Position and velocity in one dimension

Describing motion along a line with a position function, and defining velocity as the rate of change of position.

Follows MIT 8.01SC lesson 1MIT 8.01SC Lesson 1: 1D Kinematics – Position and Velocity (1.1–1.7)On MIT OpenCourseWare: the lecture and its readings

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Vectors8.01 · Lesson 0

On this page · 9 sections

Choosing a coordinate system
Position
Time interval and displacement
Average velocity
Instantaneous velocity
Derivatives you will use
Average velocity is not always the mean of two velocities
Explore derivatives
Summary

Kinematics is the description of motion, without yet asking what causes it. We start with the simplest case: an object moving back and forth along a straight line, treated as a single point.

Choosing a coordinate system#

As in Vectors, the first step is a coordinate system. In one dimension that means one axis: an origin, a positive direction, and a unit vector $\hat i$ pointing toward increasing $x$ . Together with a clock, this is a reference frame.

Position#

The object's location at time $t$ is given by its position coordinate $x(t)$ , a function of time that can be positive, zero or negative. Because the location relative to the origin has both a size and a direction, it is a vector, the position vector:

Position vector

\vec r(t) = x(t)\,\hat i

In words

vector r (t) equals x(t) i hat

Source: 8.01SC Course Notes, Chapter 4: One Dimensional Kinematics, Sections 4.1–4.3, Eq. (4.2.1)

The position at $t = 0$ gets its own symbol, $x_0 \equiv x(0)$ . Position is measured in metres.

Time interval and displacement#

A time interval $[t_1, t_2]$ has length $\Delta t = t_2 - t_1$ . The displacement during that interval is the change in position:

Displacement

\Delta \vec r = \vec r(t_2) - \vec r(t_1) = \big(x(t_2) - x(t_1)\big)\,\hat i \equiv \Delta x\,\hat i

In words

capital delta vector r equals vector r (t sub 2 ) minus vector r (t sub 1 ) equals (x(t sub 2 ) minus x(t sub 1 ) ) i hat is defined as capital delta x i hat

Source: 8.01SC Course Notes, Chapter 4: One Dimensional Kinematics, Sections 4.1–4.3, Eq. (4.2.3)

Displacement only depends on where the motion starts and ends, not on the path in between. A car that drives 3 km east and then 3 km back west has travelled 6 km, but its displacement is zero.

Average velocity#

The average velocity over an interval is the displacement divided by the time it took:

Average velocity

v_\text{ave} = \frac{\Delta x}{\Delta t}, \qquad \vec v_\text{ave} = v_\text{ave}\,\hat i

In words

v sub ave equals capital delta x over capital delta t , vector v sub ave equals v sub ave i hat

Source: 8.01SC Course Notes, Chapter 4: One Dimensional Kinematics, Sections 4.1–4.3, Eqs. (4.3.1)–(4.3.3)

This is not the same as distance divided by time (that is the average speed). On a graph of $x$ against $t$ , $v_\text{ave}$ is the slope of the straight line, the secant, joining the two points $(t_1, x(t_1))$ and $(t_2, x(t_2))$ .

Instantaneous velocity#

A speedometer does not show an average over an interval; it shows the velocity now. To get that, shrink the interval. As $\Delta t$ gets smaller, the secant line through $(t, x(t))$ and $(t + \Delta t, x(t + \Delta t))$ turns into the tangent line at $t$ . Its slope is the instantaneous velocity:

Instantaneous velocity

v(t) = \lim_{\Delta t \to 0} \frac{x(t + \Delta t) - x(t)}{\Delta t} = \frac{dx}{dt}, \qquad \vec v(t) = v(t)\,\hat i

In words

v(t) equals the limit as capital delta t goes to 0 of x(t plus capital delta t) minus x(t) over capital delta t equals dx over dt , vector v (t) equals v(t) i hat

Source: 8.01SC Course Notes, Chapter 4: One Dimensional Kinematics, Sections 4.1–4.3, Eqs. (4.3.5)–(4.3.6)

In the language of calculus, velocity is the derivative of position with respect to time. Its sign tells you the direction: positive means moving toward $+x$ , negative toward $-x$ , and zero means momentarily at rest.

Move the sliders: $t_1$ and $t_2$ set the secant (dashed line). Bring $t_2$ toward $t_1$ and the secant turns into the tangent (dotted line).

Loading graph…

x_0

v_0

b

-1.20

t_1

t_2

Position x(t) = x₀ + v₀t + ½bt². The graph is computed from the formula.

Derivatives you will use#

You rarely need to take the limit by hand. A constant has derivative zero, the derivative of a sum is the sum of the derivatives, and a constant factor just comes along. Add a short table of rules and you can handle most of mechanics:

Derivatives you will use

\frac{d}{dt}\,t^n = n\,t^{n-1}, \qquad \frac{d}{dt}\,e^{bt} = b\,e^{bt}, \qquad \frac{d}{dt}\ln t = \frac{1}{t}, \qquad \frac{d}{dt}\sin(bt) = b\cos(bt), \qquad \frac{d}{dt}\cos(bt) = -b\sin(bt)

In words

d over dt t to the power n equals n t to the power n-1 , d over dt e to the power bt equals b e to the power bt , d over dt log t equals 1 over t , d over dt sine (bt) equals b cos (bt), d over dt cos (bt) equals minus b sine (bt)

Source: MIT 8.01SC Lesson 1: 1D Kinematics – Position and Velocity (1.1–1.7), segment 1.6 Derivatives

Still, doing it once from the definition shows where the rules come from.

