Q4MechanicsStretch

Variable Acceleration

Q: When is velocity at a maximum?

When dv dt = 0, that is when the acceleration is zero. It is the ordinary stationary-point condition applied to the velocity function.

When acceleration changes with time the suvat equations are invalid, because their derivation assumed a straight velocity–time graph. What replaces them is calculus: $\textbf{differentiate}$ to go from displacement towards acceleration, and $\textbf{integrate}$ to come back.

40 min Kinematics

The big picture

This is the lesson where the two halves of the course meet. The gradient and area you read off graphs in the first kinematics lesson are exactly differentiation and integration, and here the connection is made explicit and used. Everything you know about differentiating polynomials and evaluating definite integrals transfers directly, with the physical meaning supplying the interpretation: a stationary point of the displacement function is where the object is instantaneously at rest, and a definite integral of velocity is the displacement over that interval. Nothing new is being learned about calculus — only about what it is describing.

What you'll be able to do

Differentiate to obtain velocity and acceleration from displacement
Integrate to obtain velocity and displacement, using initial conditions
Find times of instantaneous rest and maximum or minimum velocity
Distinguish distance from displacement when the motion reverses

The chain of relationships

Velocity is the rate of change of displacement, and acceleration the rate of change of velocity. So differentiating moves you one step along the chain $s \to v \to a$ , and integrating moves you back.

The direction matters for what you need. Differentiation needs nothing extra; integration produces an arbitrary constant, which must be found from an $\textbf{initial condition}$ such as "the particle starts from rest" or "at $t = 0$ the displacement is 2 m".

Forgetting the constant of integration is the characteristic error of this topic, and it makes every subsequent value wrong.

v = \dfrac{\mathrm{d}s}{\mathrm{d}t}

Differentiate displacement.

a = \dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{\mathrm{d}^{2}s}{\mathrm{d}t^{2}}

Differentiate again.

v = \int a \,\mathrm{d}t, \quad s = \int v \,\mathrm{d}t

Integrate to reverse — with a constant.

Example 1A particle has displacement

s = 2t^{3} - 9t^{2} + 12t

metres. Find expressions for

v

and

a

1Differentiate:

v = 6t^{2} - 18t + 12

2Differentiate again:

a = 12t - 18

Answer

v = 6t^{2}-18t+12\ \mathrm{m\,s^{-1}}

;

a = 12t-18\ \mathrm{m\,s^{-2}}

Why this works

The acceleration here is $12t - 18$ , which changes with time — confirming immediately that no suvat equation could be used. Checking whether $a$ is constant is a quick way to decide which toolkit a question needs.

Integrating with initial conditions

Given acceleration, integrating gives velocity plus a constant. The initial condition pins the constant down, and only then can you integrate again for displacement — with a second constant needing a second condition.

Read the conditions carefully. "Starts from rest" gives $v = 0$ at $t = 0$ ; "starts at the origin" gives $s = 0$ at $t = 0$ . A question may give one, both, or a condition at some other time.

Example 1A particle has

a = 6t - 4\ \mathrm{m\,s^{-2}}

. At

t = 0

it is at the origin with velocity

3\ \mathrm{m\,s^{-1}}

. Find

s

in terms of

t

1Integrate:

v = 3t^{2} - 4t + c

2At

t = 0

v = 3

, so

c = 3

and

v = 3t^{2}-4t+3

3Integrate again:

s = t^{3} - 2t^{2} + 3t + k

4At

t = 0

s = 0

, so

k = 0

Answer

s = t^{3} - 2t^{2} + 3t

metres

Tip — Find the first constant before integrating a second time. Carrying an unknown $c$ through the second integration produces a $ct$ term and doubles the algebra.

Instantaneous rest and maximum velocity

A particle is $\textbf{instantaneously at rest}$ when $v = 0$ . Solving that equation gives the times, and there may be several — each one a moment when the particle changes direction.

Velocity is at a maximum or minimum when $\dfrac{\mathrm{d}v}{\mathrm{d}t} = 0$ , which is when the $\textbf{acceleration}$ is zero. That is the standard stationary-point technique applied to $v$ rather than to $s$ .

Note the distinction: $v = 0$ locates rest, while $a = 0$ locates extreme velocity. Confusing the two is a common error, and the physical meanings are quite different.

Example 1With

v = 6t^{2} - 18t + 12

, find when the particle is at rest and when its velocity is minimum.

1Rest:

6t^{2}-18t+12 = 0

, so

t^{2}-3t+2 = 0

, giving

(t-1)(t-2)=0

2Minimum velocity:

a = 12t - 18 = 0

, so

t = 1.5

3Check:

v(1.5) = 6(2.25) - 27 + 12 = -1.5\ \mathrm{m\,s^{-1}}

AnswerAt rest at

t = 1

s and

t = 2

s; velocity minimum

-1.5\ \mathrm{m\,s^{-1}}

t = 1.5

Why this works

The minimum sits exactly midway between the two rest times, which is what you would expect for a quadratic velocity — and it is negative, meaning the particle is moving backwards fastest at that instant.

