Position Change Over a Time Interval
Position describes where an object is relative to a chosen origin along a line. If \(x(t)\) is the object’s position at time \(t\), then its position at two times tells us how its location changed between them. Average velocity summarizes that change over the whole time interval.
The word average matters: the object need not move at a constant rate. Its position may change unevenly, but the average velocity is determined by its starting and ending positions and the elapsed time between them.
For a position table, the same calculation uses the row at the starting time and the row at the ending time. Intermediate rows may describe what happened along the way, but they are not needed to calculate the average velocity across those endpoints.
Why Displacement Divided by Time?
Displacement records the net change in position: final position minus initial position. Elapsed time records how long that change took: final time minus initial time. Dividing the first change by the second gives the position change per unit of elapsed time over the interval.
This calculation does not assume that the object covered equal amounts of distance in equal amounts of time. It is a summary of the whole interval, not a claim that the object moved steadily throughout it.
Worked Example: Average Velocity from a Position Function
Worked Example: Average Velocity from a Position Function
An object moves along a straight line with position \(x(t)=4+3t-t^2\) metres, where \(t\) is measured in seconds. Find its average velocity from \(t=1\) to \(t=3\).
First calculate the position at each endpoint:
The displacement is final position minus initial position:
The elapsed time is:
Therefore, the average velocity is:
Check: The endpoint positions differ by \(4-6=-2\) metres, and \((-1\text{ m/s})(2\text{ s})=-2\text{ m}\). The calculated average velocity gives the correct displacement over the interval. Its negative sign says the net position change was in the negative direction.
Reading the Sign and Units
Position is measured relative to an origin and a chosen positive direction. Consequently, the sign of average velocity describes the direction of the net displacement relative to that positive direction:
- A positive average velocity means the final position is greater than the initial position.
- A negative average velocity means the final position is less than the initial position.
- A zero average velocity means the final and initial positions are the same.
These statements concern the start and end of the interval. A negative average velocity does not mean the object moved in the negative direction at every moment. It means only that its net change in position was negative over that interval.
Units come directly from the quotient. If position is recorded in metres and time in seconds, average velocity is measured in metres per second, written \(\text{m/s}\). If position is recorded in kilometres and time in hours, the result is in kilometres per hour, written \(\text{km/h}\). The units should be carried through the division, not added afterward as a guess.
Worked Example: Average Velocity from a Table
Worked Example: Average Velocity from a Table
The table gives an object's position at several times. Position is measured in metres and time in seconds. Find its average velocity from \(t=2\) seconds to \(t=8\) seconds.
| Time \(t\) (s) | Position \(x(t)\) (m) |
|---|---|
| 0 | 12 |
| 2 | 18 |
| 5 | 9 |
| 8 | 6 |
Read the starting and ending positions from the rows for \(t=2\) and \(t=8\): \(x(2)=18\text{ m}\) and \(x(8)=6\text{ m}\). Thus the displacement is:
The elapsed time is:
Divide displacement by elapsed time:
Check: Over \(6\) seconds, an average velocity of \(-2\text{ m/s}\) corresponds to a displacement of \((-2)(6)=-12\text{ m}\), matching the table’s endpoint change. Although the table includes a measurement at \(t=5\), that intermediate position does not change the average for the interval from \(2\) to \(8\) seconds.
Position Depends on the Origin; Average Velocity Does Not
A position value can change if the origin is moved. For example, describing the same locations from an origin seven metres farther along the line subtracts seven metres from every position. But both endpoint positions shift by the same amount, so their difference—and therefore the average velocity—stays unchanged.
Worked Example: Changing the Position Origin
Suppose \(x(t)=2t^2-3t+7\) metres, with \(t\) in seconds. Find the average velocity from \(t=1\) to \(t=3\), then check the result after using the shifted position coordinate \(q(t)=x(t)-7\).
For the original coordinate, calculate both endpoint positions:
The displacement is \(16-6=10\text{ m}\), and the elapsed time is \(3-1=2\text{ s}\). Hence:
Now use \(q(t)=x(t)-7\). The shifted endpoint positions are \(q(1)=6-7=-1\text{ m}\) and \(q(3)=16-7=9\text{ m}\). Their displacement is \(9-(-1)=10\text{ m}\), so:
Check: Both coordinate choices give the same \(10\text{ m}\) change over \(2\text{ s}\). Moving the origin changes the position values, but not their difference.
A Common Sign Mistake
A frequent error is to subtract the positions in the wrong order while keeping the elapsed time in the forward order. For the table example, this incorrect calculation uses the starting position minus the ending position:
The positive answer conflicts with the fact that position fell from \(18\text{ m}\) to \(6\text{ m}\). The correction is to use final minus initial in both changes:
Keep the endpoint order consistent: the position in the numerator must correspond to the time in the same place in the denominator. For a forward interval, subtract the starting values from the ending values.
Choosing the Data and Reporting the Result
To find average velocity reliably, use a short sequence of checks:
Record the starting time \(t_1\) and the later ending time \(t_2\).
Find \(x(t_1)\) and \(x(t_2)\) from the function or table.
Use displacement \(x(t_2)-x(t_1)\) and elapsed time \(t_2-t_1\).
State the result with position-per-time units and explain what its sign says about the net change.
As a final reasonableness check, multiply the reported average velocity by the elapsed time. The result should equal the displacement. Also compare the sign with the endpoint positions: increasing position should give a positive result, decreasing position a negative result, and equal endpoint positions zero.
Average velocity is useful whenever position data are known at two times and the question concerns the overall motion across that interval. It captures net change without requiring a detailed account of every position in between. The endpoints and their time separation are the essential information.
Check Your Understanding
Use displacement divided by elapsed time in each question, and include units where appropriate.
- An object is at \(5\text{ m}\) at \(t=1\text{ s}\) and at \(17\text{ m}\) at \(t=4\text{ s}\). Find its average velocity.
- A position table shows \(x(2)=-3\text{ m}\) and \(x(7)=12\text{ m}\). What is the average velocity from \(t=2\) to \(t=7\)?
- If a position function gives the same position at the start and end of an interval, what is the average velocity over that interval? Explain using the definition.
- Why do intermediate table entries not affect average velocity when the two endpoint times are fixed?
- An object’s average velocity is negative over an interval. What does this tell you about its net displacement relative to the chosen positive direction?
