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Displacement vs. Total Distance

MATH162
Reference: Stewart §4.4  •  Chapter: 4  •  Section: 4

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Displacement vs. Total Distance

When Direction Matters

Imagine you walk 5 blocks east, then 3 blocks west. Where did you end up? 2 blocks east of where you started. That's your displacement—the net change in position.

But how far did you walk? 8 blocks total. That's your total distance traveled—it counts every step regardless of direction.

Calculus captures this distinction perfectly:

Prerequisite Map

Prerequisites
Net Change TheoremAbsolute Value
→
This skill
Displacement vs Distance
→
Unlocks
Area Between CurvesPhysics Applications

Quick Reference

Property Value
Concept Integration
Course MATH161
Difficulty Intermediate
Time ~15 minutes

Key Concepts

The Two Formulas

Let $s(t)$ be position and $v(t) = s'(t)$ be velocity. Then:

$$\boxed{\text{Displacement} = \int_{t_1}^{t_2} v(t)\,dt = s(t_2) - s(t_1)}$$

$$\boxed{\text{Total Distance} = \int_{t_1}^{t_2} \vert v(t)\vert \,dt}$$

Visual Interpretation

When $v(t) \geq 0$: Object moves in positive direction (e.g., right or up) When $v(t) < 0$: Object moves in negative direction (e.g., left or down)

    v(t)
    │   ╱╲    A₁ (positive area)
    │  ╱  ╲
────┼─╱────╲────────► t
    │       ╲ ╱
    │        V   A₂ (negative area)
    │

Computing Total Distance

To compute $\int_a^b \vert v(t)\vert \,dt$:

  1. Find where $v(t) = 0$ (these are the turning points)
  2. Split the interval at each root
  3. On each subinterval, determine if $v(t) > 0$ or $v(t) < 0$
  4. Integrate accordingly:
  5. Add up all the pieces (all positive)

Example: The Setup

A particle moves with velocity $v(t) = t^2 - 4$ for $0 \leq t \leq 3$.

Step 1: Find roots. $v(t) = 0$ when $t^2 = 4$, so $t = 2$ (only root in $[0,3]$)

Step 2: Check signs:

Step 3: $$\text{Displacement} = \int_0^3 (t^2 - 4)\,dt = \left[\frac{t^3}{3} - 4t\right]_0^3 = (9 - 12) - 0 = -3$$

$$\text{Distance} = \int_0^2 \vert t^2 - 4\vert \,dt + \int_2^3 \vert t^2 - 4\vert \,dt$$ $$= \int_0^2 -(t^2 - 4)\,dt + \int_2^3 (t^2 - 4)\,dt$$ $$= \int_0^2 (4 - t^2)\,dt + \int_2^3 (t^2 - 4)\,dt$$ $$= \left[4t - \frac{t^3}{3}\right]_0^2 + \left[\frac{t^3}{3} - 4t\right]_2^3$$ $$= \left(8 - \frac{8}{3}\right) + \left[(9 - 12) - \left(\frac{8}{3} - 8\right)\right]$$ $$= \frac{16}{3} + \left[-3 + \frac{16}{3}\right] = \frac{16}{3} + \frac{7}{3} = \frac{23}{3} \approx 7.67$$

Interpretation: The particle ended up 3 units to the left of where it started, but traveled about 7.67 units total.

Practice Problems

Level 1 Conceptual Distinction

A car drives 30 miles north, then 50 miles south. Find:

  1. The displacement
  2. The total distance traveled
Thought Process

No calculus needed here—just the concept:

  • Displacement is the net change in position (can be negative if south is negative)
  • Distance is the sum of all movements (always positive)
Show Answer
  1. **Displacement:** $30 + (-50) = -20$ miles

    The car ends up 20 miles south of where it started.</li>

  2. **Total distance:** $30 + 50 = 80$ miles

    The car traveled 80 miles total.</li>

Level 2 Positive Velocity

A particle moves with velocity $v(t) = 3t + 2$ m/s for $0 \leq t \leq 4$ seconds.

  1. Find the displacement.
  2. Find the total distance traveled.
  3. Why are your answers the same?
Thought Process

First check: Is $v(t) = 3t + 2$ ever negative on $[0, 4]$?

At $t = 0$: $v(0) = 2 > 0$ At $t = 4$: $v(4) = 14 > 0$

Since $v(t)$ is linear with positive slope and positive at $t = 0$, it's always positive on $[0, 4]$.

When velocity is always positive (or always negative), displacement and distance are the same (or negatives of each other).

