Displacement vs Total Distance

6 min read

Reference: Stewart §4.4

Not All Movement Is Created Equal

When a particle moves along a line, there are two natural questions:

Where did it end up? (relative to where it started)
How far did it travel? (total ground covered)

These sound similar but have very different answers when the particle changes direction!

Prerequisite Skills

Before You Start

Prerequisite Check: Can you answer these?

From Net Change Theorem:

If $v(t)$ is velocity and $s(t)$ is position, what does $\int_a^b v(t)\, dt$ represent according to the Net Change Theorem?
Why is the result called “net” change rather than just “change”?

From Computing Indefinite Integrals:

Find $\int (3t^2 - 6t)\, dt$.
Evaluate $\int_0^2 (3t^2 - 6t)\, dt$.

Check Your Answers

$\int_a^b v(t)\, dt = s(b) - s(a)$, the displacement (net change in position) from time $a$ to time $b$.
“Net” because if the quantity goes up and down, the integral captures only the overall difference, not the total movement in both directions.
$\int (3t^2 - 6t)\, dt = t^3 - 3t^2 + C$
$\int_0^2 (3t^2 - 6t)\, dt = [t^3 - 3t^2]_0^2 = (8 - 12) - (0) = -4$

If these feel unfamiliar, review the prerequisite pages before continuing.

Quick Reference

Property	Value
Concept	Indefinite Integrals & Net Change
Chapter	Chapter 4, Section 4
Difficulty	Intermediate
Time	~20 minutes

Key Concepts

The Two Formulas

For a particle with velocity $v(t)$ on the interval $[t_1, t_2]$:

Quantity	Formula	Meaning
Displacement	$\displaystyle\int_{t_1}^{t_2} v(t)\, dt$	Net position change (signed)
Total Distance	$\displaystyle\int_{t_1}^{t_2} \vert v(t)\vert \, dt$	Total ground covered (always ≥ 0)

Why the Difference?

Displacement counts movement to the right as positive and movement to the left as negative. They can cancel out.

Total distance counts ALL movement as positive. Nothing cancels.

Visual Example:

Start: ─────●                    (position 0)
Move right 5: ──────────●        (position 5)
Move left 8:  ●──────────        (position -3)

Displacement: -3 (ended 3 units left of start)
Total distance: 5 + 8 = 13 (traveled 13 units total)

Computing Total Distance: The Procedure

Since $\vert v(t)\vert = v(t)$ when $v(t) \geq 0$ and $\vert v(t)\vert = -v(t)$ when $v(t) < 0$:

Step 1: Find where $v(t) = 0$ (the particle changes direction)

Step 2: Split the integral at these points

Step 3: On intervals where $v(t) > 0$: integrate $v(t)$ On intervals where $v(t) < 0$: integrate $-v(t)$

Step 4: Add all the pieces (all positive)

Graphical Interpretation

On a velocity-time graph:

Displacement = (Area above $t$-axis) $-$ (Area below $t$-axis)
Total Distance = (Area above $t$-axis) $+$ (Area below $t$-axis)

    v(t) ↑
         │    ╱╲
         │   ╱  ╲      A₁ (positive area)
         │  ╱    ╲
    ─────┼─╱──────╲─────────→ t
         │        ╲    ╱
         │         ╲  ╱   A₂ (negative area)
         │          ╲╱

Displacement = A₁ - A₂
Total Distance = A₁ + A₂

Practice Problems

Level 1 Conceptual Understanding

A car drives 30 miles east, then 10 miles west.

(a) What is the car’s displacement?

(b) What is the total distance traveled?

Thought Process

(a) Displacement = net change in position = 30 - 10 = 20 miles east

(b) Total distance = all movement = 30 + 10 = 40 miles

Show Answer

(a) Displacement: $30 - 10 = \boxed{20 \text{ miles east}}$

(b) Total Distance: $30 + 10 = \boxed{40 \text{ miles}}$

(c) Averages:

Average velocity $= \frac{20 \text{ mi}}{1 \text{ hr}} = \boxed{20 \text{ mph east}}$
Average speed $= \frac{40 \text{ mi}}{1 \text{ hr}} = \boxed{40 \text{ mph}}$

Note: Speed is always ≥ |velocity| when there’s direction change.

