Reading the Story in a Position-Time Graph
A position-time graph displays a particle’s position as time passes. Its horizontal coordinate represents time, and its vertical coordinate represents position. The graph therefore shows both where the particle is and how its position changes over time.
These are different kinds of information. The graph’s height gives position. The slope of the graph at a point gives instantaneous velocity. A particle can be far from the origin while momentarily at rest, or at the origin while moving quickly. Reading the graph correctly means keeping height and slope distinct.
The units follow from the axes. If position is measured in meters and time in seconds, graph height is measured in meters, while tangent slope is measured in meters per second. Always use the axis scales; a graph’s visual steepness alone does not give a numerical velocity.
Height Gives Position
To find the particle’s position at a particular time, locate that time on the horizontal axis and read the graph’s vertical coordinate. A position of \(8\) meters means the particle is at coordinate \(8\) meters. It does not mean the particle has traveled \(8\) meters, and it does not tell you by itself whether the particle is moving.
Position is a signed coordinate relative to the chosen origin. A negative position lies on the negative side of that coordinate system. The value \(x(t)=0\) means the particle is at the origin at time \(t\); it does not mean the particle is at rest. Rest is about whether position is changing, which is read from the graph’s slope.
Worked Example: Read Position and Velocity from a Tangent
A position-time graph has time in seconds and position in meters. At \(t=3\), the curve passes through \((3,8)\). Its tangent line at that point also passes through \((2,6)\) and \((4,10)\). Find the position and instantaneous velocity at \(t=3\), and interpret them.
Step 1: Read the height. The point on the curve at \(t=3\) has vertical coordinate \(8\), so \(x(3)=8\) meters.
Step 2: Calculate the tangent slope. Using the two stated points on the tangent line,
Because this is the tangent slope, the instantaneous velocity is \(x'(3)=2\) meters per second.
Interpretation and check: At \(3\) seconds the particle is at position \(8\) meters, and its position is increasing at \(2\) meters per second at that instant. The tangent line through \((3,8)\) with slope \(2\) has value \(8+2(4-3)=10\) at \(t=4\), and \(8+2(2-3)=6\) at \(t=2\), agreeing with both given tangent points.
Tangent Slope Describes Instantaneous Velocity
At a time \(t=a\) where the position function is differentiable, instantaneous velocity is the derivative \(x'(a)\). On the graph, that derivative is the slope of the tangent line at \((a,x(a))\). The slope is computed as change in vertical coordinate divided by change in horizontal coordinate, using points on the tangent line and the axis scales.
A positive tangent slope means position is increasing at that instant; a negative tangent slope means it is decreasing. A horizontal tangent has slope zero, so the instantaneous velocity is zero at that instant. These statements concern the local slope, not the height: a graph can be high, low, or at position zero while having any of these slopes.
Local Maxima and Minima Show Turning Points
A local maximum is a point whose position is greater than or equal to nearby positions on the graph. A local minimum is a point whose position is less than or equal to nearby positions. When the graph rises before a local maximum and falls afterward, the particle’s position changes from increasing to decreasing. It has reversed direction along the coordinate line. At a local minimum, a fall followed by a rise indicates the opposite reversal.
If the graph is smooth at an interior local maximum or minimum, its tangent is horizontal and its instantaneous velocity is zero there. This follows from the derivative’s definition. At a differentiable local maximum \(t=a\), for small positive \(h\), \(x(a+h)-x(a)\leq 0\), so the right-hand difference quotient is nonpositive. For small negative \(h\), \(x(a+h)-x(a)\leq 0\) while \(h<0\), so the left-hand difference quotient is nonnegative. If the two-sided derivative exists, both one-sided limits must agree; the only value that is both nonpositive and nonnegative is zero. The same argument with inequalities reversed applies at a local minimum.
The converse does not hold: a horizontal tangent alone does not establish a local maximum, a local minimum, or a direction change. The graph must actually turn. For instance, it might flatten briefly and continue rising. Also, at a sharp corner the tangent may not exist; the graph’s trend on either side can still show a reversal.
Worked Example: Identify a Reversal at a Local Maximum
A graph shows position in meters against time in seconds. It rises as time goes from \(3\) to \(4\), reaches a smooth local maximum at \((4,12)\), then falls as time goes from \(4\) to \(5\). The graph also passes through \((3,9)\) and \((5,8)\). Describe the particle’s position at \(t=4\) and its change in direction.
Step 1: Read the position. The graph’s height at \(t=4\) is \(12\), so \(x(4)=12\) meters.
