Motion Rates: From Secants to Tangents
In a motion problem, position changes as time passes. The slope between two points on a position-time graph measures the average velocity over that time interval. At a particular instant, the tangent slope gives the instantaneous velocity, or at least an estimate of it when the graph or data are approximate.
These quantities are related, but they are not interchangeable. An average velocity describes an interval; an instantaneous velocity describes one moment. Secant tables, graph-based tangent estimates, and acceleration information provide practical ways to move from one to the other.
If position is measured in metres and time in seconds, a position-time slope has units \(\text{m}/\text{s}\), or metres per second. The units are part of the meaning: a slope measured in metres per second is a velocity, while a position value alone is not.
Why a Secant Slope Is Average Velocity
A secant line joins two points on the position-time graph. The horizontal change between those points is the elapsed time, and the vertical change is the displacement. The slope of the secant is therefore displacement divided by elapsed time.
This establishes why secant slopes describe average velocity. To estimate the velocity at one time, use secants whose time intervals approach that time. Their slopes may settle toward the tangent slope. The earlier material on tangent slopes treats that geometric idea; here the emphasis is on interpreting and calculating motion quantities correctly.
The sign also matters. A positive velocity means position is increasing as time increases, and a negative velocity means position is decreasing. A zero average velocity means the net displacement over the interval is zero; it does not necessarily mean the object stayed still throughout that interval.
Worked Example: Using Nearby Secants
Worked Example: Using Nearby Secants
Suppose an object moves along a straight track, with position \(s(t)=t^2+3t\) metres at time \(t\) seconds. Estimate its instantaneous velocity at \(t=2\) seconds using secants on both sides.
First, the position at the target time is $$ s(2)=2^2+3(2)=4+6=10\text{ m}. $$ For a time \(2+h\), the position is $$ s(2+h)=(2+h)^2+3(2+h)=4+4h+h^2+6+3h=10+7h+h^2. $$ Thus the average velocity from \(2\) to \(2+h\) is $$ \frac{s(2+h)-s(2)}{h} = \frac{(10+7h+h^2)-10}{h} = 7+h\text{ m/s}, $$ for \(h\ne0\).
For \(h=0.1\) seconds, the secant slope is \(7+0.1=7.1\text{ m/s}\). Directly checking the positions gives \(s(2.1)=2.1^2+3(2.1)=4.41+6.3=10.71\text{ m}\), so the average velocity is \((10.71-10)/0.1=0.71/0.1=7.1\text{ m/s}\).
For \(h=0.01\) seconds, the slope is \(7+0.01=7.01\text{ m/s}\). Here \(s(2.01)=2.01^2+3(2.01)=4.0401+6.03=10.0701\text{ m}\), and \((10.0701-10)/0.01=0.0701/0.01=7.01\text{ m/s}\).
Now use intervals to the left. At \(t=1.9\), \(s(1.9)=1.9^2+3(1.9)=3.61+5.7=9.31\text{ m}\). The average velocity from \(1.9\) to \(2\) is \((10-9.31)/(2-1.9)=0.69/0.1=6.9\text{ m/s}\). At \(t=1.99\), \(s(1.99)=1.99^2+3(1.99)=3.9601+5.97=9.9301\text{ m}\), giving \((10-9.9301)/(2-1.99)=0.0699/0.01=6.99\text{ m/s}\).
The secant slopes from the left and right approach \(7\text{ m/s}\). The tangent-based estimate of instantaneous velocity at \(t=2\) is therefore \(7\text{ m/s}\). As a check on the arithmetic, the right-hand formula \(7+h\) gives \(7.1\) and \(7.01\) for the two positive increments, matching the calculated averages; the left-hand slopes are \(6.9\) and \(6.99\), as shown.
A table makes the approach visible. The intervals become shorter, and the average velocities move closer to the same value from either side.
| Interval (seconds) | Average velocity (m/s) |
|---|---|
| From 2 to 2.1 | 7.1 |
| From 2 to 2.01 | 7.01 |
| From 1.9 to 2 | 6.9 |
| From 1.99 to 2 | 6.99 |
Estimating Velocity from a Position-Time Graph
A graph-based estimate uses the same slope idea, but the information may come from a drawn curve rather than a formula. At the time of interest, sketch or identify the tangent line. Choose two well-separated points on that tangent line, read their coordinates from the graph, and calculate rise over run.
The chosen points should be on the tangent line, not merely on the curve. A tangent line touches the curve locally, but it usually does not pass through other points on the curve. Use the graph’s scale carefully: a horizontal grid step may represent a different number of seconds from the vertical grid step’s number of metres.
Worked Example: Reading a Tangent from a Graph
A position-time graph is drawn with time in seconds on the horizontal axis and position in metres on the vertical axis. At \(t=2\) seconds, the tangent line passes through the plotted coordinates \((1.5,8)\) and \((2.5,13)\), where each coordinate is written as \((\text{time},\text{position})\). Estimate the instantaneous velocity at \(t=2\).
