Calculus I · Unit 3A · lesson

Displacement, Distance, and Signed Accumulation

Concept

Learning objectives

Use the sign of velocity to distinguish displacement from total distance traveled.

Displacement, Distance, and Signed Accumulation

Explanation

Why signed accumulation and total travel differ

Velocity contains direction as well as speed. Positive velocity contributes positive displacement, while negative velocity contributes negative displacement. Consequently, $\int_a^b v(t)\,dt$ records the net change in position: motion in opposite directions can cancel. A particle can travel a long way and still have zero displacement if it returns to its starting point.

Total distance asks a different question and therefore uses $|v(t)|$ . Before integrating for distance, locate every time at which velocity changes sign and split the interval there. This is not ceremonial bookkeeping. It prevents a segment traveled to the left from being subtracted from a segment traveled to the right, and it keeps the mathematical result aligned with the everyday meaning of distance traveled.

The integral of velocity gives displacement:

\int_a^b v(t)\,dt=s(b)-s(a).

If velocity is negative, the object moves in the negative direction and contributes negative displacement. Total distance ignores direction and instead uses speed:

\text{distance}=\int_a^b |v(t)|\,dt.

Worked example

A particle changes direction

Let $v(t)=t-2$ on $[0,5]$ . Then

\int_0^5(t-2)\,dt=\left[\frac12t^2-2t\right]_0^5=\frac52.

The displacement is $2.5$ . Since velocity changes sign at $t=2$ , total distance is

-\int_0^2(t-2)\,dt+\int_2^5(t-2)\,dt=2+\frac92=\frac{13}{2}.

The object ends $2.5$ units to the right of its start after traveling $6.5$ units in all.

Interactive checku3a-displacement-01

If $v(t)=2t-4$ on $[0,4]$ , what is the displacement?

Your work stays on this device. No account or AI grader is used.

Show hint

Integrate the velocity over the full interval.

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Common mistake

A negative definite integral does not mean negative distance. It means the net signed change points in the negative direction. Distance requires absolute value or interval-by-interval sign analysis.

After the explanation

Use the section idea

Reading lens

Connect every antiderivative to a derivative check, and every varying rate to a sum of rate-times-width contributions.

Mental model

Indefinite integration recovers a family of functions; definite accumulation combines signed local changes into one net change.

Decision

Ask whether the task wants a general antiderivative, an initial-condition solution, displacement, distance, or a numerical total from data.

Common trap

Omitting the arbitrary constant, confusing displacement with distance, or multiplying one changing rate by the entire interval.

Check yourself

Can you differentiate your antiderivative and interpret the sign and units of a rate-based total?

Source & rights

Original instruction with traceable references.

BetterGrades-original; no direct adaptation declared in the verified handoff.

Reference textbooks remain rights-separated and are not published as application assets. Any direct adaptation requires separate identification and attribution.