From Velocity Samples to Displacement
Velocity describes how position changes over time. If velocity is known throughout a time interval, it can be used to determine the change in position. In practice, however, measurements may be available only at selected times. A useful estimate is to treat each sampled velocity as representative of a short interval, calculate the displacement associated with that interval, and add the contributions.
The key calculation is the same one used for constant velocity: velocity multiplied by elapsed time gives change in position. When velocity is negative, the resulting contribution is negative, so it records motion in the opposite direction rather than simply a smaller positive amount.
The Calculation and Its Units
Suppose an interval begins at time \(a\) and ends at time \(b\). Its duration is \(b-a\). If the velocity sample used for that interval is \(v\), the interval's estimated displacement is \(v(b-a)\). For several consecutive intervals, add these products, retaining their signs.
Here, each \(\Delta t_i\) is the duration of its corresponding interval, and \(v_i\) is the velocity sample assigned to that interval. The units confirm the kind of quantity calculated: for velocity in metres per second and time in seconds, each product is in metres.
A positive contribution represents displacement in the positive direction; a negative contribution represents displacement in the negative direction. Their sum is net displacement, the change from initial position to final position. It is not necessarily the total distance traveled.
Why Multiplication and Addition Work
First consider an interval on which velocity is constant. The definition of constant velocity gives change in position as velocity multiplied by elapsed time. Thus, on an interval of duration \(\Delta t_i\), the displacement is \(v_i\Delta t_i\).
For consecutive intervals, the total change in position is the sum of the changes on each interval. To see this directly, let the positions at the successive interval boundaries be \(x_0,x_1,\ldots,x_n\). The total change is
Every intermediate position appears once with a positive sign and once with a negative sign, so those terms cancel. If velocity is constant on each interval, each difference \(x_i-x_{i-1}\) equals \(v_i\Delta t_i\). Therefore the total displacement equals the sum of the interval products.
This distinction matters. A table of samples alone does not show exactly how velocity behaved between measurement times. Shorter intervals can make the estimate more informative, but the calculation itself does not guarantee a particular accuracy.
Worked Example: Motion in Both Directions
Worked Example: Motion in Both Directions
A cart has the following representative velocity samples. Estimate its displacement from \(0\) to \(6\) seconds.
| Time interval (s) | Sample velocity (m/s) |
|---|---|
| 0 to 2 | 4 |
| 2 to 5 | \(-2\) |
| 5 to 6 | \(-3\) |
First calculate each duration: \(2-0=2\) seconds, \(5-2=3\) seconds, and \(6-5=1\) second. Multiply each sample by its interval duration:
Add the signed contributions:
The estimated displacement is \(-1\) metre. To check, the positive-direction contribution is \(8\) metres and the combined negative-direction contribution is \(-9\) metres; \(8-9=-1\), as calculated. The negative result means the cart's final position is estimated to be one metre in the negative direction from its initial position.
Unequal Time Intervals
Time intervals need not have the same duration. Always calculate each width from that interval's own start and end times. Using a single duration for all samples would give incorrect weights whenever the intervals differ.
Worked Example: Unequal Durations and a Negative Net Change
A vehicle's representative velocity samples are \(2\) metres per second from \(0\) to \(1.5\) seconds, \(1\) metre per second from \(1.5\) to \(4\) seconds, and \(-3\) metres per second from \(4\) to \(6\) seconds. Estimate its displacement.
The interval durations are \(1.5-0=1.5\) seconds, \(4-1.5=2.5\) seconds, and \(6-4=2\) seconds. The separate contributions are
Adding them gives
The estimated displacement is \(-0.5\) metre. Check the arithmetic by grouping the positive and negative contributions: the positive contributions total \(3+2.5=5.5\) metres, and \(5.5-6=-0.5\) metres. The units also check: every velocity in metres per second is multiplied by a duration in seconds, leaving metres.
Displacement Is Not Distance
Displacement keeps track of direction. If an object moves forward and then backward, the contributions can partly cancel. Distance traveled, by contrast, adds the lengths of the paths and does not count direction. For the same intervals and samples as in the first example, adding the magnitudes of the contributions gives \(8+6+3=17\) metres as an estimate of distance, while the signed sum gives \(-1\) metre as the estimated displacement. The two answers describe different quantities.
Worked Example: Finding a Final Position from Estimated Displacement
An object starts at position \(-3\) metres. Its representative velocity is \(5\) metres per second over \(0\) to \(2\) seconds, \(0\) metres per second over \(2\) to \(3\) seconds, and \(-2\) metres per second over \(3\) to \(6\) seconds. Estimate its final position.
The interval durations are \(2-0=2\) seconds, \(3-2=1\) second, and \(6-3=3\) seconds. The displacement contributions are
The estimated net displacement is \(10+0-6=4\) metres. Final position equals initial position plus displacement, so
The estimated final position is \(1\) metre. Check by reversing the position change: \(1-(-3)=4\) metres, which agrees with the estimated displacement. The negative contribution describes the later motion in the negative direction; it does not mean the final position must be negative.
A Common Sign Error
A frequent mistake is to replace negative velocities with their absolute values before adding. For the first example, that incorrect calculation would be
This calculation removes the direction of motion and gives an estimate of total distance, not displacement. The correct displacement calculation retains the original signs:
Another error is to multiply by the time at which a sample was recorded rather than by the duration it represents. For an interval from \(2\) to \(5\) seconds, the width is \(5-2=3\) seconds, not \(5\) seconds. Use the interval's elapsed time, then check that the product has units of displacement.
How to Organize a Calculation
A careful layout helps prevent sign, interval, and unit errors:
Record each interval's start and end times, and calculate its duration by subtracting start time from end time.
Use the velocity specified as representative of that interval; do not assume a sample represents a different time span.
Multiply each velocity by its own duration, keeping negative signs.
Add the contributions to estimate net displacement, then state its units and what its sign means.
Check Your Understanding
For each question, keep track of interval widths, signs, and units.
- A runner travels at a representative velocity of \(3\) metres per second for \(4\) seconds, then \(-2\) metres per second for \(3\) seconds. What is the estimated displacement?
- Why does a negative velocity produce a negative displacement contribution when multiplied by a positive time duration?
- What interval duration should be used for a sample representing the time from \(2.5\) seconds to \(5\) seconds?
- Under what condition is adding velocity-times-duration products exact for displacement?
- How does the sum of the magnitudes of the contributions differ in meaning from their signed sum?
