Did you know that integrating velocity gives displacement, not necessarily total distance travelled? If a moving particle reverses direction, some signed movements cancel in the velocity integral. To find total distance, first solve v(t) = 0, check where the sign of velocity changes, then integrate the magnitude |v(t)| over each relevant time interval.
For example, let v(t) = t² − 4t + 3 metres per second for 0 ≤ t ≤ 4 seconds. Velocity is positive until t = 1, negative from t = 1 to 3, and positive again from t = 3 to 4. The signed displacement over four seconds is only 4/3 metres, but the total distance travelled is 4 metres. This guide works through the exact integration, position changes, average speed, common exam errors and the distinction between a zero velocity and a true direction reversal.
Displacement and distance measure different things
Displacement is the signed difference between the particle’s final position and its starting position. Along a chosen straight axis, movement in the positive direction contributes positively and movement in the negative direction contributes negatively.
Total distance is how far the particle actually travelled along its path. Every portion of the journey contributes a non-negative length, regardless of direction.
For velocity v(t), displacement over a ≤ t ≤ b is ∫_a^b v(t) dt. Total distance is ∫_a^b |v(t)| dt, provided velocity is integrable.
This distinction is essential whenever v(t) changes sign. If a particle travels three metres forward and two metres backward, its displacement is one metre forward but its total distance is five metres. One quantity cannot replace the other.
A velocity sign tells us the current direction
On a straight line with the positive axis chosen in advance, v(t) > 0 means movement in the positive direction and v(t) < 0 means movement in the negative direction.
When v(t) = 0, the particle is instantaneously at rest. It may then reverse, resume in the same direction or remain stationary for a while.
Therefore zeros of velocity are candidate times for a direction change, but a sign test is needed to establish whether the sign actually changes across each zero.
A graph of v(t) against t can help visualise the intervals. Regions above the time axis contribute positive displacement and regions below contribute negative displacement. The total distance adds the magnitudes of all those signed regions.
Factor the worked velocity polynomial
Let v(t) = t² − 4t + 3 for 0 ≤ t ≤ 4. Factor it as (t − 1)(t − 3).
Thus velocity is zero at t = 1 and t = 3. Both lie inside the observed four-second interval.
These are the important points at which the velocity sign may change. Divide the time interval into [0,1], [1,3] and [3,4].
A student should not immediately integrate from 0 to 4 and take one absolute value. That would discard the possibility of direction reversals in the middle. First identify the sign of velocity in each part of the journey.
Make the velocity sign chart
For 0 < t < 1, both factors t − 1 and t − 3 are negative, so their product is positive.
For 1 < t < 3, the first factor is positive and the second negative, so velocity is negative. For 3 < t < 4, both factors are positive, so velocity is positive.
Therefore the particle moves in the positive direction on [0,1], reverses to the negative direction on [1,3], and reverses again to positive on [3,4].
The sign chart is not decorative. It tells us which definite integrals can be used directly as distances and which must have their signs reversed when calculating the total travelled length.
Find an antiderivative of velocity
Integrate v(t) = t² − 4t + 3 term by term. A suitable antiderivative is F(t) = t³/3 − 2t² + 3t.
Its derivative is F′(t) = t² − 4t + 3, confirming that the integration is correct.
For definite integrals, the arbitrary integration constant would cancel between the upper and lower limits, so we may use this F without writing a constant at every evaluation.
This one antiderivative will be reused for all three time intervals. Calculating it once and recording its values at the sign-change boundaries is more efficient than starting a fresh integration in every region.
Evaluate F at the relevant times
At t = 0, F(0) = 0. At t = 1, F(1) = 1/3 − 2 + 3 = 4/3.
At t = 3, F(3) = 27/3 − 2(9) + 3(3) = 9 − 18 + 9 = 0.
At t = 4, F(4) = 64/3 − 2(16) + 12 = 64/3 − 20 = 4/3.
Thus the boundary values are 0, 4/3, 0 and 4/3 for t = 0,1,3,4 respectively. A table of these values makes the next displacement and distance calculations short and reduces repeated fraction arithmetic.
