Acceleration in one dimension

Acceleration as the rate of change of velocity, motion with constant acceleration, and getting velocity and position back by integration.

Follows MIT 8.01SC lesson 2MIT 8.01SC Lesson 2: 1D Kinematics – Acceleration (2.1–2.5)On MIT OpenCourseWare: the lecture and its readings

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Position and velocity in one dimension8.01 · Lesson 1

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Vectors (8.01)
Position and velocity in one dimension (8.01)

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Average and instantaneous acceleration
Constant acceleration
When the acceleration is not constant: integration
Summary

Velocity tells us how fast position changes. Acceleration tells us how fast velocity changes. It is the quantity that Newton's second law will connect to forces, and the last ingredient we need to describe motion along a line.

Average and instantaneous acceleration#

We follow exactly the same steps as for velocity. Over a time interval $\Delta t$ the velocity changes by $\Delta v = v(t + \Delta t) - v(t)$ . The average acceleration is

Average acceleration

a_\text{ave} = \frac{\Delta v}{\Delta t}, \qquad \vec a_\text{ave} = a_\text{ave}\,\hat i

In words

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

Source: 8.01SC Course Notes, Chapter 4: One Dimensional Kinematics, Sections 4.4–4.6, Eq. (4.3.24)

measured in metres per second squared, $\text{m/s}^2$ . On a graph of $v$ against $t$ , it is the slope of a secant line. Shrinking the interval to zero gives the slope of the tangent, the instantaneous acceleration:

Instantaneous acceleration

a(t) = \lim_{\Delta t \to 0} \frac{v(t + \Delta t) - v(t)}{\Delta t} = \frac{dv}{dt} = \frac{d^2 x}{dt^2}

In words

a(t) equals the limit as capital delta t goes to 0 of v(t plus capital delta t) minus v(t) over capital delta t equals dv over dt equals d squared x over dt squared

Source: 8.01SC Course Notes, Chapter 4: One Dimensional Kinematics, Sections 4.4–4.6, Eqs. (4.3.25)–(4.3.27)

Since velocity is the derivative of position, acceleration is the second derivative of position. So if you know $x(t)$ , you get $a(t)$ by differentiating twice. For example, $x(t) = 3 + 2t - 4t^3$ gives $v(t) = 2 - 12t^2$ and $a(t) = -24t$ , all in SI units.

Constant acceleration#

The most important special case is constant acceleration $a$ . A dropped stone near Earth's surface is close to it, with $a = -g \approx -9.8\ \text{m/s}^2$ if $+x$ points up. When $a$ is constant, the velocity changes by the same amount every second, so $v(t)$ is a straight line:

Velocity, constant acceleration

v(t) = v_0 + a\,t

In words

v(t) equals v sub 0 plus a t

Source: 8.01SC Course Notes, Chapter 4: One Dimensional Kinematics, Sections 4.4–4.6, Sec. 4.5, Eq. (4.4.2)

Here is a useful way to see it. The change in velocity, $a\,t$ , is the area under the $a(t)$ graph, a rectangle of height $a$ and width $t$ .

The same idea gives the position. The displacement is the area under the $v(t)$ graph. For constant acceleration that region is a rectangle ( $v_0 t$ ) plus a triangle ( $\tfrac12 \cdot t \cdot a t$ ), so

Position, constant acceleration

x(t) = x_0 + v_0\,t + \tfrac12\,a\,t^2

In words

x(t) equals x sub 0 plus v sub 0 t plus one half a t squared

Source: 8.01SC Course Notes, Chapter 4: One Dimensional Kinematics, Sections 4.4–4.6, Sec. 4.5.2, Eq. (4.4.9)

Differentiating this gives back $v = v_0 + at$ , and differentiating again gives $a$ . The two descriptions agree.

Loading graph…

x_0

1.50

v_0

a

-9.80

A ball thrown upward at 12 m/s from 1.5 m, with a = −9.8 m/s². Where v(t) crosses zero, x(t) has its peak. Each curve is the slope of the one above it.

Worked example

A ball thrown upward

A ball is thrown straight up at $12\ \text{m/s}$ from a hand $1.5$ m above the ground. Ignore air resistance and take $g = 9.8\ \text{m/s}^2$ . How high does it rise, and how fast is it moving when it hits the ground?

Choose coordinates

Origin at the ground, $+x$ upward, $t = 0$ at the throw. Then $x_0 = 1.5$ m, $v_0 = +12\ \text{m/s}$ and $a = -9.8\ \text{m/s}^2$ .

Time to the top

At the top, $v = 0$ : $0 = 12 - 9.8\,t_\text{top}$ , so $t_\text{top} = 1.22$ s.

Maximum height

$x(t_\text{top}) = 1.5 + 12(1.22) - \tfrac12 (9.8)(1.22)^2 = 1.5 + 7.35 = 8.85$ m above the ground.

When it lands

Set $x(t) = 0$ : $4.9\,t^2 - 12\,t - 1.5 = 0$ . The quadratic formula gives $t = \dfrac{12 + \sqrt{144 + 29.4}}{9.8} = 2.57$ s. The other root is negative, before the throw, so we discard it.

