Why Integrate Velocity?
A particle’s velocity describes how its position changes with time. When velocity varies, a single velocity value does not tell us the particle’s overall change in position during an interval. A definite integral combines the velocity values across that interval to give the particle’s net displacement.
The word “net” matters. Velocity can be positive at some times and negative at others, so contributions in opposite directions can cancel. The integral records the resulting change in position, not the particle’s position by itself.
Definition and Units
Let \(x(t)\) denote a particle’s position at time \(t\), and let \(v(t)\) denote its velocity. Displacement over the interval from \(t=a\) to \(t=b\) is the change in position, \(x(b)-x(a)\). Since velocity is the derivative of position, integrating velocity gives that change.
The units follow from the meaning of velocity. If \(v\) is measured in meters per second and \(t\) in seconds, then the integral is measured in meters:
The sign of the integral indicates the direction of the net change along the number line. A positive displacement means the final position is greater than the initial position; a negative displacement means it is less. A displacement of zero means the particle ends where it began, not necessarily that it remained at rest.
Deriving the Displacement Formula
The formula comes directly from the relationship between position and velocity. Suppose \(x'(t)=v(t)\) throughout the interval. Then \(x\) is an antiderivative of \(v\), so the Fundamental Theorem of Calculus applies.
At each time \(t\), velocity is the derivative of position: \(x'(t)=v(t)\).
Because \(x\) is an antiderivative of \(v\), \(\int_a^b v(t)\,dt=\int_a^b x'(t)\,dt\).
\(\int_a^b x'(t)\,dt=x(b)-x(a)\), which is the change in position by definition.
This argument explains why the integral gives displacement: it accumulates the rate of change of position, and the accumulated change is the final value minus the initial value. If velocity is continuous on \([a,b]\), it is integrable there, so the integral is well-defined and this reasoning applies whenever \(x'=v\).
Worked Example: Positive Net Displacement
Worked Example: Positive Net Displacement
A particle moves along a line with velocity \(v(t)=3t+1\) meters per second, where \(t\) is measured in seconds. Find its displacement from \(t=0\) to \(t=4\).
Integrate the velocity over the given time interval:
An antiderivative of \(3t+1\) is \(\frac{3}{2}t^2+t\). Evaluate it at the bounds:
The displacement is \(28\) meters. It is positive, so the particle’s final position is 28 meters greater than its initial position.
Check: Differentiate the antiderivative: \(\frac{d}{dt}\left(\frac{3}{2}t^2+t\right)=3t+1\), the given velocity. The evaluated bounds give \(28-0=28\) meters.
Negative Displacement and Initial Position
A negative integral is not an error. It means that the net change in position is in the negative direction. To find a final position, however, displacement alone is not enough: the starting position must also be included.
Worked Example: Negative Net Displacement
Worked Example: Negative Net Displacement
A particle has velocity \(v(t)=2-t\) meters per second for \(0\leq t\leq 5\), with time in seconds. Its position at \(t=0\) is \(7\) meters. Find its displacement and its position at \(t=5\).
First calculate the displacement:
The displacement is \(-\frac{5}{2}\) meters, or \(-2.5\) meters. Now add it to the initial position:
Interpretation: The particle’s final position is \(4.5\) meters. Its displacement is negative because it ends \(2.5\) meters in the negative direction from where it started; its position is still positive because it began at \(7\) meters.
Check: The derivative of \(2t-\frac{t^2}{2}\) is \(2-t\), as required. The positions satisfy \(x(5)-x(0)=4.5-7=-2.5\) meters, matching the integral.
Worked Example: Displacement Can Be Small Even When Velocity Changes Sign
Worked Example: Displacement Can Be Small Even When Velocity Changes Sign
A particle has velocity \(v(t)=t^2-4t+3\) meters per second, with \(t\) in seconds. Its position at \(t=0\) is \(-1\) meter. Find its displacement and position at \(t=2\).
Integrate the velocity:
Use the initial position to find the position at \(t=2\):
Check: Differentiating \(\frac{t^3}{3}-2t^2+3t\) gives \(t^2-4t+3\). Also, the position change is \(-\frac{1}{3}-(-1)=\frac{2}{3}\) meter, equal to the integral.
Reading the Sign Correctly
The definite integral is a signed accumulation. On portions of the interval where \(v(t)>0\), the position increases and the integral receives a positive contribution. Where \(v(t)<0\), the position decreases and the contribution is negative. Thus, contributions in opposite directions can partly or completely cancel.
This is why displacement and position must not be confused. Displacement is a change, \(x(b)-x(a)\), and can be positive, negative, or zero. Position is a location relative to a chosen origin. It depends on the initial position as well as the displacement.
| Quantity | Meaning | How to find it |
|---|---|---|
| Velocity | Rate and direction of position change | \(v(t)=x'(t)\) |
| Displacement | Net change in position over an interval | \(\int_a^b v(t)\,dt\) |
| Final position | Position at the end of the interval | Initial position plus displacement |
Common Mistake: Reporting Displacement as Final Position
A frequent error is to calculate the integral correctly, then report it as the final position even when an initial position is given. In the negative-displacement example, the integral is \(-2.5\) meters and the initial position is \(7\) meters. Saying that the final position is \(-2.5\) meters confuses the change with the location.
The correction is to label the integral as \(\Delta x\), then use \(x(b)=x(a)+\Delta x\) whenever a final position is requested. If the question asks only for displacement, do not add an initial position: displacement is already the answer.
A Reliable Calculation Process
For a velocity-integral question, identify the requested quantity before calculating. Then use the bounds and units carefully.
Displacement is a change in position; final position also requires the initial position.
Write \(\int_a^b v(t)\,dt\), using the given start and end times in the correct order.
Find an antiderivative, evaluate upper bound minus lower bound, and state the displacement units.
Describe the net direction. If final position is requested, add the displacement to the initial position and check that their difference is the integral.
Why the Result Matters
Velocity is a rate, while displacement is the accumulated effect of that rate. The integral connects the two and provides a precise way to determine how far a particle’s position changes overall, even when its velocity varies throughout the interval.
Always preserve the distinction between a rate, a change, and a location. A velocity integral gives displacement. Adding that displacement to a known starting position gives the final position. Units and signs help verify that the result is being interpreted correctly.
Check Your Understanding
Answer each question. For calculation problems, include units and state whether the result is displacement or position.
- A particle has velocity \(v(t)=4\) meters per second from \(t=1\) to \(t=6\) seconds. What is its displacement?
- If a particle begins at position \(-3\) meters and has displacement \(5\) meters, what is its final position?
- What does a negative value of \(\int_a^b v(t)\,dt\) tell you about the particle’s initial and final positions?
- A particle has velocity \(v(t)=t\) meters per second for \(0\leq t\leq 3\), with time in seconds. Find its displacement and check the antiderivative.
- Can a particle have zero displacement over an interval without remaining at rest? Explain what the integral says.
