Variable Acceleration Motion

Variable acceleration motion occurs when acceleration changes during the interval being studied. The acceleration may depend on time, position, velocity, or several variables at once. In such problems the familiar constant acceleration equations are generally not valid. Instead, one returns to the differential definitions

dv- a = dt ,

(1)

and

dr v = ---, dt

(2)

and integrates or solves the resulting differential equations using the supplied acceleration law and initial conditions.

The essential idea is simple: acceleration tells how velocity changes, and velocity tells how position changes. When acceleration is variable, those changes must be accumulated with calculus rather than represented by one constant slope.

Figure 1. Variable acceleration is accumulated by integration. Integrating the acceleration law gives velocity, and integrating the velocity gives position. Initial conditions determine the constants of integration.

1 Why the constant acceleration equations stop working

For constant acceleration one may write

v = v0 + at.

(3)

If acceleration varies with time, however, the quantity at no longer represents the accumulated change in velocity. The correct relation is

∫ t v (t) − v0 = a (τ )dτ. t0

(4)

Likewise, the displacement is not generally

Δx = v0Δt + 1-a(Δt)2. 2

(5)

Instead,

∫ t x (t) − x0 = v (τ)dτ. t0

(6)

The constant acceleration equations are therefore special cases of the more general integral relations.

2 Time dependent acceleration

Suppose one-dimensional acceleration is known explicitly as a function of time:

a = a(t).

(7)

Since

dv a(t) = --, dt

(8)

we have

dv = a(t)dt.

(9)

Integrating from an initial time t₀ with velocity v₀ to a later time t gives

∫ t v (t) = v0 + a(τ )dτ. t0

(10)

The dummy variable τ is used inside the integral so that the upper limit t remains distinct from the integration variable.

Once v(t) is known,

∫ t x (t) = x0 + v (τ)dτ. t0

(11)

Thus a known a(t) is integrated twice to obtain position.

3 A double integral form

Substituting the velocity integral into the position integral gives

[ ] ∫ t ∫ τ x(t) = x0 + v0(t − t0) + a(s)ds dτ. t0 t0

(12)

For most introductory problems it is clearer to integrate in two separate steps, but the double integral form makes the structure explicit: variable acceleration is accumulated once into velocity and again into position.

4 Geometric interpretation on motion graphs

On an acceleration time graph,

∫ Δv = a(t)dt

(13)

is the signed area under the acceleration curve.

On a velocity time graph,

∫ Δx = v(t)dt

(14)

is the signed area under the velocity curve.

The graphical and calculus viewpoints are the same statement in different forms.

Figure 2. Example of time dependent acceleration. The changing acceleration produces a curved velocity graph; integrating again produces the position history.

5 Acceleration given as a function of position

Sometimes acceleration is specified as a function of position rather than time:

a = a(x ).

(15)

The definition

a = dv- dt

(16)

contains a time derivative, but the chain rule can eliminate time:

dv- dvdx- dt = dx dt.

(17)

Since

dx-= v, dt

(18)

we obtain the fundamental one-dimensional identity

dv a = v --. dx

(19)

Therefore

v dv = a(x )dx.

(20)

Integrating from (x₀,v₀) to (x,v) gives

1 ∫ x -(v2 − v20) = a (ξ )dξ. 2 x0

(21)

This relation is especially useful when time is absent from the problem.

Figure 3. If acceleration is supplied as a function of position, the chain rule converts the time derivative into the useful form a = v dv∕dx.

6 Connection with the constant acceleration no time equation

If a(x) = a₀ is constant, then

∫ x 1-(v2 − v20) = a0dξ = a0(x − x0). 2 x0

(22)

Multiplying by two gives

v2 = v2 + 2a0Δx. 0

(23)

Thus the familiar no time constant acceleration equation is a special case of the variable position relation.

7 Acceleration given as a function of velocity

A prescribed kinematic model may also have

a = a (v ).

(24)

Because

dv- dt = a(v),

(25)

one may separate variables when a(v)≠0:

dv dt = ----. a(v)

(26)

Hence

∫ v t − t = -du--. 0 v0 a(u)

(27)

Using a = v dv∕dx also gives

v dv dx = ----. a(v)

(28)

These forms become important later when forces such as aerodynamic drag produce acceleration that depends on speed. In the present article the emphasis remains kinematic: the acceleration law is assumed to be given.