The time change along the tangent is \(2.5-1.5=1.0\text{ s}\). Its position change is \(13-8=5\text{ m}\). Therefore the tangent slope is $$ \frac{13-8}{2.5-1.5} = \frac{5\text{ m}}{1.0\text{ s}} = 5\text{ m/s}. $$
The estimated instantaneous velocity is \(5\text{ m/s}\). Check the result by confirming that the position change divided by the time change is \(5\text{ m}/1.0\text{ s}=5\text{ m/s}\). This is a graph-based estimate; its precision depends on how accurately the tangent and its coordinates can be read.
Using Acceleration to Find Velocity
Acceleration describes how velocity changes over time. Its average value on an interval is the change in velocity divided by elapsed time:
If velocity is measured in metres per second and time in seconds, acceleration has units \((\text{m/s})/\text{s}=\text{m/s}^2\). When acceleration is constant throughout an interval, the average acceleration on that interval equals the constant acceleration. This gives a direct way to calculate velocity at a later time from a starting velocity.
To justify the result, constant acceleration means the average acceleration between \(t_0\) and any later time \(t\) is \(a\). Substitute \(v(t)\) and \(v_0\) into the definition of average acceleration, then solve for \(v(t)\):
The condition that acceleration is constant is essential to this calculation. If acceleration varies, one average acceleration over an interval does not in general determine the velocity at every time in that interval.
Worked Example: Velocity from Constant Acceleration
A cyclist is moving at \(4\text{ m/s}\) at \(t=0\). The cyclist’s acceleration is constant at \(3\text{ m/s}^2\). Find the velocity at \(t=2\) seconds.
The elapsed time is \(2-0=2\text{ s}\). Using the constant-acceleration result, $$ v(2)=4\text{ m/s}+(3\text{ m/s}^2)(2\text{ s}). $$ The acceleration contribution is \(3\times2=6\), with units \(\text{m/s}^2\times\text{s}=\text{m/s}\). Therefore, $$ v(2)=4\text{ m/s}+6\text{ m/s}=10\text{ m/s}. $$
Check the result by computing the average acceleration over the two-second interval: \((10-4)\text{ m/s}/2\text{ s}=6/2=3\text{ m/s}^2\), which agrees with the stated constant acceleration. Because the acceleration is constant, the velocity at \(t=2\) is \(10\text{ m/s}\), not merely an interval average.
A Reliable Workflow for Motion Problems
A short sequence of checks helps prevent common interpretation and unit errors.
Determine whether the problem asks for position, average velocity, instantaneous velocity, or acceleration. The same numerical slope can have different meaning depending on the axes and interval.
Use two position-time points for average velocity, nearby secants or a tangent for an instantaneous-velocity estimate, and velocity-time information for average acceleration.
Write the change in the vertical quantity over the change in time. Keep units through the calculation so that the result can be interpreted.
Ask whether the answer is an interval average or an estimate at one instant, and verify that its units match that quantity.
Common Pitfall: Treating an Average as Instantaneous
Suppose a vehicle travels \(60\text{ m}\) in \(10\text{ s}\). Its average velocity over that interval is $$ \frac{60\text{ m}}{10\text{ s}}=6\text{ m/s}. $$ It is incorrect to conclude from this alone that the vehicle’s instantaneous velocity was \(6\text{ m/s}\) at every moment. The calculation uses only the total displacement and total elapsed time. The vehicle could have moved at varying velocities, as long as the net displacement over the interval was \(60\text{ m}\).
The correction is to state exactly what the data support: the average velocity over those ten seconds is \(6\text{ m/s}\). To estimate instantaneous velocity at a particular time, use nearby secant slopes or a tangent estimate at that time. To conclude that velocity follows a constant-acceleration model, verify that the problem gives or justifies constant acceleration.
Why the Distinctions Matter
In applied motion problems, a result is only useful if its interpretation is clear. An average velocity can summarize a trip, but it may hide changes in direction or speed. A tangent estimate describes local motion near a chosen time, but its accuracy depends on the graph or data. An acceleration value tells how velocity changes, and a constant-acceleration formula can turn that information into a velocity at a specified time only when its assumption holds.
Always report units and, when appropriate, indicate whether an answer is exact or estimated. For graph readings, avoid implying more precision than the graph supports. For table estimates, compare values from both sides where possible. These habits keep the numerical calculation connected to the physical meaning.
Check Your Understanding
For each question, state what information is needed and include the relevant units in your reasoning.
- An object’s position changes from \(12\text{ m}\) to \(27\text{ m}\) over \(3\text{ s}\). What is its average velocity over that interval?
- Why do secant slopes from shorter intervals near a time help estimate instantaneous velocity?
- A tangent to a position-time graph rises \(8\text{ m}\) while time advances \(2\text{ s}\). What instantaneous-velocity estimate does that slope give?
- An object has velocity \(5\text{ m/s}\) and constant acceleration \(2\text{ m/s}^2\) for \(4\text{ s}\). What is its final velocity?
- Why does knowing an object’s average velocity over an interval not usually establish its velocity at every instant?
