Reading Motion from Straight Segments
A position-time graph gives an object’s position as time passes. When a graph consists of straight segments, each segment has a constant slope. That slope tells us the object’s velocity throughout the corresponding time interval.
The interpretation needs care at the points where the segments meet. If the slope changes abruptly at a vertex, the graph has a corner: the velocity just before the vertex differs from the velocity just after it. The position at the vertex can still be well defined, even though there is no single derivative there.
In this tutorial, assume the graph is a continuous polyline: adjoining segments meet at the same position, and each segment covers a nonzero time interval. Time is measured in seconds and position in metres in the examples, so slopes are measured in metres per second. The same reasoning applies with other units.
Velocity on One Straight Segment
Suppose an object’s position is \(x_1\) at time \(t_1\) and \(x_2\) at time \(t_2\), and its position-time graph is straight between those times. The slope is the change in position divided by the change in time. On a straight position graph, this slope is also the constant velocity on the segment.
The numerator is signed displacement, not total distance. A positive slope means position increases as time increases; a negative slope means position decreases. A horizontal segment has slope zero, so the object’s position is constant during that interval.
The slope has units of position divided by time. For example, if position is measured in metres and time in seconds, then \((x_2-x_1)/(t_2-t_1)\) is measured in \(\mathrm{m/s}\). Keeping the signs and units in the calculation helps distinguish velocity from the graph’s position values.
Worked Example: Find the Velocity on Each Segment
Worked Example: Find the Velocity on Each Segment
A continuous position graph passes through the points \((0,1)\), \((2,7)\), \((5,1)\), and \((7,5)\), where time is in seconds and position is in metres. Find the velocity on each straight segment, write a position formula on each interval, and identify the velocity changes at the vertices.
First segment, from \(t=0\) to \(t=2\): The position changes from \(1\) m to \(7\) m, and the time changes by \(2\) s. Thus,
The line through \((0,1)\) with slope \(3\) has position formula \(x(t)=1+3t\). Checking the endpoints gives \(x(0)=1+3(0)=1\) m and \(x(2)=1+3(2)=7\) m.
Second segment, from \(t=2\) to \(t=5\): The position changes from \(7\) m to \(1\) m over \(3\) s. Therefore,
Starting from position \(7\) m at \(t=2\), a formula for this segment is \(x(t)=7-2(t-2)=11-2t\). At its endpoints, \(x(2)=11-4=7\) m and \(x(5)=11-10=1\) m, as required.
Third segment, from \(t=5\) to \(t=7\): The position changes from \(1\) m to \(5\) m over \(2\) s, so
The position formula is \(x(t)=1+2(t-5)=2t-9\). Checking gives \(x(5)=10-9=1\) m and \(x(7)=14-9=5\) m.
The adjoining slopes are \(3\) and \(-2\ \mathrm{m/s}\) at \(t=2\), then \(-2\) and \(2\ \mathrm{m/s}\) at \(t=5\). Each vertex is a corner with unequal slopes, so neither vertex has one instantaneous velocity represented by the graph. The segment velocities on either side are still well defined.
Why a Corner May Have No Single Instantaneous Velocity
The derivative at a time measures how position changes per unit time in the limit of increasingly small time intervals around that time. On a straight segment, this limiting rate is simply its slope. At a corner, approaching from the left and from the right can give different limiting rates.
To make this precise, let a vertex occur at time \(t=c\), with position \(x(c)\). Suppose the segment immediately to its left has slope \(m_L\), and the segment immediately to its right has slope \(m_R\). For sufficiently small \(h<0\), the point at \(c+h\) lies on the left segment. Because that segment is linear and ends at \((c,x(c))\),
Thus, the left-hand limit of the difference quotient is \(m_L\). Similarly, for sufficiently small \(h>0\), the point at \(c+h\) lies on the right segment, so
The right-hand limit of the difference quotient is \(m_R\). A two-sided derivative exists precisely when these one-sided limits exist and are equal. Here they exist because each adjoining segment is straight; they are equal exactly when \(m_L=m_R\).
This test does not say that position fails to exist at a corner. For a continuous polyline, the two segments meet at one position, so the position at the vertex is defined. It is the rate of change that fails to have one common value when the slopes differ.
Worked Example: Interpret a Pause and a Change of Direction
An object’s position graph passes through \((0,4)\), \((3,4)\), \((6,-2)\), and \((10,-2)\). Find the velocity on each segment and interpret the vertices at \(t=3\) and \(t=6\).