Worked example

Velocity from the definition

A cart's position is $x(t) = 2 + 3t - t^2$ , with $x$ in metres and $t$ in seconds. Find $v(t)$ from the limit definition, then find when and where the cart turns around.

Write the position a short time later

$x(t + \Delta t) = 2 + 3(t + \Delta t) - (t + \Delta t)^2 = 2 + 3t + 3\Delta t - t^2 - 2t\,\Delta t - \Delta t^2$ .

Form the displacement

$\Delta x = x(t + \Delta t) - x(t) = 3\Delta t - 2t\,\Delta t - \Delta t^2$ .

Divide by Δt and take the limit

$\dfrac{\Delta x}{\Delta t} = 3 - 2t - \Delta t$ . As $\Delta t \to 0$ the last term vanishes, so $v(t) = 3 - 2t$ m/s. The power rule gives the same thing: $\frac{d}{dt}(2 + 3t - t^2) = 3 - 2t$ .

Find the turning point

The cart turns around when it is momentarily at rest: $3 - 2t = 0$ gives $t = 1.5$ s. There, $x(1.5) = 2 + 4.5 - 2.25 = 4.25$ m. Before $t = 1.5$ s, $v > 0$ (moving toward $+x$ ); afterwards $v < 0$ .

Check with an average

Between $t = 1$ s and $t = 2$ s: $x(1) = 4$ m and $x(2) = 4$ m, so $v_\text{ave} = 0$ . That fits: the interval is centred on the turning point, and the cart goes out and comes back.

Average velocity is not always the mean of two velocities#

It is tempting to compute an average velocity as $\big(v(t_1) + v(t_2)\big)/2$ . MIT's notes are careful here. That shortcut gives the right answer only when the velocity changes linearly with time. In general, the average velocity is the displacement divided by the time interval, and nothing else.

Explore derivatives#

The simulation lets you draw any function and see its derivative (and integral) drawn live. Sketch a position curve and look at the velocity curve it produces.

Calculus Grapher. Simulation by PhET Interactive Simulations, University of Colorado Boulder, licensed under CC BY-NC 4.0 (https://phet.colorado.edu). Open full screen

Why cosmology needs this

In cosmology the distance between two distant galaxies is written $\ell(t) = a(t)\,\ell_c$ . Here $\ell_c$ is a fixed "comoving" distance and $a(t)$ is the scale factor, which says how much the universe has expanded. The galaxies separate at the rate

Hubble's law from a derivative

v = \frac{d\ell}{dt} = \frac{da}{dt}\,\ell_c = \frac{\dot a}{a}\,\ell \equiv H(t)\,\ell

In words

v equals d ell over dt equals da over dt ell sub c equals a dot over a ell is defined as H(t) ell

Source: MIT 8.286 The Early Universe (Fall 2013), Lecture 4: The Kinematics of the Homogeneous Expanding Universe

This is just the derivative from this page, with $\dot a$ standing for $da/dt$ . The Hubble parameter $H(t) = \dot a / a$ measures how fast the universe is expanding at time $t$ ; its value today is called $H_0$ .

Summary#

Position along a line: $\vec r(t) = x(t)\,\hat i$ . Displacement $\Delta x$ is the change in position.
Average velocity $v_\text{ave} = \Delta x / \Delta t$ is the slope of a secant on the $x(t)$ graph.
Instantaneous velocity $v = dx/dt$ is the slope of the tangent, the limit of the average as $\Delta t \to 0$ .
The sign of $v$ gives the direction of motion. Where $v = 0$ the object is momentarily at rest; it turns around there if $v$ changes sign.

Symbols on this page

Show the 14 symbols used here

$\Delta$ capital delta: A change or difference in a quantity, e.g. Δx = x₂ − x₁. Say it: “capital delta”.
$\ell$ script l: A length or distance; ℓ_p is a physical distance and ℓ_c a comoving one. Say it: “ell”.
$\equiv$ identical to: Is defined as (or is identically equal to). Say it: “is defined as”.
$\to$ arrow: Goes to (tends to), or becomes. Say it: “goes to”.
$\lim$ limit: Limit: the value approached as the variable written below approaches its target. Say it: “the limit”.
$\sin$ sine: Sine of an angle. Say it: “sine”.
$\cos$ cosine: Cosine of an angle. Say it: “cos”.
$\ln$ natural logarithm: Natural logarithm (base e ≈ 2.718). Say it: “log”.
$\vec v$ vector arrow: A vector: a quantity with a size and a direction. Say it: “vector”.
$\hat\imath$ hat: A unit vector: a vector of length 1 that only gives a direction (î, r̂). Say it: “hat”.
$\dot a$ dot: Rate of change with time: ȧ = da/dt. Say it: “dot”.
$H$ H: The Hubble parameter H = ȧ/a: the expansion rate of the universe. Say it: “H”.
$H_{0}$ Hubble constant: The Hubble constant: today's value of H, about 67–73 km/s/Mpc. Say it: “H nought”.
$\ell_{c}$ comoving distance: A comoving distance: fixed as the universe expands, measured in coordinate units (Guth's notches). Say it: “ell c”.

Tap or click a symbol in a formula to see the same explanation. All symbols: symbols and notation.

Where this leads

In cosmology

42 more cosmology pages that build on this

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Check your understanding

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On a graph of $x(t)$ , the instantaneous velocity at time $t$ is…
If $x(t) = 4 + 2t^2$ (in metres, with $t$ in seconds), what is $v(3\ \text{s})$ ?
The velocity component $v(t)$ is negative. This means the object…

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