivative exists matters. If the left-hand and right-hand difference quotients approach different values, there is no single derivative value at that instant, and the graph does not have one tangent slope in the derivative sense. A tangent estimate is meaningful when the graph is sufficiently smooth near the point and its tangent direction is clear.
Worked Example: Finding Instantaneous Velocity from a Position Formula
Worked Example: Finding Instantaneous Velocity from a Position Formula
An object's position is \(s(t)=t^2+2t\) meters, where \(t\) is measured in seconds. Find its instantaneous velocity at \(t=3\) seconds using the difference quotient.
First calculate the position at \(t=3\):
For a nonzero time change \(h\), calculate the nearby position:
Now form the average velocity from \(3\) to \(3+h\):
Take the limit as \(h\) approaches zero:
Therefore, the instantaneous velocity at \(3\) seconds is \(8\) meters per second. As a check, differentiating the position function gives \(s'(t)=2t+2\), so \(s'(3)=2(3)+2=8\text{ m/s}\), agreeing with the difference-quotient calculation.
Estimating Velocity from Nearby Position Data
In some problems, position is provided in a table rather than by a formula. A natural estimate uses two nearby values on opposite sides of the time of interest. This is a centered difference quotient: it approximates the tangent slope using a short secant through points on both sides. It is still an estimate unless the problem provides enough information to determine the exact derivative.
Worked Example: Estimating from a Table
Worked Example: Estimating from a Table
A particle's position is \(s(t)=t^2+3t\) meters. Estimate its instantaneous velocity at \(t=2\) seconds from the positions at \(1.9\), \(2.0\), and \(2.1\) seconds.
Compute the three positions:
The average velocity from \(1.9\) to \(2.0\) seconds is
From \(2.0\) to \(2.1\) seconds it is
The centered estimate, using the two outer values, is
Thus, the table suggests an instantaneous velocity near \(7.0\text{ m/s}\). Check the estimate against the exact derivative: from the difference quotient at \(t=2\),
Taking \(h\to0\) gives \(s'(2)=7\text{ m/s}\), which matches the centered estimate. The left and right estimates, \(6.9\) and \(7.1\), lie on opposite sides of that value.
For actual measured data, the exact derivative is usually unknown, and position measurements may be rounded or noisy. A centered estimate can be useful, but it does not turn approximate data into an exact velocity. The time spacing and precision of the given values affect how reliable the estimate is.
Estimating Velocity from a Position-Time Graph
To estimate a tangent slope from a graph, first identify the time of interest and the point on the position curve. Then use the drawn tangent line, not an arbitrary line through two curve points. Choose two well-separated points on that tangent line, read their coordinates using the axis scales, and calculate rise divided by run. Using points farther apart on the tangent can reduce errors from reading a very small change, but the points must lie on the tangent line.
Worked Example: Reading a Tangent Slope from a Graph
Worked Example: Reading a Tangent Slope from a Graph
A position-time graph shows a tangent line at \(t=4\) seconds. The line passes through the plotted coordinates \((3,10)\) and \((5,14)\), where time is in seconds and position is in meters. Estimate the instantaneous velocity at \(t=4\).
The vertical change between those tangent-line points is
The horizontal change is
Therefore the tangent slope is
The estimated instantaneous velocity at \(4\) seconds is \(2\text{ m/s}\). Check the calculation with the slope formula using the same points: \(\frac{14-10}{5-3}=\frac{4}{2}=2\). Because both selected points are on the tangent line, this is the tangent's slope and hence the graph-based estimate of the derivative at the point of tangency.
Average Velocity Is Not Instantaneous Velocity
A common error is to calculate an average velocity over a noticeable interval and report it as the instantaneous velocity at one endpoint. The two quantities may happen to match in a particular situation, but that agreement should not be assumed. The interval average uses two distinct times; the instantaneous value requires a limit or a tangent slope at the specified time.
The correction is to match the method to the question. For a finite time interval, calculate average velocity. For one specified instant, use the derivative definition, a derivative value supplied by the problem, or a tangent-slope estimate from the graph.
When the Instantaneous Velocity Does Not Exist
The derivative-based definition requires the difference quotient to approach one finite value. A position function can fail that test even when the position itself is defined. For example, consider \(s(t)=|t|\) at \(t=0\). For \(h>0\), the quotient is \(\frac{|h|-|0|}{h}=1\). For \(h<0\), it is \(\frac{|h|-|0|}{h}=-1\). The one-sided limits disagree, so the two-sided derivative does not exist at \(0\). There is no single derivative-based instantaneous velocity there.
This is why graph estimates require care near a sharp corner or another point where the curve does not have a unique tangent direction. A graph can show a position at a time without having a well-defined tangent slope at that time.
What to Keep in Mind
Instantaneous velocity links three descriptions of the same rate: a limit of average velocities, the derivative of position, and the tangent slope on a position-time graph. In calculations, keep track of which one is exact and which one is estimated. A derivative found from a formula can be exact; a slope read from a graph or estimated from a table is generally approximate.
Always include units and identify the time at which the velocity is being reported. For position measured in meters and time in seconds, a velocity is expressed in meters per second. If the calculation uses two separated times, label it an average velocity over that interval. If the question asks for an instant, explain how the derivative or tangent slope provides the instantaneous value.
Check Your Understanding
Use the definitions and methods in this tutorial to answer the following questions.
- In the difference quotient for \(s'(t)\), what does \(\frac{s(t+h)-s(t)}{h}\) represent when \(h\ne0\)?
- A position-time graph has a tangent line that rises \(6\) meters over a run of \(3\) seconds. What instantaneous velocity does its slope estimate?
- For \(s(t)=t^2\), find the average velocity from \(t=2\) to \(t=2.5\), and explain why this is not automatically the instantaneous velocity at \(t=2\).
- Why can a centered difference quotient from nearby position data estimate an instantaneous velocity?
- For \(s(t)=|t|\), why does the derivative-based instantaneous velocity fail to exist at \(t=0\)?