Level 2 Given v(t), No Direction Change

A particle moves with velocity $v(t) = 3t^2 - 6t + 5$ m/s for $0 \leq t \leq 2$.

(a) Verify that $v(t) > 0$ for all $t$ in $[0, 2]$.

(b) Find the displacement.

Thought Process

(a) Check if $3t^2 - 6t + 5 > 0$ on $[0,2]$. Discriminant: $b^2 - 4ac = 36 - 60 = -24 < 0$ Parabola opens up with no real roots → always positive

(b) Since $v(t) > 0$ everywhere, displacement = total distance $\int_0^2 (3t^2 - 6t + 5)\, dt$

Show Answer

(a) Verify $v(t) > 0$:

For $v(t) = 3t^2 - 6t + 5$, check discriminant: $$\Delta = (-6)^2 - 4(3)(5) = 36 - 60 = -24 < 0$$

Since the parabola opens upward (coefficient of $t^2$ is positive) and has no real roots, $v(t) > 0$ for all $t$. ✓

(b) Displacement: $$\int_0^2 (3t^2 - 6t + 5)\, dt = \left[t^3 - 3t^2 + 5t\right]_0^2$$ $$= (8 - 12 + 10) - (0) = \boxed{6 \text{ meters}}$$

(c) Total distance:

Since $v(t) > 0$ on $[0, 2]$, the particle never changes direction.

$$\text{Total distance} = \text{Displacement} = \boxed{6 \text{ meters}}$$

Level 3 Classic Direction Change Problem

A particle moves with velocity $v(t) = t^2 - t - 6$ m/s for $1 \leq t \leq 4$ seconds.

(a) Find the displacement during this time period.

(b) Find the total distance traveled.

Thought Process

(a) Displacement: $\int_1^4 (t^2 - t - 6)\, dt$ (straightforward integration)

(b) Total distance requires finding where $v(t) = 0$: $t^2 - t - 6 = (t-3)(t+2) = 0$ $t = 3$ or $t = -2$

Only $t = 3$ is in $[1, 4]$.

Check signs:

At $t = 2$: $v(2) = 4 - 2 - 6 = -4 < 0$
At $t = 3.5$: $v(3.5) = 12.25 - 3.5 - 6 = 2.75 > 0$

Split: $\int_1^3 \vert v\vert \, dt + \int_3^4 \vert v\vert \, dt$

Show Answer

(a) Displacement: $$\int_1^4 (t^2 - t - 6)\, dt = \left[\frac{t^3}{3} - \frac{t^2}{2} - 6t\right]_1^4$$

At $t = 4$: $\frac{64}{3} - 8 - 24 = \frac{64}{3} - 32 = \frac{64 - 96}{3} = -\frac{32}{3}$

At $t = 1$: $\frac{1}{3} - \frac{1}{2} - 6 = \frac{2 - 3 - 36}{6} = -\frac{37}{6}$

$$= -\frac{32}{3} - \left(-\frac{37}{6}\right) = -\frac{64}{6} + \frac{37}{6} = -\frac{27}{6} = \boxed{-\frac{9}{2} \text{ m}}$$

The particle is 4.5 meters to the LEFT of its starting position.

