Acceleration in Mechanics

Acceleration describes how velocity changes with time. Because velocity is a vector, acceleration measures changes in either the magnitude of velocity, its direction, or both.

For a finite time interval, average acceleration is

aavg = Δv--. Δt

(1)

At an instant, acceleration is the time derivative of velocity,

dv- a = dt .

(2)

Since velocity is the derivative of position,

d2r a = --2. dt

(3)

Acceleration is therefore a second time derivative of position. These definitions are kinematic: they describe motion without yet asking what forces produce it. Dynamics enters later through Newton’s laws.

1 Average acceleration

Suppose a particle has velocity $v1$ at time $t1$ and velocity $v2$ at time $t2$ . Define

Δv = v2 − v1

(4)

and

Δt = t2 − t1.

(5)

The average acceleration is

v2 − v1 aavg = --------. t2 − t1

(6)

The direction of average acceleration is the direction of the velocity change $Δv$ , not necessarily the direction of motion.

Figure 1. Acceleration is associated with the change in the velocity vector. The vector $Δv = v2 − v1$ can be nonzero because the speed changes, the direction changes, or both.

In one dimension,

v − v ax,avg = -x,2----x,1. t2 − t1

(7)

The SI unit is

[a] = m ∕s2.

(8)

The dimensions of acceleration are

[a] = LT −2.

(9)

2 Instantaneous acceleration as a limit

Average acceleration describes a finite interval. To obtain the acceleration at one instant, shrink the interval around that instant.

Let the velocity at time $t$ be $v(t)$ . A short time $Δt$ later, the velocity is $v (t + Δt )$ . The velocity change is

Δv = v (t + Δt ) − v(t).

(10)

The instantaneous acceleration is

a(t) = lim v(t-+-Δt-) −-v-(t). Δt→0 Δt

(11)

Thus

a (t) = dv-. dt

(12)

Using

dr- v = dt,

(13)

we also have

d2r- a = dt2.

(14)

In one-dimensional motion,

dvx d2x ax = ----= --2 . dt dt

(15)

3 Acceleration components in Cartesian coordinates

For motion described by

r(t) = x(t)ex + y(t)ey + z (t)ez,

(16)

with a fixed Cartesian basis,

v = ˙xex + ˙yey + ˙zez.

(17)

Differentiating once more gives

a = ¨xex + ¨yey + ¨zez.

(18)

Therefore

ax = ¨x, ay = y¨, az = ¨z.

(19)

The magnitude is

∘ ------------ a = |a| = a2 + a2 + a2. x y z

(20)

4 Acceleration from a velocity-time graph

For one-dimensional motion, acceleration is the slope of a velocity-time graph.

Over a finite interval,

Δvx- ax,avg = Δt ,

(21)

which is the secant slope of the $v x$ -versus- $t$ graph.

At an instant,

a = dvx-, x dt

(22)

which is the tangent slope.

Figure 2. On a velocity-time graph, average acceleration is a secant slope and instantaneous acceleration is the tangent slope. The area under an acceleration-time graph will later give the change in velocity.

The inverse relation follows from integration:

∫ t2 Δvx = ax(t)dt. t1

(23)

This relationship becomes central in M01-04 on motion graphs and M01-05 on constant-acceleration motion.

5 The sign of acceleration in one dimension

A common misconception is that positive acceleration means “speeding up” and negative acceleration means “slowing down.” The sign of acceleration instead tells how the velocity component changes.

In one dimension, compare the signs of $vx$ and $ax$ :

if $v x$ and $a x$ have the same sign, the speed increases;
if $vx$ and $ax$ have opposite signs, the speed decreases;
if $vx = 0$ , the sign of $ax$ determines which way the velocity begins to change.

Figure 3. In one-dimensional motion, speeding up or slowing down depends on the relative signs of velocity and acceleration, not on the sign of acceleration alone.

For example, an object moving left has $vx < 0$ . If it also has $ax < 0$ , its velocity becomes more negative and its speed increases.

6 Zero velocity does not imply zero acceleration

A particle can be instantaneously at rest while still having nonzero acceleration.

A vertically thrown ball provides the standard example. At its highest point,

v = 0, y

(24)

but, neglecting air resistance, its acceleration is still approximately

ay = − g.

(25)

The zero value of velocity is only an instantaneous statement about motion. Acceleration describes how that velocity is changing.

