Acceleration Measures How Velocity Changes
Velocity describes both how quickly an object’s position changes and its direction of motion. Acceleration describes how velocity changes over time. That distinction matters: an object moving in the negative direction can have positive acceleration, and positive acceleration does not by itself mean that the object is moving forward or speeding up.
Acceleration may be found from a formula for velocity, estimated from a table of velocity values, or interpreted as the slope of a tangent line on a velocity-versus-time graph. In every case, the quantity of interest is the rate of change of velocity with respect to time.
When velocity comes from a position function \(s(t)\), the earlier result that velocity is \(s'(t)\) gives \(a(t)=v'(t)=s''(t)\). Here the emphasis is on finding and interpreting acceleration using velocity information directly.
Calculating Acceleration from a Formula
For a velocity function, differentiate with respect to time to obtain an acceleration function. Then substitute the time at which acceleration is requested. Keep track of units: differentiating a quantity in meters per second with respect to seconds produces meters per second squared.
Worked Example: Differentiate a Velocity Function
An object’s velocity is \(v(t)=3t^2-12t+5\) meters per second, where \(t\) is in seconds. Find its acceleration at \(t=3\), and interpret the signs of velocity and acceleration there.
Differentiate each term with respect to \(t\):
Evaluate acceleration at \(t=3\):
Evaluate velocity at the same time:
The velocity is negative, so the object is moving in the negative direction at that instant. Acceleration is positive, so velocity is increasing—that is, becoming less negative at that instant. Since velocity and acceleration have opposite signs, the object’s speed is decreasing at that instant.
Check the differentiation by differentiating \(6t-12\) back through the original velocity terms: \(\frac{d}{dt}(3t^2)=6t\), \(\frac{d}{dt}(-12t)=-12\), and \(\frac{d}{dt}(5)=0\). These terms give \(a(t)=6t-12\), as calculated.
Estimating Acceleration from a Table
A table gives velocity at selected times, not necessarily a formula for every time. To estimate acceleration near a time \(t\), calculate a change in velocity divided by the corresponding change in time. This is the slope of a secant line through two data points on a velocity-versus-time graph.
If the data include times on both sides of the time of interest, a centered difference is often a useful estimate. For a time \(t\) with equally spaced neighboring times \(t-h\) and \(t+h\), use \(\frac{v(t+h)-v(t-h)}{2h}\). This estimates the instantaneous rate at the middle time; it is not automatically the exact derivative.
Worked Example: Estimate Acceleration from Velocity Data
A table records an object’s velocity in meters per second:
| Time \(t\) (s) | Velocity \(v(t)\) (m/s) |
|---|---|
| 1 | 4 |
| 2 | 9 |
| 3 | 16 |
| 4 | 25 |
Estimate the acceleration at \(t=2\) using the data on either side. The neighboring times are \(t=1\) and \(t=3\), so the time interval between them is \(3-1=2\) seconds. The change in velocity is \(16-4=12\) meters per second. Therefore, the centered estimate is
The units follow from dividing meters per second by seconds. The table also shows that velocity increases from \(4\) to \(9\) to \(16\) m/s over these times, consistent with positive acceleration near \(t=2\). Because the calculation uses a finite interval, \(6\text{ m/s}^2\) is an estimate of the instantaneous acceleration, not a guaranteed exact value.
Check the arithmetic: the difference quotient uses the velocity change \(12\text{ m/s}\) over \(2\text{ s}\), and \(12/2=6\). The table provides no exact formula for \(v(t)\), so an exact derivative cannot be confirmed from these entries alone.
Reading Acceleration from a Graph
On a graph with time on the horizontal axis and velocity on the vertical axis, the slope represents the rate of change of velocity. The slope of a secant line gives average acceleration over an interval. The slope of the tangent line at a particular time gives instantaneous acceleration there, provided the graph has a tangent with a defined slope at that time.
Worked Example: Use a Tangent-Line Slope
A velocity-versus-time graph has a tangent line at \(t=4\) seconds. Two points on that tangent line are \((3,2)\) and \((5,10)\), with time in seconds and velocity in meters per second. Estimate the acceleration at \(t=4\).