Calculate the first forward distance
On 0 ≤ t ≤ 1, velocity is non-negative. Therefore the distance travelled equals the signed integral ∫_0^1 v(t) dt.
Using the antiderivative, this is F(1) − F(0) = 4/3 − 0 = 4/3 metres.
The particle begins at its starting position and moves forward by 4/3 metres before velocity reaches zero at t = 1.
No absolute-value reversal is required here because v(t) remains positive inside the interval. The sign chart establishes the correct interpretation of the definite integral.
Calculate the backward distance
On 1 ≤ t ≤ 3, velocity is non-positive. The signed displacement is F(3) − F(1) = 0 − 4/3 = −4/3 metres.
The negative sign means that the particle moved in the negative direction. The actual distance travelled in this segment is the positive magnitude 4/3 metres.
Thus the distance contribution is −∫_1^3 v(t) dt = 4/3.
A common mistake is to add the negative signed integral directly to the other positive distance contributions. That computes a mixture of displacement and distance, not total distance travelled. Every interval contribution to total distance must be non-negative.
Calculate the final forward distance
On 3 ≤ t ≤ 4, the velocity becomes non-negative again. The signed displacement is F(4) − F(3) = 4/3 − 0 = 4/3 metres.
Because this is forward motion, its distance contribution is also 4/3 metres.
The journey thus consists of three signed stages: forward 4/3, backward 4/3 and forward 4/3.
The backward section cancels one of the forward contributions when calculating net displacement, but it still counts in full toward the distance travelled. This is the central conceptual distinction in the problem.
Add the three non-negative distance contributions
Total distance is 4/3 + 4/3 + 4/3 = 12/3 = 4 metres.
This is the exact travelled length over the four-second interval. The answer does not require approximating any fractions.
The integral expression is ∫_0^4 |t² − 4t + 3| dt = ∫_0^1 v(t) dt − ∫_1^3 v(t) dt + ∫_3^4 v(t) dt.
The minus sign in the middle is intentional: velocity is negative there, so subtracting its signed integral converts the contribution to positive distance. The sign chart tells us which pieces need that treatment.
Calculate displacement in one integral
The net displacement is ∫_0^4 v(t) dt = F(4) − F(0) = 4/3 metres.
We can also add the three signed displacements: +4/3 − 4/3 + 4/3 = +4/3.
Therefore the particle finishes 4/3 metres in the positive direction from its starting point, despite having travelled a total distance of 4 metres.
This is the result a student would get by integrating once over the full interval without absolute values. That calculation is correct for displacement; it becomes wrong only if labelled as total distance.
The magnitude of displacement is not total distance
The magnitude of displacement is |4/3| = 4/3 metres. Total distance is 4 metres.
The general inequality total distance ≥ |displacement| always holds for motion along a line, because reversing direction adds travelled length without increasing net separation in the same way.
Equality occurs when the velocity does not reverse direction apart from possible stationary instants. When reversals happen, the inequality is usually strict.
This provides a useful final check. If a calculated total distance is smaller than the magnitude of displacement, a sign was probably mishandled, a segment omitted or the integration limits copied incorrectly.
Recover actual position if a starting point is given
Suppose the particle begins at position s(0) = 5 metres on the chosen axis. Since s′(t) = v(t), its position function is s(t) = 5 + F(t).
Therefore s(0) = 5, s(1) = 5 + 4/3 = 19/3, s(3) = 5 and s(4) = 19/3.
The position first rises to 19/3, returns to 5, and rises again to 19/3. That gives the same total distance: each of the three position changes has magnitude 4/3.
Notice that the starting coordinate changes absolute positions but not the total distance or net displacement during the specified time interval. Motion differences depend on velocity, not the arbitrary origin of position.
A return to the starting position need not mean no movement
Over 0 ≤ t ≤ 3, the same velocity function has displacement F(3) − F(0) = 0.