Conversely, zero acceleration does not imply zero velocity. A particle moving with constant nonzero velocity has

a = 0.

(26)

7 Acceleration can change direction without changing speed

Because velocity is a vector, acceleration need not change speed. It can change only the direction of motion.

Uniform circular motion is the simplest example. A particle moves with constant speed $v$ , but its tangent velocity vector continually changes direction. The acceleration points inward toward the center of the circle.

For a circle of radius $R$ , the magnitude is

v2- ac = R .

(27)

The detailed derivation belongs to M01-08, but the kinematic lesson is already important:

constant speed ⁄⇒ zero acceleration.

(28)

Figure 4. In uniform circular motion the speed can remain constant while the tangent velocity changes direction. The resulting acceleration is directed toward the center.

More generally, acceleration can be decomposed into a part that changes speed and a part that changes direction. That decomposition is developed later in curvilinear kinematics.

8 Acceleration and reference frames

Acceleration is not frame-independent in every possible reference frame. The precise Newtonian statement is narrower.

For two Cartesian frames whose origins move with constant relative velocity $V$ ,

v′ = v − V.

(29)

Differentiating gives

a′ = a

(30)

because $dV ∕dt = 0$ .

Thus acceleration is invariant under a Galilean transformation between inertial frames related by constant translational velocity.

If the second frame has translational acceleration $A$ , then instead

a′ = a − A.

(31)

Rotating frames introduce additional terms. Those effects are treated later in the mechanics sequence.

9 Acceleration and Newtonian dynamics

Kinematics defines acceleration without reference to force. Dynamics explains how forces determine acceleration.

For a constant-mass particle in an inertial frame, Newton’s second law is

Fnet = ma.

(32)

This does not define acceleration; it relates the kinematically defined acceleration to the net force and mass.

The distinction is useful:

dv- a = dt

(33)

is a kinematic statement, while

Fnet = ma

(34)

is a dynamical law.

10 Constant acceleration as an important special case

If acceleration is constant,

a = constant,

(35)

then integrating

dv- dt = a

(36)

gives

v (t) = v + a(t − t ). 0 0

(37)

The full constant-acceleration equations, including position as a function of time and velocity as a function of displacement, are derived systematically in M01-05.

11 Gravitational acceleration near Earth’s surface

Near Earth’s surface, freely falling objects experience an approximately downward gravitational acceleration when air resistance is neglected. A commonly used reference value is the standard gravity

gn = 9.80665 m ∕s2.

(38)

Actual local gravitational acceleration varies with location and altitude. In introductory mechanics, it is often approximated as

g ≃ 9.8 m ∕s2.

(39)

The sign assigned to gravitational acceleration depends on the chosen coordinate direction. If upward is positive,

ay = − g.

(40)

12 Worked example 1: acceleration from a position function

A particle moves along the $x$ axis according to

x(t) = 2 + 5t − 3t2 + t3,

(41)

with $x$ in metres and $t$ in seconds. Find $vx(t)$ , $ax(t)$ , and the acceleration at $t = 2 s$ .

Differentiate the position once:

2 vx(t) = 5 − 6t + 3t .

(42)

Differentiate again:

ax(t) = − 6 + 6t.

(43)

At $t = 2 s$ ,

a (2) = − 6 + 12 = 6m ∕s2. x

(44)

Thus

ax(2) = 6 m ∕s2.

(45)

13 Worked example 2: vector average acceleration

A particle’s velocity changes from

v1 = 4ex − 2ey m ∕s

(46)

v2 = − 2ex + 4ey m ∕s

(47)

over $3.0s$ . Find the average acceleration.

The velocity change is

Δv = v2 − v1 = − 6ex + 6ey m ∕s.

(48)

Therefore

a = Δv--= − 2e + 2e m ∕s2. avg Δt x y

(49)

Its magnitude is

∘ ----2------2- √ -- 2 |aavg| = (− 2 ) + (2) = 2 2 ≈ 2.83m ∕s .

(50)

14 Worked example 3: constant speed with nonzero acceleration

A particle moves around a circle of radius

R = 3.0 m

(51)

at constant speed

v = 6.0m ∕s.

(52)

Although the speed is constant, the velocity changes direction. The centripetal acceleration magnitude is

v2- 36- 2 ac = R = 3 = 12m ∕s .

(53)

The acceleration points radially inward.

