From a Motion Graph to an Average Rate
A graph can show how an object's position or velocity changes over time. To estimate an average motion rate, choose the beginning and ending times, read the corresponding values from the graph, and compare the change in the vertical quantity with the elapsed time.
The straight line joining two selected points on a graph is called a secant line. Its slope is the vertical change divided by the horizontal change. For a position-time graph, that slope estimates average velocity. For a velocity-time graph, it estimates average acceleration.
The graph's axes matter. A visual rise of two grid squares does not necessarily represent two units: read the tick labels and units before calculating. The result is an estimate when the graph values themselves must be read approximately.
Read the Axes Before Finding a Slope
On a position-time graph, time is on the horizontal axis and position is on the vertical axis. The secant slope is therefore position change divided by time change, with units such as metres per second, \( \text{m/s} \).
On a velocity-time graph, time is again horizontal, but velocity is vertical. The secant slope is velocity change divided by time change, with units such as metres per second per second, written \( \text{m/s}^2 \). That slope estimates average acceleration, not average velocity.
In either case, use the same endpoint order in the numerator and denominator: value at the second time minus value at the first time. If the interval runs from \(t_1\) to \(t_2\), calculate the change as final minus initial. Reversing both differences leaves the ratio unchanged, but reversing only one changes its sign.
Worked Example: Estimating Average Velocity from a Position Graph
A position-time graph has time measured in seconds and position measured in metres. The selected points are \((1,3)\) and \((5,15)\): at \(1\) second the position is \(3\) metres, and at \(5\) seconds it is \(15\) metres. Estimate the average velocity over this interval.
The elapsed time is \( \Delta t=5-1=4\text{ s} \).
The change in position is \( \Delta x=15-3=12\text{ m} \).
Divide the position change by the elapsed time: \( \frac{\Delta x}{\Delta t}=\frac{12\text{ m}}{4\text{ s}}=3\text{ m/s} \).
The estimated average velocity is \(3\text{ m/s}\). The positive sign indicates that position increased over the interval. To check the calculation, multiply the rate by the elapsed time: \( (3\text{ m/s})(4\text{ s})=12\text{ m} \), which matches the position change \(15-3=12\text{ m}\). The units also reduce to metres per second.
Why the Secant Slope Gives an Average Rate
The graph records an input time and an output value at each selected point. A secant slope compares the output difference with the input difference. For a position graph, the output difference is displacement and the input difference is elapsed time. That is exactly the definition of average velocity. For a velocity graph, the output difference is change in velocity, and dividing by elapsed time gives average acceleration.
Justification. The slope between two points \((a_1,b_1)\) and \((a_2,b_2)\), where \(a_1\ne a_2\), is defined as \( (b_2-b_1)/(a_2-a_1) \). For the position graph, substitute \(a_1=t_1\), \(a_2=t_2\), \(b_1=x(t_1)\), and \(b_2=x(t_2)\). The resulting slope is \( [x(t_2)-x(t_1)]/[t_2-t_1] \), the average-velocity formula. For the velocity graph, substitute \(b_1=v(t_1)\) and \(b_2=v(t_2)\) instead. The slope becomes \( [v(t_2)-v(t_1)]/[t_2-t_1] \), the average-acceleration formula. Thus the graph interpretation follows directly from the definition of slope and the definitions of these average rates.
This reasoning also explains why the two graph types must not be confused. The vertical quantity differs: position on one graph, velocity on the other. The same geometric calculation, rise over run, represents a different physical rate depending on what the vertical axis measures.
Average Acceleration from a Velocity-Time Graph
Worked Example: Reading Average Acceleration from a Velocity Graph
A velocity-time graph has time in seconds and velocity in metres per second. At \(t=2\text{ s}\), the graph gives \(v=6\text{ m/s}\). At \(t=8\text{ s}\), it gives \(v=-3\text{ m/s}\). Find the average acceleration from \(2\) to \(8\) seconds.
The elapsed time is \( \Delta t=8-2=6\text{ s} \).
The velocity change is \( \Delta v=-3-6=-9\text{ m/s} \).
Therefore, \( \frac{\Delta v}{\Delta t}=\frac{-9\text{ m/s}}{6\text{ s}}=-1.5\text{ m/s}^2 \).
The estimated average acceleration is \(-1.5\text{ m/s}^2\). The negative sign means velocity decreased over the interval; it does not, by itself, say that the object was moving in the negative direction throughout. Check: \((-1.5\text{ m/s}^2)(6\text{ s})=-9\text{ m/s}\), matching the change from \(6\text{ m/s}\) to \(-3\text{ m/s}\). The units reduce to metres per second squared.