The particle returns to its starting position at t = 3, but it has moved forward by 4/3 metres and backward by 4/3 metres.
Thus its total distance on [0,3] is 8/3 metres, even though displacement is zero.
This is one of the clearest counterexamples to equating total distance with the absolute value of the overall signed integral. A round trip can have no net displacement but substantial movement.
A shorter interval can have a different answer
Take the same velocity v(t) = t² − 4t + 3 but restrict the question to 0 ≤ t ≤ 2.
Only the velocity zero at t = 1 lies inside this interval. The sign is positive from 0 to 1 and negative from 1 to 2.
The antiderivative value at t = 2 is F(2) = 8/3 − 8 + 6 = 2/3.
The forward distance is 4/3, while the backward distance is |F(2) − F(1)| = |2/3 − 4/3| = 2/3. Total distance is 2 metres, and displacement is 2/3 metre. The time interval is part of the question, not an optional detail.
Average velocity uses signed displacement
Average velocity over [a,b] is total signed displacement divided by elapsed time b − a.
For the four-second example, displacement is 4/3 metres and elapsed time is 4 seconds. Therefore average velocity is (4/3)/4 = 1/3 metre per second.
This agrees with the average value of the velocity function: (1/4)∫_0^4 v(t) dt = 1/3.
The result has a positive sign because the final position is ahead of the starting one on the chosen axis. It does not describe the average of the instantaneous velocity magnitudes.
Average speed uses total distance
Average speed is total distance travelled divided by elapsed time, regardless of direction.
For the main journey, total distance is 4 metres and elapsed time is 4 seconds. Therefore average speed is 1 metre per second.
Average velocity is only 1/3 metre per second. The two averages differ because part of the journey goes backward.
When no direction changes occur, the magnitude of average velocity can equal average speed. But in a reversing journey, average speed accounts for all movement while signed average velocity includes cancellations.
A linear velocity with zero displacement
Consider a second example v(t) = 6t − 12 for 0 ≤ t ≤ 4. Velocity is zero at t = 2, negative before that and positive after.
An antiderivative is F(t) = 3t² − 12t. Its values at t = 0,2,4 are 0, −12 and 0.
The first signed displacement is −12 metres, and the second is +12 metres. Net displacement is zero, but total distance is 12 + 12 = 24 metres.
Average velocity is zero, while average speed is 24/4 = 6 metres per second. The example makes the difference especially visible without fractions.
A zero of velocity may not be a reversal
Now consider v(t) = (t − 1)² over 0 ≤ t ≤ 2. Velocity equals zero at t = 1, but is positive on both sides.
The particle slows to an instantaneous rest and resumes in the same positive direction. There is no direction reversal.
Therefore total distance is simply ∫_0^2 (t − 1)² dt, which is non-negative throughout. It equals [(t − 1)³/3]_0^2 = 1/3 − (−1/3) = 2/3.
This demonstrates why solving v(t) = 0 is only the beginning. A sign chart distinguishes a simple root that changes sign from a repeated root at which velocity may merely touch zero.
Multiplicity helps predict sign changes
For a polynomial velocity, a factor (t − c) raised to an odd positive power changes sign as t crosses c, provided other nonzero factors do not change sign simultaneously.
A factor raised to an even positive power does not change sign merely by crossing c. For example, (t − 1)² remains non-negative on both sides of t = 1.
This algebraic pattern helps forecast reversals, but direct test points on the appropriate intervals remain a reliable practical method.
A student should not count every velocity zero as an additional direction change or automatically split distance into alternating positive and negative segments without checking actual signs.
Acceleration zero is not the same as velocity zero
For v(t) = t² − 4t + 3, acceleration is a(t) = v′(t) = 2t − 4.
Acceleration vanishes at t = 2. But velocity at that time is v(2) = 4 − 8 + 3 = −1, not zero.
The particle is still moving in the negative direction at t = 2, even though its acceleration is instantaneously zero. A change in acceleration sign does not by itself create a reversal in motion.
Reversals are identified from velocity signs, not acceleration signs. Confusing these quantities can lead a student to split the total-distance integral at the wrong times.