(b) Total Distance:

Step 1: Find where $v(t) = 0$: $$t^2 - t - 6 = (t-3)(t+2) = 0 \Rightarrow t = 3 \text{ (in } [1,4])$$

Step 2: Determine signs:

On $[1, 3]$: $v(2) = -4 < 0$ (moving left)
On $[3, 4]$: $v(3.5) = 2.75 > 0$ (moving right)

Step 3: Compute: $$\int_1^3 \vert v(t)\vert \, dt = \int_1^3 -(t^2 - t - 6)\, dt = \int_1^3 (-t^2 + t + 6)\, dt$$

$$= \left[-\frac{t^3}{3} + \frac{t^2}{2} + 6t\right]_1^3 = \left(-9 + \frac{9}{2} + 18\right) - \left(-\frac{1}{3} + \frac{1}{2} + 6\right)$$

$$= \frac{27}{2} - \frac{37}{6} = \frac{81 - 37}{6} = \frac{44}{6} = \frac{22}{3}$$

$$\int_3^4 v(t)\, dt = \left[\frac{t^3}{3} - \frac{t^2}{2} - 6t\right]_3^4 = -\frac{32}{3} - \left(-\frac{27}{2}\right) = -\frac{32}{3} + \frac{27}{2} = \frac{17}{6}$$

Total distance: $$\frac{22}{3} + \frac{17}{6} = \frac{44 + 17}{6} = \frac{61}{6} \approx \boxed{10.17 \text{ m}}$$

Level 4 Velocity with Multiple Direction Changes

A particle has velocity $v(t) = 3t - 5$ m/s for $0 \leq t \leq 3$ seconds.

(a) Find the displacement.

(b) Find the total distance traveled.

Thought Process

(a) Displacement: $\int_0^3 (3t - 5)\, dt$

(b) Find where $v(t) = 0$: $3t - 5 = 0 \Rightarrow t = \frac{5}{3}$

On $[0, \frac{5}{3}]$: $v(0) = -5 < 0$ (moving left)
On $[\frac{5}{3}, 3]$: $v(2) = 1 > 0$ (moving right)

Show Answer

(a) Displacement: $$\int_0^3 (3t - 5)\, dt = \left[\frac{3t^2}{2} - 5t\right]_0^3 = \frac{27}{2} - 15 = \frac{27 - 30}{2} = \boxed{-\frac{3}{2} \text{ m}}$$

(b) Total Distance:

Find direction change: $3t - 5 = 0 \Rightarrow t = \frac{5}{3}$

On $[0, \frac{5}{3}]$: $v < 0$ (particle moves left)
On $[\frac{5}{3}, 3]$: $v > 0$ (particle moves right)

$$\int_0^{5/3} \vert 3t - 5\vert \, dt = \int_0^{5/3} (5 - 3t)\, dt = \left[5t - \frac{3t^2}{2}\right]_0^{5/3}$$ $$= \frac{25}{3} - \frac{25}{6} = \frac{50 - 25}{6} = \frac{25}{6}$$

$$\int_{5/3}^3 (3t - 5)\, dt = \left[\frac{3t^2}{2} - 5t\right]_{5/3}^3 = -\frac{3}{2} - \left(-\frac{25}{6}\right) = -\frac{9}{6} + \frac{25}{6} = \frac{16}{6} = \frac{8}{3}$$

$$\text{Total distance} = \frac{25}{6} + \frac{8}{3} = \frac{25 + 16}{6} = \boxed{\frac{41}{6} \approx 6.83 \text{ m}}$$

(c) Position graph:

$s(t) = \frac{3t^2}{2} - 5t + 2$ is a parabola opening upward.

Key points:

$s(0) = 2$ (starting position)
Minimum at $t = \frac{5}{3}$: $s(\frac{5}{3}) = \frac{25}{6} - \frac{25}{3} + 2 = -\frac{25}{6} + 2 = -\frac{13}{6} \approx -2.17$
$s(3) = \frac{27}{2} - 15 + 2 = \frac{1}{2}$ (ending position)

The particle starts at $s = 2$, moves left to $s \approx -2.17$, then moves right to $s = 0.5$.

Level 5 Acceleration to Distance

A particle starts from rest and has acceleration $a(t) = t + 4$ m/s² for $0 \leq t \leq 10$ seconds.

(a) Find the velocity function $v(t)$.

(b) Show that $v(t) > 0$ for all $t > 0$, so the particle always moves in the positive direction.

