Position And Average Velocity
Master position and average velocity for Queensland Year 11 Mathematical Methods (QCAA). Average velocity is the average rate of change of position — an object's displacement divided by the time taken.
You will learn to find average velocity from a position rule, tell displacement apart from distance, and compare average velocity with average speed, reading each as the gradient of a secant line on a position-time graph.
Every question with a fully worked solution.
- Position And Average Velocity - Video - Average Velocity Watch
Theory
In Year 11 Mathematical Methods (QCAA, Unit 2), for straight-line motion the average velocity is the average rate of change of position: displacement divided by time taken, which equals the gradient of the secant of the position-time graph. This page shows how to read position and displacement, compute average velocity from a graph or a position rule \(x(t)\), and tell displacement and average velocity (signed) from distance and average speed.
The position \(x(t)\) of an object gives where it is on a line at time \(t\). The displacement over an interval is the change in position, \(x(t_2)-x(t_1)\); it is a signed quantity (it can be negative).
The average velocity is the displacement divided by the time taken, \(\dfrac{x(t_2)-x(t_1)}{t_2-t_1}\). This is the average rate of change of position, so it is the gradient of the secant joining the two points on the position-time graph.
Distance is the total path length travelled (always \(\ge 0\)), and average speed is distance divided by time. When the object reverses direction, distance is greater than the size of the displacement, so average speed differs from the size of the average velocity.
For a position \(x(t)\), the average velocity over \([t_1,t_2]\) is
Average speed uses the total distance (path length) instead of the signed displacement:
How to find an average velocity over \([t_1,t_2]\)
- Positions: read or evaluate \(x(t_1)\) and \(x(t_2)\) from the graph, table or rule.
- Displacement: subtract to get \(x(t_2)-x(t_1)\), keeping the sign.
- Divide: compute \(\dfrac{x(t_2)-x(t_1)}{t_2-t_1}\), the gradient of the secant, and state the units (m/s) and direction.
Displacement is the change in position:
| \(\Delta x\) | \(=\) | \(x(5)-x(1)\) |
| \(=\) | \(50-10\) | |
| \(=\) | \(40\text{ m}\) |
Average velocity is displacement over time (the secant gradient):
| \(\bar v\) | \(=\) | \(\dfrac{40}{5-1}\) |
| \(=\) | \(\dfrac{40}{4}\) | |
| \(=\) | \(10\text{ m/s}\) |
Average velocity \(=10\) m/s.
Position at each endpoint:
| \(x(1)\) | \(=\) | \((1)^2+2(1)=3\) |
| \(x(4)\) | \(=\) | \((4)^2+2(4)=24\) |
Average velocity is displacement over time:
| \(\bar v\) | \(=\) | \(\dfrac{x(4)-x(1)}{4-1}\) |
| \(=\) | \(\dfrac{24-3}{3}\) | |
| \(=\) | \(\dfrac{21}{3}\) | |
| \(=\) | \(7\text{ m/s}\) |
Average velocity \(=7\) m/s.
Key positions — substitute \(t=0,3,5\):
| \(x(0)\) | \(=\) | \(0\) |
| \(x(3)\) | \(=\) | \((3)^2-6(3)=-9\) |
| \(x(5)\) | \(=\) | \((5)^2-6(5)=-5\) |
Average velocity uses the displacement \(x(5)-x(0)\):
| \(\bar v\) | \(=\) | \(\dfrac{-5-0}{5-0}\) |
| \(=\) | \(\dfrac{-5}{5}\) | |
| \(=\) | \(-1\text{ m/s}\) |
Distance adds each leg: \(0\to-9\) then \(-9\to-5\):
| \(\text{distance}\) | \(=\) | \(|-9-0|+|-5-(-9)|\) |
| \(=\) | \(9+4\) | |
| \(=\) | \(13\text{ m}\) |
Average speed uses the distance:
| \(\text{avg speed}\) | \(=\) | \(\dfrac{13}{5}\) |
| \(=\) | \(2.6\text{ m/s}\) |
Average velocity \(=-1\) m/s; average speed \(=2.6\) m/s.
Average velocity over \([1,3]\):
| \(\bar v_1\) | \(=\) | \(\dfrac{x(3)-x(1)}{3-1}\) |
| \(=\) | \(\dfrac{9-1}{2}\) | |
| \(=\) | \(4\text{ m/s}\) |
Average velocity over \([3,5]\):
| \(\bar v_2\) | \(=\) | \(\dfrac{x(5)-x(3)}{5-3}\) |
| \(=\) | \(\dfrac{25-9}{2}\) | |
| \(=\) | \(8\text{ m/s}\) |
Both are positive (moving forwards) and \(8>4\), so the second secant is steeper: the particle is speeding up.
\([1,3]\): \(4\) m/s; \([3,5]\): \(8\) m/s — the average velocity is greater on the later interval.
Common pitfalls
Frequently asked questions
What is average velocity?
The average rate of change of position: displacement divided by time taken, \(\dfrac{x(t_2)-x(t_1)}{t_2-t_1}\), which is the gradient of the secant of the position-time graph.
What is the difference between distance and displacement?
Displacement is the signed change in position \(x(t_2)-x(t_1)\); distance is the total path length travelled, which is never negative.
What is the difference between average speed and average velocity?
Average velocity uses displacement (signed); average speed uses distance (always \(\ge0\)). They differ whenever the object changes direction.
How do you find average velocity from a position rule?
Evaluate the position at both times, subtract for the displacement, then divide by the time taken. No differentiation is needed.
Why is average velocity the gradient of a secant?
Because it is the change in position over the change in time between two points, which is exactly the gradient of the line joining those points on the position-time graph.
Can average velocity be negative?
Yes. If the final position is less than the starting position the displacement is negative, so the average velocity is negative, meaning net motion in the negative direction.