Position and Time: What a Graph’s Slope Represents
A position-versus-time graph shows how an object’s position changes as time passes. Time is plotted on the horizontal axis, and position is plotted on the vertical axis. The graph’s slope therefore compares a change in position with a change in time.
Over a time interval, that slope describes an average rate of change. At a particular instant, the relevant slope is the slope of the graph’s tangent line at that point. This tangent slope is the object’s instantaneous velocity.
This interpretation connects a geometric feature of a graph with a physical quantity. The tangent’s steepness tells us the velocity’s magnitude, its upward or downward direction gives the velocity’s sign, and the axes determine the physical units.
Here, \(x(t)\) denotes position, not distance traveled. Position can be positive, zero, or negative according to the chosen coordinate system. The velocity is the rate at which that position changes. Its units are position units divided by time units: for example, metres per second or kilometres per hour.
Why a Tangent Slope Gives Instantaneous Velocity
Take two points on the position-time graph: one at time \(t\), and another at time \(t+h\). The slope of the line through those points is the change in position divided by the elapsed time. It describes the average rate of change over that interval.
As the second point moves closer to the first, the secant line through the two points approaches the tangent line, provided the graph has a tangent there. The corresponding slopes approach the tangent slope. The derivative is precisely this limiting rate of change, so the tangent slope equals the instantaneous velocity when the derivative exists.
This relationship explains why the tangent matters: it captures the local change at one instant, rather than averaging the motion over an interval of positive length. The limit must exist for the instantaneous velocity to be defined by this derivative.
Reading the Sign and Magnitude of the Slope
The sign of the tangent slope indicates whether position is increasing or decreasing at that instant. An upward-sloping tangent has positive slope; a downward-sloping tangent has negative slope. A horizontal tangent has slope zero. These are statements about the graph’s local trend in position.
The magnitude of the slope describes how steep the tangent is. A tangent with a larger absolute slope represents a greater change in position per unit of time than one with a smaller absolute slope, using the same units on both graphs. The sign and magnitude together give the signed velocity value. For instance, \(5\text{ m/s}\) and \(-5\text{ m/s}\) have equal magnitudes but opposite signs.
Compare slopes using the graph’s numerical axis scales, not just how steep a line looks on the page. The same physical slope can appear different if the graph is stretched or compressed horizontally or vertically. To calculate a slope from plotted coordinates, use the coordinate values and their stated units.
Worked Examples
Worked Example: Finding Velocity from a Position Function
An object moves along a straight line. Its position in metres after \(t\) seconds is \(x(t)=t^2+2t\). Find its instantaneous velocity at \(t=3\) seconds, and write the tangent line at that point.
Differentiate the position function with respect to time:
Substitute \(t=3\):
Find the position at that time to locate the point of tangency:
The point on the graph is \((3,15)\), with time measured in seconds and position in metres. A line of slope \(8\text{ m/s}\) through this point has equation \(x-15=8(t-3)\), or \(x=15+8(t-3)\). Its slope is \(8\text{ m/s}\), matching the derivative.
Check: At \(t=2.5\), the function gives \(x(2.5)=2.5^2+2(2.5)=6.25+5=11.25\text{ m}\). The tangent line gives \(15+8(2.5-3)=15-4=11\text{ m}\). These values are close, consistent with the tangent describing the graph locally; the line need not equal the curved graph away from the contact point.
Worked Example: Interpreting a Negative Tangent Slope
A position function, measured in metres, is \(x(t)=20+6t-t^2\), where time is measured in seconds. Find the instantaneous velocity at \(t=4\), and interpret the tangent’s sign and magnitude.
Differentiate:
Evaluate at \(t=4\):
The position at that time is
The tangent line at \((4,28)\) is \(x-28=-2(t-4)\), or \(x=28-2(t-4)\). Its negative slope means the position is decreasing at that instant. The slope’s magnitude is \(2\text{ m/s}\): the position changes at a rate of two metres per second at that instant, with a negative sign indicating the downward trend on the graph.
Check: The tangent line gives \(x=30\text{ m}\) at \(t=3\) and \(x=26\text{ m}\) at \(t=5\). Its slope between those points is \((26-30)/(5-3)=-4/2=-2\text{ m/s}\), as required. These are points on the tangent line; they need not lie on the position curve.
Worked Example: Reading Velocity from a Tangent on a Graph
Suppose a position-time graph has time in hours on its horizontal axis and position in kilometres on its vertical axis. At the point \((2,5)\), the tangent line passes through the additional points \((1,2)\) and \((3,8)\). Find the instantaneous velocity at \(t=2\) hours.
Use the two stated points on the tangent to calculate its slope. The change in position is \(8-2=6\) kilometres, and the change in time is \(3-1=2\) hours:
Because this is the slope of the tangent at \((2,5)\), it is the instantaneous velocity at \(t=2\) hours. It is positive because the tangent rises from left to right. The position is increasing at a rate of \(3\) kilometres per hour at that instant.
Check: The line through \((1,2)\) with slope \(3\) has equation \(x-2=3(t-1)\), or \(x=3t-1\). At \(t=2\), it gives \(x=3(2)-1=5\), the stated point of tangency. Its slope is \(3\text{ km/h}\), confirming the result.
A Common Mistake: Treating an Interval Slope as Instantaneous
A secant slope can be a useful rate of change, but it is not generally the instantaneous velocity at either endpoint or at a time between them. Confusing the two gives an incorrect answer when the rate changes over the interval.
For example, let \(x(t)=t^2\) metres, with \(t\) measured in seconds. A calculation from \(t=3\) to \(t=5\) gives
That value is the average rate over the two-second interval. A mistake would be to report \(8\text{ m/s}\) as the instantaneous velocity at \(t=3\). The derivative instead gives \(v(t)=2t\), so \(v(3)=2(3)=6\text{ m/s}\).
Check: The average-rate calculation uses positions \(x(5)=25\text{ m}\) and \(x(3)=9\text{ m}\), so it correctly gives \(16\text{ m}/2\text{ s}=8\text{ m/s}\). The instantaneous value is the tangent slope, \(6\text{ m/s}\), not that secant slope. As the interval shrinks toward \(t=3\), the secant slopes approach \(6\text{ m/s}\).
Units and Graph Reading
Velocity units come from dividing the vertical-axis units by the horizontal-axis units. If position is measured in metres and time in seconds, the slope has units \(\text{m/s}\). If position is in kilometres and time in hours, the slope has units \(\text{km/h}\). Write the units with the slope; a bare numerical value does not fully describe a physical velocity.
When reading a graph, identify the time and position coordinates carefully, then use the tangent line’s rise divided by its run. A rising tangent has positive velocity, a falling tangent has negative velocity, and a horizontal tangent has zero velocity at that instant, provided the tangent exists. Do not substitute the slope of the whole curve or an arbitrary secant for the tangent slope.
Check Your Understanding
Use the relationship between tangent slope and instantaneous velocity to answer these questions.
- A position-time graph has a downward-sloping tangent at a particular time. What does the sign of the instantaneous velocity tell you about position at that instant?
- Position is measured in metres and time in seconds. What units should a tangent slope have?
- At a point on a position-time graph, the tangent passes through \((1,4)\) and \((3,10)\), where time is in seconds and position is in metres. What is the instantaneous velocity at the point of tangency?
- For \(x(t)=3t^2\) metres, with time in seconds, find the instantaneous velocity at \(t=2\) seconds.
- Why is a secant slope over a nonzero time interval not generally the instantaneous velocity at one endpoint?