8 Vector form for time dependent acceleration

If the acceleration vector is known as a function of time,

a = a(t),

(29)

then

∫ t v (t) = v0 + a(τ)dτ, t0

(30)

and

∫ t r (t) = r0 + t v (τ )dτ. 0

(31)

In a fixed Cartesian basis these vector integrals are evaluated component by component.

9 Initial conditions and constants of integration

Indefinite integration introduces constants. For example, if

a(t) = 6t,

(32)

then

v(t) = 3t2 + C1,

(33)

and

x (t) = t3 + C1t + C2.

(34)

The initial conditions determine C₁ and C₂. Using definite integrals usually makes this bookkeeping more transparent because the initial values enter directly.

10 Turning points and sign information

A turning point in one-dimensional motion occurs when

v = 0

(35)

and the velocity changes sign. Variable acceleration can produce several such points.

The equation v = 0 identifies candidate turning times or positions, but one should check the velocity on either side. A zero of velocity need not be a reversal if the velocity only touches zero and retains the same sign.

Similarly, the sign of acceleration alone does not determine whether the particle speeds up or slows down. Speed increases when velocity and acceleration have the same sign and decreases when their signs are opposite.

11 Piecewise acceleration

Realistic simplified models often specify different acceleration laws on different intervals. The solution must then be propagated continuously from one interval to the next.

If the first interval ends at t = t₁, compute

x(t1), v(t1),

(36)

and use those values as the initial conditions for the next interval. Position and velocity are normally continuous unless the idealized model explicitly introduces an instantaneous impulse.

12 Numerical integration preview

Not every acceleration law has a convenient elementary antiderivative. Even when the acceleration is known exactly, one may need numerical integration.

Over a small time step Δt, a basic approximation is

vn+1 ≃ vn + anΔt,

(37)

xn+1 ≃ xn + vnΔt.

(38)

This is the forward Euler method. It is easy to understand but may require very small time steps for good accuracy. Later computational mechanics articles can introduce more accurate schemes such as midpoint and Runge–Kutta methods.

Figure 4. Choosing the constant acceleration method begins by asking what the acceleration is given as a function of: time, position, velocity, or a more general state.

13 Worked example 1: acceleration varying linearly with time

A particle moves along the x axis with

a(t) = 4 − 2t m ∕s2.

(39)

At t = 0,

v = 3 m∕s, x = 1 m. 0 0

(40)

Find v(t) and x(t), then evaluate them at t = 3 s.

Integrate the acceleration:

∫ t v(t) = 3 + (4 − 2τ)dτ. 0

(41)

Thus

2 v(t) = 3 + 4t − t .

(42)

Integrate velocity:

∫ t 2 x(t) = 1 + (3 + 4τ − τ ) dτ, 0

(43)

1 x(t) = 1 + 3t + 2t2 − --t3. 3

(44)

At t = 3 s,

v(3) = 6 m ∕s,

(45)

and

x (3 ) = 19 m.

(46)

The acceleration has changed from +4 m∕s² initially to −2 m∕s² at t = 3 s, yet the velocity is still positive because velocity depends on the accumulated acceleration history.

14 Worked example 2: acceleration proportional to time

A particle has

a(t) = 1.5t m ∕s2,

(47)

with

v0 = 2.0 m ∕s, x0 = 0.

(48)

Find the velocity and position at t = 4 s.

Integrating,

∫ t v(t) = 2.0 + 1.5τ dτ = 2.0 + 0.75t2. 0

(49)

Therefore

v (4) = 14.0 m ∕s.

(50)

Now integrate velocity:

∫ t 2 3 x(t) = 0 (2.0 + 0.75τ )d τ = 2.0t + 0.25t .

(51)

Thus

x (4) = 24.0 m.

(52)

15 Worked example 3: position dependent acceleration

A particle moves along the x axis with

a(x) = − 4x,

(53)

where the numerical coefficient has units of s⁻². At

x0 = 1.0 m,

(54)

it is momentarily at rest. Find its velocity when it reaches

x = 0.60 m.

(55)

Use

dv a = v --. dx

(56)

Then

v dv = − 4x dx.

(57)

Integrate from (x₀,v₀) = (1, 0) to (0.60,v):

∫ 1-2 0.60 2v = − 4x dx. 1

(58)

Therefore

1 -v2 = − 2[(0.60 )2 − 12] = 1.28, 2

(59)

v2 = 2.56.

(60)

The speed is 1.60 m∕s. Because the particle began at positive x with acceleration toward decreasing x, its velocity on the first inward pass is

v = − 1.60 m ∕s.

(61)

The sign is supplied by the direction of motion; the integrated equation determines v².