Velocity and position graphs encode different information
On a velocity–time graph, the signed area between the curve and time axis equals displacement. The total area after reflecting below-axis regions upward equals distance.
On a position–time graph, the gradient equals velocity. A direction reversal appears as a local maximum or minimum of position, when the gradient changes sign.
For the main example, position has a local maximum at t = 1 and a local minimum at t = 3. The velocity graph crosses the time axis at those times.
Students who can switch between these two views understand why the same definite integral gives signed displacement and why a positive distance needs separate interval contributions.
Why the absolute value belongs inside the integral
For distance, the correct mathematical expression is ∫_a^b |v(t)| dt, where each instantaneous velocity contributes its non-negative magnitude.
In general, this is not equal to |∫_a^b v(t) dt|. The latter takes the magnitude only after positive and negative contributions have already cancelled.
For the main example, ∫_0^4 |v| = 4, whereas |∫_0^4 v| = 4/3.
This distinction is an instance of the general integral inequality |∫f| ≤ ∫|f|. It captures the simple geometric fact that a total journey length cannot be smaller than the direct separation between its starting and ending positions.
A short solution format for timed examination questions
First record the time interval and solve v(t) = 0. Keep only zero times inside the permitted interval, including endpoints where relevant.
Second build a sign chart for the subintervals. Third find an antiderivative F of v, and evaluate F at every boundary.
Fourth calculate each signed displacement F(t₂) − F(t₁), then add the magnitudes to obtain total distance. Report the net signed sum separately if displacement is also requested.
Finally divide by the total elapsed time for average speed or average velocity as appropriate. This workflow makes the meaning of every number explicit and avoids accidental sign cancellation.
Practice questions with checked answers
Question one: v(t) = t² − 4t + 3 on 0 ≤ t ≤ 4. Displacement: 4/3 metres. Total distance: 4 metres.
Question two: for the same journey, find average velocity and average speed. Answer: 1/3 m/s and 1 m/s.
Question three: same velocity on 0 ≤ t ≤ 3. Displacement: 0. Distance: 8/3 metres.
Question four: same velocity on 0 ≤ t ≤ 2. Displacement: 2/3 metre. Distance: 2 metres.
Question five: v(t) = 6t − 12 on 0 ≤ t ≤ 4. Displacement: 0. Total distance: 24 metres.
Question six: v(t) = (t − 1)² on 0 ≤ t ≤ 2. Number of direction reversals: zero. Distance: 2/3 unit.
Before looking at the answers, mark the sign of velocity on each interval and identify which integrals must be negated for distance.
Common errors and targeted teaching
Some students integrate the velocity correctly but label the signed result as total distance. Others locate the velocity zeros but never check signs, so a non-reversing zero creates an unnecessary negative region.
Another learner knows how to split the integral but writes an incorrect antiderivative for the polynomial. That is an integration prerequisite, not a conceptual motion problem.
A strong correction starts at the first invalid decision: distinguish displacement from distance, establish sign intervals, then apply the power rule and exact bounds.
Changing the observation interval or velocity polynomial after the repair tests whether the student can transfer the method rather than memorise a three-part calculation.
Syllabus suitability and connected learning
Velocity, displacement, distance and average value link calculus to applied motion. The precise assessed scope depends on the learner’s Mathematics programme; some polynomial-velocity extensions are enrichment for certain secondary pathways.
Singapore families can consult the relevant SEAB references for 2026 O-Level Additional Mathematics, 2027 SEC G3 or 2027 SEC G2, as appropriate.
Related guides include How to Find a Function’s Average Value from a Definite Integral, How to Find an Unknown Parameter from a Definite Integral, How to Check the Meaning of a Calculus Answer and the Additional Mathematics archive.
Families may enquire about eduKate Singapore’s small-group Mathematics tutorials near Sixth Avenue MRT. The lasting skill is translating velocity’s sign into movement direction before choosing what an integral should measure. Properly taught kids shine a bright light into the future.