Using Tick Marks and Physical Units
When a graph is drawn on a grid, count the labelled intervals, not merely the squares. Suppose each horizontal grid interval represents \(2\) seconds while each vertical interval represents \(4\text{ m/s}\). A secant that moves right by two horizontal intervals and down by two vertical intervals represents a time change of \(4\text{ s}\) and a velocity change of \(-8\text{ m/s}\). Its slope is \(-8/4=-2\text{ m/s}^2\).
Worked Example: Converting a Grid Slope into Acceleration
On a velocity-time graph, each horizontal grid interval represents \(2\text{ s}\), and each vertical grid interval represents \(4\text{ m/s}\). The secant runs from a point at \(t=1\text{ s}\), \(v=6\text{ m/s}\), to a point at \(t=5\text{ s}\), \(v=-2\text{ m/s}\). Find its slope and interpret it.
First check the horizontal change: \(5-1=4\text{ s}\), which is two horizontal intervals of \(2\text{ s}\) each.
Next check the vertical change: \(-2-6=-8\text{ m/s}\), which is a downward change of two vertical intervals, each representing \(4\text{ m/s}\).
The secant slope is \( \frac{-8\text{ m/s}}{4\text{ s}}=-2\text{ m/s}^2 \). So the average acceleration is \(-2\text{ m/s}^2\).
Check the endpoint change using the computed rate: \((-2\text{ m/s}^2)(4\text{ s})=-8\text{ m/s}\), as required. This confirms both the scale reading and the sign. Counting the grid squares without using their labels would not be enough to determine the physical rate.
Keep the Time Order Consistent
Graph points can be read in either order, but the subtraction order must match. The interval from \(t=3\) to \(t=7\) has positive elapsed time \(7-3\). If the endpoints are listed in reverse order, both the time difference and the position difference are negative, so their ratio still gives the same average velocity. The following example makes this explicit.
Worked Example: A Position Graph Read Backwards
A position-time graph shows position \(20\text{ m}\) at \(t=3\text{ s}\) and position \(12\text{ m}\) at \(t=7\text{ s}\). Find the average velocity from \(3\) to \(7\) seconds.
Using the stated time order, the elapsed time is \(7-3=4\text{ s}\). The position change is \(12-20=-8\text{ m}\). Thus the average velocity is \( \frac{-8\text{ m}}{4\text{ s}}=-2\text{ m/s} \).
Now check by reversing both endpoint differences: the position difference is \(20-12=8\text{ m}\), and the time difference is \(3-7=-4\text{ s}\). The ratio is \(8/(-4)=-2\text{ m/s}\), the same result. Finally, \((-2\text{ m/s})(4\text{ s})=-8\text{ m}\), which matches the change from \(20\text{ m}\) to \(12\text{ m}\). The negative average velocity indicates decreasing position over this interval.
Common Pitfall: Reversing Rise and Run
A frequent error is to divide the time change by the position change. For the first position-graph example, this incorrect calculation would be \( \frac{4\text{ s}}{12\text{ m}}=\frac{1}{3}\text{ s/m} \). This is not average velocity: its units are seconds per metre, and it is the reciprocal of the desired slope.
The correction is to use vertical change divided by horizontal change. On a position-time graph, that means position change divided by time change. On a velocity-time graph, it means velocity change divided by time change. Then check that the units are appropriate: \(\text{m/s}\) for average velocity and \(\text{m/s}^2\) for average acceleration.
Another useful check is to multiply the calculated rate by the elapsed time. The product should equal the change in the graph's vertical quantity. This will not detect every possible reading error, but it quickly reveals many sign, arithmetic, and unit mistakes.
What an Average Slope Tells You
A secant summarizes the change between two selected times. It does not claim that the object's motion had that rate at every moment in the interval. A position graph may curve while its secant still gives the average velocity between its endpoints. Likewise, a velocity graph may vary while its secant gives the average acceleration across the chosen interval.
For graph-based estimates, record the interval and the units along with the answer. If the plotted values can only be read approximately, report the rate as an estimate rather than implying greater precision than the graph supports. The rate's sign comes from the changes in the plotted values, while its physical meaning comes from the axis labels.
Check Your Understanding
Use the endpoint changes and axis units in each question. Treat graph readings as exact unless the question says they are estimates.
- A position graph shows \(x=4\text{ m}\) at \(t=2\text{ s}\) and \(x=16\text{ m}\) at \(t=6\text{ s}\). What is the average velocity?
- A velocity graph shows \(v=10\text{ m/s}\) at \(t=1\text{ s}\) and \(v=4\text{ m/s}\) at \(t=4\text{ s}\). What is the average acceleration, including its units?
- On a velocity-time graph, the secant moves right by three grid intervals and up by one. Each horizontal interval is \(2\text{ s}\) and each vertical interval is \(3\text{ m/s}\). What average acceleration does this represent?
- Why must both the numerator and denominator use the same endpoint order when calculating a secant slope?
- A student divides elapsed time by position change when estimating average velocity. What is wrong with this calculation, and what is the correct order of division?