Step 2: Read the trend on each side. From \(t=3\) to \(t=4\), position rises from \(9\) meters to \(12\) meters. From \(t=4\) to \(t=5\), position falls from \(12\) meters to \(8\) meters. The stated graph is rising up to the local maximum and falling afterward, so this is a reversal, not merely a comparison of endpoint values.
Step 3: Interpret the tangent. Because the maximum is smooth and interior, its tangent is horizontal. Thus \(x'(4)=0\) meters per second at the turning instant.
Check: The maximum’s height, \(12\), is greater than the neighboring listed heights \(9\) and \(8\), consistent with the stated local maximum. The tangent slope \(0\) is consistent with a smooth peak. The rise before and fall after show why the particle reverses direction.
Worked Example: Identify a Reversal at a Local Minimum
A graph of position in meters against time in seconds falls to a smooth local minimum at \((2,-3)\), then rises. It passes through \((1,-1)\) before the minimum and \((3,-1)\) after it. What is the position at \(t=2\), and what does the graph show about the motion?
Step 1: Read the height. At \(t=2\), the vertical coordinate is \(-3\), so \(x(2)=-3\) meters. The particle is three meters on the negative side of the origin.
Step 2: Compare the trend on each side. Before the minimum, the graph falls from position \(-1\) meter to \(-3\) meters. After the minimum, it rises from \(-3\) meters to \(-1\) meter. Thus the position changes from decreasing to increasing: the particle reverses direction at the minimum.
Step 3: Interpret the tangent. The minimum is smooth, so the tangent is horizontal and \(x'(2)=0\) meters per second at that instant.
Check: The minimum value \(-3\) is less than both neighboring listed values \(-1\). The graph’s downward trend before it and upward trend afterward agree with a local minimum and a reversal.
A Horizontal Tangent Does Not Always Mean a Turn
At a smooth interior local extremum, zero tangent slope accompanies a direction change. But zero slope by itself is not enough. A particle can momentarily have zero velocity without reversing direction: on a graph, this can look like a flattening point where the curve continues rising on both sides, or continues falling on both sides.
Likewise, the particle being at the origin is not evidence of rest. These two statements refer to different graph features: “at the origin” concerns height, while “at rest at that instant” concerns tangent slope. Before identifying a reversal, inspect the graph’s trend immediately before and after the point.
Worked Example: Separate Position from Motion
A position-time graph passes through \((2,0)\), and the tangent at that point passes through \((1,-3)\) and \((3,3)\). Time is measured in seconds and position in meters. Is the particle at rest at \(t=2\)?
Step 1: Read position. The point \((2,0)\) gives \(x(2)=0\) meters, so the particle is at the origin.
Step 2: Compute tangent slope.
The instantaneous velocity is \(x'(2)=3\) meters per second, not zero.
Check: The tangent line through \((2,0)\) with slope \(3\) has value \(0+3(1-2)=-3\) at \(t=1\) and \(0+3(3-2)=3\) at \(t=3\), matching the stated points. Therefore the particle is at the origin but is moving at that instant.
A Reliable Reading Process
When interpreting a position-time graph, answer separate questions in order. This prevents the common error of treating vertical height as velocity or treating a zero velocity as automatic proof of a reversal.
At the specified time, locate the curve’s vertical coordinate and include its units.
Find the tangent slope, accounting for the horizontal and vertical axis scales; report position units per time unit.
Inspect the graph immediately before and after a candidate point. A rise followed by a fall, or a fall followed by a rise, shows a reversal.
This process matters in contextual questions because an answer must distinguish the particle’s location from its rate of motion. A statement such as “the particle is at position \(12\) meters” describes height. A statement such as “its velocity is \(2\) meters per second” describes tangent slope. A statement such as “it reverses direction at the peak” describes the change in trend around a local maximum.
Check Your Understanding
Use the graph features described in each question. Keep position and tangent slope separate in your answers.
- A particle’s position-time graph passes through \((5,-2)\). What is its position at \(t=5\), and what information would you need to determine its instantaneous velocity then?
- A smooth graph rises to an interior local maximum at \(t=4\) and falls afterward. What is the tangent slope at the maximum, and what does the change in trend indicate?
- A particle is at position \(0\) meters at a time when the tangent slope is negative. Is it at rest? Explain using the graph’s height and slope.
- A graph has a horizontal tangent but continues rising on both sides. Does the graph show a direction reversal at that point? Explain.
- A position-time graph has position in feet and time in seconds. What units must its tangent slope have?
