From Velocity to Acceleration
A velocity-time graph shows how an object's velocity changes as time passes. Its vertical coordinate gives velocity, and its horizontal coordinate gives time. The graph's slope therefore describes a rate of change of velocity with respect to time.
At a particular instant, the relevant slope is the slope of the tangent line to the graph at that instant. This slope is the object's instantaneous acceleration. As with other derivatives, the tangent slope can be positive, negative, or zero. Its sign describes the local direction in which the velocity graph changes vertically; it is not, by itself, the sign of the velocity.
On a velocity-time graph, \(a(t)\) is the slope of the tangent to the velocity graph at time \(t\). Acceleration is a rate of change of velocity, not the velocity value itself.
Tangent Slope and Secant Slope
Two points on a velocity-time graph determine a secant line. If their times are \(t_1\) and \(t_2\), their velocities are \(v(t_1)\) and \(v(t_2)\), and \(t_1\ne t_2\), the secant slope is the average rate of change of velocity over that interval:
This is the average acceleration between those times. It is not generally the instantaneous acceleration at either endpoint. To obtain the instantaneous value at \(t\), consider secant slopes over intervals that shrink toward \(t\). If the limit exists, it is the tangent slope and gives \(v'(t)\).
The distinction matters on a curved graph: the secant slope depends on the chosen interval, while the instantaneous acceleration is the slope of the tangent at one specified time.
Why the Tangent Slope Gives Instantaneous Acceleration
The connection follows from the definition of the derivative. For a short time change \(h\), the velocity change is \(v(t+h)-v(t)\), and the elapsed time is \(h\). Their quotient is the average rate of change of velocity over that short interval. As \(h\) approaches zero, these secant slopes approach the tangent slope, when the derivative exists.
At time \(t\), locate the point \((t,v(t))\) on the velocity graph.
For a nearby time \(t+h\), calculate the slope \(\frac{v(t+h)-v(t)}{h}\).
As \(h\) approaches zero, the secant slope approaches the tangent slope, which is \(a(t)\).
This also explains a graph-reading limitation. A graph with a sharp corner at the time of interest may have different limiting slopes from the left and right, so there is no single tangent slope there. In that case, the graph does not give a unique instantaneous acceleration at that instant.
Reading the Sign of the Acceleration
The horizontal axis is time and the vertical axis is velocity. Consequently, a tangent that rises from left to right has positive slope, one that falls from left to right has negative slope, and a horizontal tangent has slope zero. These statements concern the change in velocity indicated by the graph, not whether the velocity itself is positive or negative.
| Tangent on the velocity-time graph | Instantaneous acceleration |
|---|---|
| Rises from left to right | Positive |
| Falls from left to right | Negative |
| Horizontal | Zero |
The sign is determined by the vertical change in velocity divided by the horizontal change in time. For a downward-sloping tangent, for example, the velocity change is negative when time increases, so the slope is negative.
Units and Graph Scales
Acceleration units come from dividing velocity units by time units. If velocity is measured in metres per second and time in seconds, then acceleration is measured in metres per second per second, written \(\text{m/s}^2\). If velocity is in kilometres per hour and time is in hours, the acceleration unit is \(\text{km/h}^2\).
When estimating slope from a graph, use the values represented by its axes, not merely the number of grid squares. If each horizontal square represents \(0.5\) seconds and each vertical square represents \(2\) metres per second, then a change of two vertical squares over three horizontal squares corresponds to a velocity change of \(4\text{ m/s}\) over \(1.5\text{ s}\), not a change of two units over three units.
Worked Examples
Worked Example: Finding Acceleration from a Velocity Formula
Suppose \(v(t)=2t^2-8t+3\), where \(t\) is measured in seconds and \(v\) in metres per second. Find the instantaneous acceleration at \(t=1\text{ s}\), and check the tangent slope.
Differentiate velocity with respect to time:
Evaluate at \(t=1\):
The velocity at that time is \(v(1)=2(1)^2-8(1)+3=2-8+3=-3\text{ m/s}\). Thus the point of tangency is \((1,-3)\). The tangent line with slope \(-4\) is \(v+3=-4(t-1)\), or \(v=1-4t\). At \(t=0\), this tangent has \(v=1\); at \(t=2\), it has \(v=-7\). Its slope from these two points is
This agrees with the derivative. The negative acceleration is the negative slope of the velocity graph at \(t=1\).
Worked Example: Estimating Acceleration from a Graphed Tangent
Suppose the axes of a velocity-time graph are labelled in seconds and metres per second. At \(t=2\text{ s}\), a tangent line passes through the graph coordinates \((1,6)\) and \((3,-2)\). Find the instantaneous acceleration at \(t=2\text{ s}\).
Use the two points on the tangent to calculate its slope. The change in velocity is \(-2-6=-8\text{ m/s}\), and the change in time is \(3-1=2\text{ s}\). Therefore,
The tangent's slope is negative because the line falls as time increases. As a check that the stated instant lies on this tangent, at \(t=2\) the line has velocity \(6+(-4)(2-1)=2\text{ m/s}\). Thus its point there is \((2,2)\), consistent with a tangent through the specified instant. The instantaneous acceleration is \(-4\text{ m/s}^2\).
Worked Example: Positive Acceleration with Negative Velocity
Let \(v(t)=-5+3t\), with time in seconds and velocity in metres per second. Find the velocity and instantaneous acceleration at \(t=1\text{ s}\), and identify each sign.
First evaluate the velocity:
Differentiate the velocity function and evaluate at the same time:
The velocity is negative, while the acceleration is positive. This is possible because they represent different quantities: velocity is the graph's vertical coordinate, whereas acceleration is its slope. The graph is a straight line with slope \(\frac{v(2)-v(1)}{2-1}=\frac{1-(-2)}{1}=3\text{ m/s}^2\), confirming the derivative value.
A Common Graph-Reading Mistake
A common mistake is to calculate a slope using grid-square counts while ignoring the units and scale labels. Suppose a tangent rises by three vertical squares, each representing \(2\text{ m/s}\), over two horizontal squares, each representing \(0.5\text{ s}\). It is wrong to report the slope as \(3/2=1.5\text{ m/s}^2\): that calculation treats each square as one unit.
The actual velocity change is \(3(2)=6\text{ m/s}\), and the actual time change is \(2(0.5)=1\text{ s}\). Therefore the correct slope is
Always convert changes in grid squares into axis values before dividing. Also preserve the sign: if the tangent falls as time increases, its velocity change and acceleration are negative.
What the Graph Tells You
A velocity-time graph gives instantaneous acceleration at a chosen time when the tangent slope is defined. The value may be read exactly from a supplied tangent or model, or estimated from a carefully drawn tangent. A graph alone may not provide an exact numerical value if its scales or tangent are imprecise; in that situation, report an estimate rather than implying unwarranted precision.
Keep the roles of the graph's two features separate: the height at a time gives velocity, while the tangent's slope gives instantaneous acceleration. The slope's sign records whether the velocity graph is rising, falling, or momentarily horizontal, and its units are velocity per unit time.
Check Your Understanding
Answer each question using tangent slopes and the graph's labelled units.
- What feature of a velocity-time graph represents instantaneous acceleration at a specified time?
- A tangent falls by \(6\text{ m/s}\) over \(2\text{ s}\). What is its slope, including units?
- If a velocity-time graph has a horizontal tangent at one instant, what is the instantaneous acceleration there?
- A graph shows velocity in kilometres per hour and time in hours. What units should be used for acceleration?
- Can a velocity be negative at an instant when the acceleration is positive? Explain what the two signs describe.