Show Answer
  1. $$\text{Displacement} = \int_0^4 (3t + 2)\,dt = \left[\frac{3t^2}{2} + 2t\right]_0^4 = (24 + 8) - 0 = 32 \text{ m}$$
  2. Since $v(t) = 3t + 2 > 0$ for all $t \in [0, 4]$, we have $\vert v(t)\vert = v(t)$. $$\text{Distance} = \int_0^4 \vert v(t)\vert \,dt = \int_0^4 (3t + 2)\,dt = 32 \text{ m}$$
  3. The answers are the same because **the particle never reverses direction**. When $v(t) \geq 0$ throughout the interval, every bit of motion is in the positive direction. Displacement equals total distance.
Level 3 Finding the Turning Point

A particle moves along a line with velocity $v(t) = t^2 - 5t + 4$ m/s for $0 \leq t \leq 5$ seconds.

  1. Find the displacement of the particle.
  2. Find the total distance traveled.
Thought Process

Factor the velocity: $v(t) = t^2 - 5t + 4 = (t-1)(t-4)$

Roots at $t = 1$ and $t = 4$. Both are in $[0, 5]$.

Sign analysis:

  • $t \in [0, 1]$: Choose $t = 0.5$. $v(0.5) = (0.5)(−3.5) < 0$? Wait, let me recalculate: $v(0.5) = 0.25 - 2.5 + 4 = 1.75 > 0$
  • $t \in [1, 4]$: Choose $t = 2$. $v(2) = 4 - 10 + 4 = -2 < 0$
  • $t \in [4, 5]$: Choose $t = 4.5$. $v(4.5) = 20.25 - 22.5 + 4 = 1.75 > 0$

So: positive on $[0,1]$, negative on $[1,4]$, positive on $[4,5]$.

Show Answer

Step 1: Factor: $v(t) = (t-1)(t-4)$, so $v(t) = 0$ at $t = 1, 4$.

Step 2: Sign analysis:

  • $[0, 1]$: $v(0.5) = 1.75 > 0$ ✓
  • $[1, 4]$: $v(2) = -2 < 0$ ✓
  • $[4, 5]$: $v(4.5) = 1.75 > 0$ ✓
  1. $$\text{Displacement} = \int_0^5 (t^2 - 5t + 4)\,dt = \left[\frac{t^3}{3} - \frac{5t^2}{2} + 4t\right]_0^5$$ $$= \frac{125}{3} - \frac{125}{2} + 20 = \frac{250 - 375 + 120}{6} = \frac{-5}{6} \text{ m}$$

    The particle ends up $\frac{5}{6}$ m to the left of its starting position.

  2. $$\text{Distance} = \int_0^1 v(t)\,dt + \int_1^4 \vert v(t)\vert \,dt + \int_4^5 v(t)\,dt$$ $$= \int_0^1 v(t)\,dt - \int_1^4 v(t)\,dt + \int_4^5 v(t)\,dt$$

    Let $F(t) = \frac{t^3}{3} - \frac{5t^2}{2} + 4t$. Then:

    • $F(0) = 0$
    • $F(1) = \frac{1}{3} - \frac{5}{2} + 4 = \frac{2 - 15 + 24}{6} = \frac{11}{6}$
    • $F(4) = \frac{64}{3} - 40 + 16 = \frac{64 - 72}{3} = -\frac{8}{3}$
    • $F(5) = \frac{125}{3} - \frac{125}{2} + 20 = -\frac{5}{6}$

    $$\text{Distance} = [F(1) - F(0)] - [F(4) - F(1)] + [F(5) - F(4)]$$ $$= \frac{11}{6} - \left(-\frac{8}{3} - \frac{11}{6}\right) + \left(-\frac{5}{6} + \frac{8}{3}\right)$$ $$= \frac{11}{6} + \frac{16 + 11}{6} + \frac{-5 + 16}{6}$$ $$= \frac{11 + 27 + 11}{6} = \frac{49}{6} \approx 8.17 \text{ m}$$

Level 4 Velocity from Acceleration

A particle has acceleration $a(t) = 6t - 12$ m/s² and initial velocity $v(0) = 9$ m/s.

  1. Find the velocity function $v(t)$.
  2. Find the total distance traveled during $0 \leq t \leq 4$ seconds.
Thought Process

Since $a(t) = v'(t)$, we can find $v(t)$ by integrating $a(t)$ and using the initial condition.

Once we have $v(t)$, follow the standard procedure: find roots, check signs on each interval, integrate $\vert v(t)\vert $.

Show Answer
  1. $$v(t) = \int a(t)\,dt = \int (6t - 12)\,dt = 3t^2 - 12t + C$$

    Using $v(0) = 9$: $C = 9$

    $$v(t) = 3t^2 - 12t + 9 = 3(t^2 - 4t + 3) = 3(t-1)(t-3)$$

  2. $v(t) = 0$ when $t = 1$ or $t = 3$.

    Sign analysis:

    • $[0, 1]$: $v(0.5) = 3(0.5)(-2.5) < 0$? Let me check: $v(0.5) = 3(0.25) - 6 + 9 = 0.75 + 3 = 3.75 > 0$
    • $[1, 3]$: $v(2) = 3(1)(-1) = -3 < 0$
    • $[3, 4]$: $v(3.5) = 3(2.5)(0.5) = 3.75 > 0$

    Let $F(t) = t^3 - 6t^2 + 9t$ (antiderivative of $v(t)$).