This example demonstrates that acceleration measures change of velocity, not merely change of speed.

15 Common mistakes

Treating negative acceleration as automatically meaning that an object slows down.
Assuming $v = 0$ at one instant implies $a = 0$ .
Assuming constant speed implies zero acceleration.
Confusing acceleration with velocity or displacement because all can carry a sign in one dimension.
Reading acceleration as the height of a velocity-time graph rather than its slope.
Forgetting that acceleration is Galilean-invariant only between inertial frames with constant relative translational velocity.
Using $g$ without stating a sign convention for the vertical axis.

16 Practice problems

M01-03-P01

A CAR’s velocity changes from $8m ∕s$ east to $20m ∕s$ east in $4.0 s$ . Find its average acceleration.

M01-03-P02

A particle moves along the $x$ axis with

vx(t) = 12 − 4t.

(54)

Find $ax(t)$ , and determine whether the particle is speeding up or slowing down at $t = 1s$ and $t = 4 s$ .

M01-03-P03

A particle has

4 2 x(t) = t − 4t + 2t.

(55)

Find $vx(t)$ , $ax(t)$ , and $ax (1)$ .

M01-03-P04

Velocity changes from $v1 = 3ex + 4ey m ∕s$ to $v2 = − 3ex + 10ey m∕s$ in $2.0s$ . Find $aavg$ and its magnitude.

M01-03-P05

A ball is thrown vertically upward. At its highest point, state the velocity and acceleration when upward is positive and air resistance is neglected.

M01-03-P06

A particle moves left with $vx = − 7 m ∕s$ and has $ax = − 2m ∕s2$ . Is it speeding up or slowing down? Explain using signs.

M01-03-P07

A particle moves right with $vx = 9m ∕s$ and has $ax = − 3m ∕s2$ . Is it speeding up or slowing down?

M01-03-P08

A car travels around a circular curve of radius $50 m$ at constant speed $15m ∕s$ . Find the acceleration magnitude and state its direction.

M01-03-P09

Frame $′ S$ moves at constant velocity relative to inertial frame $S$ . A particle has acceleration $2 a = 2ex − 3ey m ∕s$ in $S$ . What acceleration is measured in $′ S$ ?

M01-03-P10

A frame has translational acceleration $A = 4ex m ∕s2$ relative to an inertial frame. A particle has inertial-frame acceleration $a = ex + 2ey m ∕s2$ . Find its acceleration in the translating frame.

17 Compact answer check

P01: $3.0m ∕s2$ east.
P02: $2 ax = − 4m ∕s$ ; at $1 s$ , $vx = 8 > 0$ , so slowing down; at $4s$ , $vx = − 4 < 0$ , so speeding up.
P03: $vx = 4t3 − 8t + 2$ ; $ax = 12t2 − 8$ ; $ax(1) = 4m ∕s2$ .
P04: $aavg = − 3ex + 3ey m ∕s2$ ; magnitude $√ -- 3 2 ≈ 4.24m ∕s2$ .
P05: $v = 0 y$ ; $a = − g ≈ − 9.8 m ∕s2 y$ .
P06: Speeding up because $vx$ and $ax$ have the same sign.
P07: Slowing down because $vx$ and $ax$ have opposite signs.
P08: $ac = v2∕R = 4.5m ∕s2$ , inward.
P09: The same acceleration: $2e − 3e m ∕s2 x y$ .
P10: $′ 2 a = a − A = − 3ex + 2ey m ∕s$ .

18 Connection to the next articles

M01-03 completes the basic position-velocity-acceleration chain:

dr dv r(t) −→ v (t) = ---−→ a(t) = ---. dt dt

(56)

M01-04 uses motion graphs to connect slopes and areas among $x(t)$ , $v(t)$ , and $a(t)$ . M01-05 then develops the special but extremely important case of constant acceleration in full.

References

[1] PhysicsLibrary, Acceleration, object id 73.

[2] PhysicsLibrary, M01-02: Velocity in Mechanics.

[3] OpenStax / cnxuniphysics, University Physics Volume 1, archived 2016 BCcampus clone, CC BY 4.0.

[4] University of California, Davis, Physics 9A: Classical Mechanics, Physics LibreTexts, CC BY-SA 4.0.

[5] J. R. Taylor, Classical Mechanics, University Science Books, 2005. Used as a scope and notation reference.

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