Distance when the motion reverses

The definite integral $\int_{a}^{b} v\,\mathrm{d}t$ gives the $\textbf{displacement}$ over that interval, with intervals of negative velocity subtracting.

For $\textbf{distance}$ you must first find where $v = 0$ , split the integral at those times, and add the magnitudes of the separate pieces. Integrating straight through cancels the outward and return journeys against each other.

This is the single most common source of lost marks in the topic, because the calculation looks complete without the split.

\text{displacement} = \int_{t_{1}}^{t_{2}} v \,\mathrm{d}t

Signed.

\text{distance} = \sum \left| \int v \,\mathrm{d}t \right|

Split at every time where

v = 0

Example 1A particle has

v = t^{2} - 4t\ \mathrm{m\,s^{-1}}

. Find the distance travelled from

t = 0

t = 6

1Find where

v = 0

t(t-4) = 0

, so

t = 0

and

t = 4

. Split at

t = 4

2From 0 to 4:

\left[\tfrac{t^{3}}{3} - 2t^{2}\right]_{0}^{4} = \tfrac{64}{3} - 32 = -\tfrac{32}{3}

3From 4 to 6:

\left[\tfrac{t^{3}}{3}-2t^{2}\right]_{4}^{6} = \left(72 - 72\right) - \left(-\tfrac{32}{3}\right) = \tfrac{32}{3}

4Distance adds magnitudes:

\tfrac{32}{3} + \tfrac{32}{3}

Answer

\tfrac{64}{3} \approx 21.3\ \mathrm{m}

(displacement would be 0).

Tip — Whenever a question asks for distance rather than displacement, check for sign changes in $v$ first. Here the displacement is exactly zero while the distance is over 21 m.

Think like an examiner

State whether you are differentiating or integrating and why, at each step.
Never omit the constant of integration, and find it before integrating again.
For distance, find where $v = 0$ , split the integral, and add magnitudes.
Check whether $a$ is constant before starting — if it is, suvat is quicker.

Common misconceptions

Using suvat when the acceleration depends on

t

Those equations require constant acceleration. Use calculus instead.

Omitting the constant of integration.

Every integration introduces one, and it must be found from an initial condition.

Integrating straight through to find distance when the velocity changes sign.

Split at each root of

v = 0

and add the magnitudes.

Setting

v = 0

to find maximum velocity.

Maximum velocity is where

a = 0

v = 0

gives instantaneous rest.

Calculus in kinematics

v = \tfrac{\mathrm{d}s}{\mathrm{d}t}

Differentiate displacement.

a = \tfrac{\mathrm{d}v}{\mathrm{d}t}

Differentiate velocity.

s = \int v\,\mathrm{d}t

Integrate, plus a constant.

v = 0

Instantaneously at rest.

a = 0

Maximum or minimum velocity.

Stretch yourself

A particle moves with acceleration $a = 6 - 6t\ \mathrm{m\,s^{-2}}$ . It starts at the origin with velocity $-4\ \mathrm{m\,s^{-1}}$ . Find the distance travelled in the first 3 seconds.

Hint — Integrate twice with the given conditions, then locate any times where $v = 0$ before integrating for distance.

Reveal solution

SolutionIntegrating,

v = 6t - 3t^{2} + c

; at

t=0

v = -4

, so

c = -4

and

v = -3t^{2} + 6t - 4

. Check for sign changes: the discriminant of

-3t^{2}+6t-4

36 - 48 = -12 < 0

, and since the leading coefficient is negative,

v

is negative for all

t

. So the particle never changes direction and no split is needed. Integrating for displacement:

s = -t^{3} + 3t^{2} - 4t

(constant zero, since

s=0

t=0

). At

t = 3

s = -27 + 27 - 12 = -12

m. The displacement is

-12

m and, because the velocity never changed sign, the distance is simply

12

m. The discriminant check is worth doing routinely — it settles in one line whether a split is needed at all.

\blacksquare

Questions students ask

How do I know whether to use suvat or calculus?

Check whether the acceleration is constant. A numerical value means suvat; an expression involving

t

means calculus.

Why do I need initial conditions?

Integration determines a function only up to a constant. The initial condition selects the one that matches the actual motion.

What is the difference between distance and displacement here?

Displacement is the signed integral of velocity. Distance requires splitting at every time where

v = 0

and summing the magnitudes, so that a return journey adds rather than cancels.

When is velocity at a maximum?

When

\tfrac{\mathrm{d}v}{\mathrm{d}t} = 0

, that is when the acceleration is zero. It is the ordinary stationary-point condition applied to the velocity function.

Key takeaways

Differentiate along $s \to v \to a$ ; integrate to reverse, adding a constant each time.
Initial conditions determine the constants and must be applied in order.
$v = 0$ gives instantaneous rest; $a = 0$ gives maximum or minimum velocity.
Definite integration of $v$ gives displacement, with negative regions subtracting.
For distance, split at each root of $v = 0$ and add the magnitudes.

How this fits the course

Build on

Newton’s Laws of Motion

Leads to

Newton’s Laws of Motion

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

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