Speed at impact

$v = 12 - 9.8(2.57) = -13.2\ \text{m/s}$ . The minus sign means it is moving downward, at $13.2\ \text{m/s}$ , a bit faster than it was thrown, because it falls 1.5 m further than it rose from the hand.

When the acceleration is not constant: integration#

If $a$ changes with time, the formulas above no longer apply, but the area idea still works. Cut the time interval into thin slices. In each slice the velocity changes by about $a\,\Delta t$ . Adding up the slices and letting them shrink to zero turns the sum into an integral:

Velocity and position by integration

v(t) = v_0 + \int_0^t a(t')\,dt', \qquad x(t) = x_0 + \int_0^t v(t')\,dt'

In words

v(t) equals v sub 0 plus the integral from 0 to t of a(t prime ) dt prime , x(t) equals x sub 0 plus the integral from 0 to t of v(t prime ) dt prime

Source: 8.01SC Course Notes, Chapter 4: One Dimensional Kinematics, Sections 4.4–4.6, Sec. 4.6, Eqs. (4.5.19)–(4.5.21)

Integration undoes differentiation. The one integral you need most often is the reverse of the power rule:

Power rule for integrals

\int_0^t t'^{\,n}\,dt' = \frac{t^{\,n+1}}{n+1} \qquad (n \neq -1)

In words

the integral from 0 to t of t prime to the power n dt prime equals t to the power n plus 1 over n plus 1 (n is not equal to minus 1)

Source: MIT 8.01SC Lesson 2: 1D Kinematics – Acceleration (2.1–2.5), segment 2.5 List of Useful Integrals

Worked example

Non-constant acceleration

An object moves with acceleration $a(t) = 6t - 2$ (in $\text{m/s}^2$ , with $t$ in seconds). At $t = 0$ it is at $x_0 = 0$ with $v_0 = 1\ \text{m/s}$ . Find its velocity and position at $t = 2$ s.

Integrate the acceleration

$v(t) = 1 + \int_0^t (6t' - 2)\,dt' = 1 + 3t^2 - 2t$ (m/s).

Integrate the velocity

$x(t) = 0 + \int_0^t (1 - 2t' + 3t'^2)\,dt' = t - t^2 + t^3$ (m).

Evaluate at t = 2 s

$v(2) = 1 + 12 - 4 = 9\ \text{m/s}$ and $x(2) = 2 - 4 + 8 = 6$ m.

Check by differentiating

$\dfrac{dx}{dt} = 1 - 2t + 3t^2 = v(t)$ ✓, and $\dfrac{dv}{dt} = 6t - 2 = a(t)$ ✓.

Loading graph…

The non-constant acceleration example, plotted from the formulas. Notice v(t) has its minimum where a(t) crosses zero.

Why cosmology needs this

Cosmologists describe the whole universe with one function, the scale factor, also written $a(t)$ (a different $a$ from the acceleration on this page). Its first derivative gives the expansion rate, and its second derivative says whether the expansion is speeding up or slowing down. Gravity from all the matter in the universe, ordinary and dark, should slow it down. In 1998 two teams measured distant Type Ia supernovae and found the opposite:

Accelerating expansion

\frac{d^2 a}{dt^2} > 0 \quad \text{(today)}

In words

d squared a over dt squared greater than 0 (today)

Source: MIT 8.286 The Early Universe (Fall 2013), Lecture 20: Supernovae Ia and Vacuum Energy Density

The expansion is accelerating. That discovery was the first direct evidence for dark energy, and it rests on exactly the idea from this page: acceleration is the rate of change of a rate of change.

Summary#

Acceleration is the rate of change of velocity: $a = dv/dt = d^2x/dt^2$ , the slope of the $v(t)$ graph.
Speed increases when $v$ and $a$ have the same sign and decreases when they differ.
With constant $a$ : $v = v_0 + at$ and $x = x_0 + v_0 t + \tfrac12 a t^2$ .
In general, velocity is the area under $a(t)$ and displacement is the area under $v(t)$ . Integration undoes differentiation.

Symbols on this page

Show the 9 symbols used here

$\Delta$ capital delta: A change or difference in a quantity, e.g. Δx = x₂ − x₁. Say it: “capital delta”.
$\approx$ approximately equal: Is approximately equal to. Say it: “is approximately”.
$\neq$ not equal: Is not equal to. Say it: “is not equal to”.
$\to$ arrow: Goes to (tends to), or becomes. Say it: “goes to”.
$\cdot$ dot: Times (multiplication). Between two vectors: the dot product. Say it: “times”.
$\int$ integral: Integral: the sum of infinitely many small pieces, such as the area under a curve. Say it: “the integral of”.
$\lim$ limit: Limit: the value approached as the variable written below approaches its target. Say it: “the limit”.
$\vec v$ 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”.

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

Where this leads

In cosmology

41 more cosmology pages that build on this

Pages that build on this

Motion in two dimensions and projectiles8.01 · Lesson 3

Check your understanding

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