Compute the slope of the tangent line by subtracting the coordinates in the same order:
The tangent line rises as time increases, so its slope and the acceleration estimate are positive. This tells us that velocity is increasing at \(t=4\); it does not, by itself, tell us whether the object is moving in the positive direction. For that, we would also need the value of velocity at \(t=4\).
Check the slope calculation: the vertical change is \(10-2=8\text{ m/s}\), and the horizontal change is \(5-3=2\text{ s}\). Their quotient is \(4\text{ m/s}^2\), the graph’s tangent-line slope in the required units.
Why the Difference Quotient and Tangent Slope Work
The derivative is defined as the limit of average rates of change over intervals that shrink toward the time of interest. This definition explains why nearby table values and tangent-line slopes provide estimates of acceleration.
For a graph, the quotient is a secant-line slope. As the second point approaches the first, the secant slope approaches the tangent-line slope, when that tangent slope exists. For table data, the time gaps cannot usually be shrunk continuously: the available difference quotients approximate the derivative rather than defining it exactly. A centered estimate uses values on both sides of the target time to represent the local change from one side to the other.
For an estimate at \(t\), select data close to \(t\); when available, choose one time before and one after it.
Subtract the earlier velocity from the later velocity, keeping the order consistent with the time subtraction.
The resulting change in velocity per unit time estimates acceleration and has units of velocity divided by time.
Acceleration, Direction, and Speed
Three ideas should be kept distinct. The sign of velocity indicates direction of motion along the chosen number line. The sign of acceleration indicates whether velocity is increasing or decreasing. Speed is the magnitude of velocity, \(|v|\), so whether speed increases depends on both signs.
| Velocity | Acceleration | Local effect on speed, when \(v\ne 0\) |
|---|---|---|
| Positive | Positive | Speed increases |
| Positive | Negative | Speed decreases |
| Negative | Positive | Speed decreases |
| Negative | Negative | Speed increases |
When velocity is nonzero, the rate of change of speed is \(\frac{d}{dt}|v(t)|=\frac{v(t)}{|v(t)|}a(t)\). Thus speed increases when velocity and acceleration have the same sign, and decreases when their signs differ. At a time when \(v=0\), this sign test does not apply; the surrounding behavior must be examined instead.
Worked Example: Positive Acceleration While Moving Backward
At a particular instant, a car has velocity \(v=-8\text{ m/s}\) and acceleration \(a=3\text{ m/s}^2\). Determine its direction of motion and whether its speed is increasing or decreasing at that instant.
The negative velocity means the car is moving in the negative direction. The positive acceleration means velocity is increasing. Since the velocity and acceleration have opposite signs, its speed is decreasing at that instant.
This can also be checked from the rate of change of speed. Since \(v=-8\), \(\frac{v}{|v|}=-1\), and so
The negative rate of change confirms that speed is decreasing. The acceleration remains positive; it is the direction of motion, not the sign of acceleration, that is negative.
A Common Sign Mistake
A common error is to conclude that positive acceleration always means an object is speeding up. For example, given \(v=-8\text{ m/s}\) and \(a=3\text{ m/s}^2\), an incorrect claim would be: “Acceleration is positive, so speed is increasing.” That conclusion confuses increasing velocity with increasing speed.
The correction is to compare the signs. Here \(v<0\) and \(a>0\), so velocity is moving toward zero locally and the speed is decreasing. Positive acceleration means the signed velocity is increasing; it does not necessarily mean the magnitude of velocity is increasing.
Check Your Understanding
Answer each question, paying attention to units and to the difference between velocity and speed.
- If \(v(t)=5t^2-2t\) meters per second, what is \(a(2)\), including units?
- A table gives \(v(2)=7\text{ m/s}\) and \(v(6)=19\text{ m/s}\). What is the average acceleration from \(t=2\) to \(t=6\)?
- On a velocity-versus-time graph, what does a negative tangent-line slope indicate about velocity?
- An object has \(v=6\text{ m/s}\) and \(a=-2\text{ m/s}^2\). Is it moving in the positive or negative direction, and is its speed increasing or decreasing?
- An object has negative velocity and negative acceleration. What happens to its speed locally, provided its velocity is not zero?