16 Worked example 4: vector variable acceleration

A particle has

a(t) = 2te + 6 e m ∕s2, x y

(62)

with

v = 1e − 2 e m ∕s, 0 x y

(63)

and

r0 = 0ex + 1 ey m.

(64)

Find v and r at t = 2 s.

Integrating componentwise,

v(t) = (1 + t2)ex + (− 2 + 6t)ey.

(65)

Thus

v (2) = 5ex + 10ey m ∕s.

(66)

Integrating again,

( ) t3 2 r(t) = t + 3 ex + (1 − 2t + 3t)ey.

(67)

At t = 2 s,

14- r(2) = 3 ex + 9ey m.

(68)

17 Practice problems

M01-06-P01

A particle has a(t) = 6t m/s², v(0) = 2 m/s, and x(0) = 1 m. Find v(t) and x(t).

M01-06-P02

A particle has a(t) = 3 − 4t m/s² and v(0) = −1 m/s. Find the velocity at t = 2 s.

M01-06-P03

For a(t) = 2t² m/s² with v(0) = 0 and x(0) = 0, find the position at t = 3 s.

M01-06-P04

A one-dimensional particle has velocity v(t) = 5 + 2t − t² m/s. Find its acceleration and identify the time at which the acceleration changes sign.

M01-06-P05

A particle begins with v₀ = 4 m/s. Its acceleration is +2 m/s² from t = 0 to 3 s and then −1 m/s² from t = 3 to 7 s. Find its final velocity.

M01-06-P06

For a(x) = 3x with x in meters and the coefficient in s⁻², a particle has v₀ = 2 m/s at x₀ = 0. Find its speed at x = 2 m.

M01-06-P07

A particle satisfies a(x) = −9x and is released from rest at x₀ = 0.50 m. Find its speed when it first reaches x = 0.

M01-06-P08

Show that if a(x) = a₀ is constant, integrating a = v dv∕dx gives v² = v₀² + 2a₀(x − x₀).

M01-06-P09

A particle has a(t) = 4e_x − 2te_y m/s² and starts from rest at the origin. Find v(t) and r(t).

M01-06-P10

A velocity dependent acceleration is a(v) = −kv, where k is a positive constant with units s⁻¹. Starting from dv∕dt = −kv, separate variables and show that v(t) = v₀e^−kt.

18 Compact answer check

M01-06-P01:

2 3 v = 2 + 3t , x = 1 + 2t + t .

(69)

M01-06-P02:

v(2) = − 3 m ∕s.

(70)

M01-06-P03:

27 x (3 ) = ---m = 13.5 m. 2

(71)

M01-06-P04:

a(t) = 2 − 2t,

(72)

so a = 0 at t = 1 s.

M01-06-P05:

v(7) = 6 m ∕s.

(73)

M01-06-P06:

v = 4 m ∕s.

(74)

M01-06-P07:

|v | = 1.5 m ∕s.

(75)

M01-06-P08: result as stated.

M01-06-P09:

v = 4tex − t2ey,

(76)

r = 2t2e − 1t3e . x 3 y

(77)

M01-06-P10:

v = v0e− kt.

(78)

19 Common mistakes

Using the four constant acceleration equations when a varies appreciably.
Forgetting the initial condition constants after indefinite integration.
Treating a(x) as though it could be integrated directly with respect to time without first relating x and t.
Using a = v dv∕dx without recognizing that it is a one-dimensional chain rule result.
Taking the positive square root of v² automatically; the sign of velocity must agree with the actual direction of motion.
Confusing zero acceleration at one instant with constant velocity over the entire motion.

20 Summary

Variable acceleration kinematics returns to the definitions

dv dr a = ---, v = --. dt dt

(79)

For time dependent acceleration,

∫ v(t) = v0 + a(t) dt,

(80)

followed by integration of velocity to obtain position. For one-dimensional position dependent acceleration, the chain rule gives

dv- a = v dx,

(81)

which removes time and relates speed directly to position. These ideas form the natural extension of constant acceleration kinematics and prepare for variable force dynamics, oscillations, drag, and numerical integration.

References

[1] J. R. Taylor, Classical Mechanics, University Science Books, 2005. Used as a scope and notation reference; prose and figures in this entry are original.

[2] J. W. Schnick, Calculus-Based Physics I, Open Textbook Library / BCcampus collection, CC BY-SA 3.0.

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

[4] OpenStax, University Physics Volume 1, archived 2016 BCcampus revision, CC BY 4.0.

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