Thought Process

(a) Velocity is antiderivative of acceleration with initial condition: $v(t) = \int a(t)\, dt$ with $v(0) = 0$ (starts from rest)

$v(t) = \frac{t^2}{2} + 4t + C$ $v(0) = 0 \Rightarrow C = 0$

(b) $v(t) = \frac{t^2}{2} + 4t = t(\frac{t}{2} + 4)$ For $t > 0$: both factors positive → $v(t) > 0$

Show Answer

(a) Find $v(t)$:

$$v(t) = \int a(t)\, dt = \int (t + 4)\, dt = \frac{t^2}{2} + 4t + C$$

Using $v(0) = 0$ (starts from rest): $$0 = 0 + 0 + C \Rightarrow C = 0$$

$$\boxed{v(t) = \frac{t^2}{2} + 4t}$$

(b) Show $v(t) > 0$ for $t > 0$:

$$v(t) = \frac{t^2}{2} + 4t = t\left(\frac{t}{2} + 4\right)$$

For $t > 0$:

$t > 0$ ✓
$\frac{t}{2} + 4 > 4 > 0$ ✓

Product of two positives is positive, so $v(t) > 0$ for all $t > 0$. ✓

This means the particle always moves in the positive direction (no direction changes).

(c) Total distance:

Since $v(t) \geq 0$ on $[0, 10]$: $$\text{Total distance} = \int_0^{10} v(t)\, dt = \int_0^{10} \left(\frac{t^2}{2} + 4t\right)\, dt$$

$$= \left[\frac{t^3}{6} + 2t^2\right]_0^{10} = \frac{1000}{6} + 200 = \frac{1000 + 1200}{6} = \frac{2200}{6} = \frac{1100}{3}$$

$$= \boxed{\frac{1100}{3} \approx 366.67 \text{ meters}}$$

Common Misconceptions

Common misconception

$\int_a^b v(t)\,dt$ gives the total distance traveled by a particle.

This is the height-vs-slope error. The integral of velocity gives displacement, the net signed change in position, not the total ground covered. When $v(t) < 0$ on part of $[a, b]$, the backward motion partially cancels the forward motion in the integral. Total distance requires integrating speed: $\int_a^b |v(t)|\,dt$. For $v(t) = t^2 - 4$ on $[0, 3]$, the displacement is $-3$ (net position change), while the total distance is $\tfrac{23}{3} \approx 7.67$ (all movement counted positively).

Mastery Checklist

I understand that displacement can be negative, but total distance is always non-negative
I can compute displacement using $\int v(t)\, dt$
I can find where $v(t) = 0$ to locate direction changes
I can split an integral at direction changes to compute total distance
I can interpret areas on a velocity-time graph as displacement vs distance

Mental Model

The Taxi Meter vs GPS

Think of a taxi meter and a GPS:

GPS (Displacement): Shows straight-line distance from start to finish. If you end up back where you started, it reads 0.
Taxi Meter (Total Distance): Counts every block you travel. Driving in circles? It keeps ticking up.

The integral $\int v(t)\, dt$ is like GPS. The integral $\int \vert v(t)\vert \, dt$ is like the taxi meter.

Common Errors to Avoid

Error	Correction
Using $\int v(t)\, dt$ for total distance	Must use $\int \vert v(t)\vert \, dt$ when direction changes
Forgetting to check for sign changes	Always solve $v(t) = 0$ first
Getting wrong sign when $v < 0$	When $v(t) < 0$, $\vert v(t)\vert = -v(t)$
Adding areas with wrong signs	For total distance, ALL areas are positive

Connections

Looking back:

Net Change Theorem: displacement IS net change in position
Velocity and Position: the relationship $v = s'$

Looking ahead:

Arc Length: total distance along a curve
Work: similar signed/unsigned distinction

Real-world connections:

Fitness trackers report “distance walked” (total), not “net displacement from home”
Odometers measure total distance; GPS measures displacement
In physics, work can be positive or negative; total energy expended is always positive

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Net Change Theorem	Skills Index	u-Substitution

Last updated: 2026-01-22