From \(t=0\) to \(t=3\), position remains \(4\) m. The velocity is
A matching position formula is \(x(t)=4\). It gives \(x(0)=4\) m and \(x(3)=4\) m.
From \(t=3\) to \(t=6\), position changes from \(4\) m to \(-2\) m. Therefore,
Using the position \(4\) m at \(t=3\), a formula is \(x(t)=4-2(t-3)=10-2t\). The endpoint checks are \(x(3)=10-6=4\) m and \(x(6)=10-12=-2\) m.
From \(t=6\) to \(t=10\), position remains \(-2\) m, so
The formula \(x(t)=-2\) checks at both endpoints: \(x(6)=-2\) m and \(x(10)=-2\) m.
At \(t=3\), the adjoining slopes are \(0\) and \(-2\ \mathrm{m/s}\), so the graph has a corner and no single derivative there. At \(t=6\), the adjoining slopes are \(-2\) and \(0\ \mathrm{m/s}\), so there is another corner with no single derivative. The negative middle slope indicates decreasing position; the horizontal outer segments indicate constant position on their intervals. The slope comparison, rather than the position value at either vertex, diagnoses the corners.
Equal Slopes at a Joining Point
A graph can be described as a sequence of line segments even when a joining point is not a corner in the sense of a change in slope. When adjoining segments have the same slope, the difference quotients from both sides agree, so the derivative exists at their joining time.
Worked Example: A Joining Point with a Derivative
A continuous graph passes through \((1,2)\), \((4,8)\), and \((7,14)\). Determine whether there is a corner at \(t=4\) and find the instantaneous velocity there.
The slope from \(t=1\) to \(t=4\) is
The slope from \(t=4\) to \(t=7\) is
Because \(m_L=m_R=2\ \mathrm{m/s}\), both one-sided difference-quotient limits equal \(2\ \mathrm{m/s}\). The derivative at \(t=4\) therefore exists and is \(x'(4)=2\ \mathrm{m/s}\). In fact, both line segments lie on \(x(t)=2t\): this formula gives \(x(1)=2\) m, \(x(4)=8\) m, and \(x(7)=14\) m. The joining point is not a corner with a change in slope.
A Common Mistake: Averaging the Two Slopes
At a corner, it is tempting to average the slope on the left and the slope on the right and call the result the instantaneous velocity. This does not work. A derivative, when it exists, must be the common limit of the difference quotient from both sides; an average of two unequal limits is neither of those limits.
Worked Example: Why an Average Does Not Fix a Corner
At a vertex, suppose the left slope is \(4\ \mathrm{m/s}\) and the right slope is \(-2\ \mathrm{m/s}\). The incorrect calculation would be
But the left-hand difference quotient remains \(4\ \mathrm{m/s}\), while the right-hand difference quotient remains \(-2\ \mathrm{m/s}\), for time intervals small enough to stay on the respective straight segments. Neither limit is \(1\ \mathrm{m/s}\), and the two limits do not agree. Therefore the derivative at the vertex does not exist; \(1\ \mathrm{m/s}\) is not an instantaneous velocity represented by this graph.
Endpoints and the Model’s Interpretation
The corner test applies to an interior joining point, where there is a segment on both sides. At the beginning or end of the time interval shown, the graph supplies information on only one side. It can determine a one-sided rate from that side, but it does not by itself provide a two-sided derivative at the endpoint.
A sharp corner in an idealized position graph means that the model changes velocity abruptly at one instant. It does not imply that the object’s position jumps: the graph can remain continuous while its slope changes. Whether a real object could follow that idealization exactly is a separate physical question. For interpreting the graph mathematically, compare the segment slopes and report what the graph represents.
These ideas matter whenever motion is represented by a simplified, piecewise linear model. Each interval gives a clear constant velocity, while the vertices signal where that description changes. Checking the two adjoining slopes prevents assigning a single instantaneous velocity where the graph does not support one.
Check Your Understanding
Use slope and the adjoining-segment test to answer the following questions.
- A position graph goes from \((1,3)\) to \((5,11)\), with time in seconds and position in metres. What velocity does this segment represent?
- Two segments meet at a vertex. Their slopes are \(-3\ \mathrm{m/s}\) and \(2\ \mathrm{m/s}\). Does the derivative exist at the vertex? Explain using the one-sided rates.
- A continuous graph has adjoining slopes of \(5\ \mathrm{m/s}\) on both sides of an interior joining point. What is the derivative at that point?
- Why is the average of unequal left- and right-hand slopes not the instantaneous velocity at a corner?
- What information about the derivative is available at the first point of a graph if the graph is shown only for later times?