    • $F(0) = 0$
    • $F(1) = 1 - 6 + 9 = 4$
    • $F(3) = 27 - 54 + 27 = 0$
    • $F(4) = 64 - 96 + 36 = 4$

    $$\text{Distance} = \vert F(1) - F(0)\vert + \vert F(3) - F(1)\vert + \vert F(4) - F(3)\vert $$ $$= \vert 4 - 0\vert + \vert 0 - 4\vert + \vert 4 - 0\vert $$ $$= 4 + 4 + 4 = 12 \text{ m}$$

Level 5 When Are They Equal?
  1. Prove that for any velocity function $v(t)$ on $[a, b]$: $$\left\vert \int_a^b v(t)\,dt\right\vert \leq \int_a^b \vert v(t)\vert \,dt$$ In other words, **|displacement| ≤ total distance**.
  2. Under what conditions does equality hold? Prove your answer.
  3. A particle has velocity $v(t) = \sin(t)$ for $0 \leq t \leq 2\pi$. Without computing any integrals, determine: - Is displacement positive, negative, or zero? - Is total distance greater than, less than, or equal to $\vert $displacement$\vert $? Justify your answers.
Thought Process

For part (a), this is related to the triangle inequality. When we integrate $v(t)$, positive and negative parts can cancel. When we integrate $\vert v(t)\vert $, everything accumulates positively.

For part (b), equality holds when there's no cancellation—i.e., when $v(t)$ doesn't change sign.

For part (c), think about the symmetry of $\sin(t)$ over one full period. Also, $\sin(t)$ changes sign at $t = \pi$, so there will be cancellation.

Show Answer
  1. **Proof:** For any $v(t)$, we have $-\vert v(t)\vert \leq v(t) \leq \vert v(t)\vert $.

    Integrating all three parts from $a$ to $b$: $$-\int_a^b \vert v(t)\vert \,dt \leq \int_a^b v(t)\,dt \leq \int_a^b \vert v(t)\vert \,dt$$

    This is exactly the definition of: $$\left\vert \int_a^b v(t)\,dt\right\vert \leq \int_a^b \vert v(t)\vert \,dt$$

    Physical interpretation: The magnitude of net displacement can never exceed total distance traveled. You can't end up farther from your starting point than the total ground you covered.

  2. **Equality holds if and only if $v(t)$ does not change sign on $[a, b]$.**

    Proof of "if": Suppose $v(t) \geq 0$ on $[a, b]$. Then $\vert v(t)\vert = v(t)$, so both integrals are identical.

    Similarly if $v(t) \leq 0$: then $\vert v(t)\vert = -v(t)$ and $\int_a^b v(t)\,dt \leq 0$, so: $$\left\vert \int_a^b v(t)\,dt\right\vert = -\int_a^b v(t)\,dt = \int_a^b (-v(t))\,dt = \int_a^b \vert v(t)\vert \,dt$$

    Proof of "only if" (contrapositive): Suppose $v(t)$ changes sign. Then there exist subintervals where $v(t) > 0$ and others where $v(t) < 0$. When computing $\int v(t)\,dt$, these partially cancel. When computing $\int \vert v(t)\vert \,dt$, they add. Since both positive and negative contributions exist, the total distance strictly exceeds $\vert $displacement$\vert $.

  3. - **Displacement:** $\int_0^{2\pi} \sin(t)\,dt = [-\cos(t)]_0^{2\pi} = (-\cos(2\pi)) - (-\cos(0)) = -1 + 1 = 0$

    The displacement is zero. Over one full period, $\sin(t)$ spends equal "time" positive and negative, and by symmetry, the positive and negative areas cancel exactly.

    • Total distance vs |displacement|: Since displacement = 0 and total distance > 0 (the particle is moving), we have:

    The inequality is strict. This is because $v(t) = \sin(t)$ changes sign at $t = \pi$, so by part (b), equality cannot hold.

    Verification: $\int_0^{2\pi} \vert \sin(t)\vert \,dt = \int_0^{\pi} \sin(t)\,dt + \int_{\pi}^{2\pi} (-\sin(t))\,dt = 2 + 2 = 4$

Mastery Checklist

Mental Model

The Pedometer vs. GPS Analogy:

Your phone's pedometer counts every step—it doesn't care which direction you're going. That's like total distance: $\int \vert v(t)\vert \,dt$.

Your phone's GPS tracks your position—if you walk in circles and return home, it shows zero displacement. That's like $\int v(t)\,dt$.

If you walk in a straight line without turning around, both give the same answer. But the moment you backtrack, the pedometer keeps counting while the GPS shows less progress.

Connections

Looking back:

Looking ahead:

Real-world connections:

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Last updated: 2026-01-22