Calculus in kinematics

Calculus describes motion when velocity or acceleration changes continuously. If a particle moves along a straight line and its displacement from a fixed origin is $s$ at time $t$, then

$$v=\frac{ds}{dt}, \qquad a=\frac{dv}{dt}=\frac{d^2s}{dt^2}.$$

Differentiation moves from displacement to velocity to acceleration. Integration reverses this chain, but introduces constants that must be found from information about the motion.

Prerequisites

You should be able to:

• interpret displacement, velocity, speed and acceleration in kinematics language;
• differentiate powers and standard functions using differentiation basics;
• find indefinite and definite integrals using integration basics;
• solve polynomial equations and use exact values where possible.

The motion in this lesson is along a straight line. A positive direction is chosen once and used consistently.

The displacement, velocity and acceleration chain

For displacement $s(t)$,

$$s(t)\xrightarrow{\frac{d}{dt}}v(t)\xrightarrow{\frac{d}{dt}}a(t).$$

In the reverse direction,

$$a(t)\xrightarrow{\int\,dt}v(t)\xrightarrow{\int\,dt}s(t).$$

The units provide a useful check. If $s$ is measured in metres and $t$ in seconds, then

$$[s]=\mathrm m, \qquad [v]=\mathrm{m\,s^{-1}}, \qquad [a]=\mathrm{m\,s^{-2}}.$$

Each differentiation with respect to time contributes one factor of $\mathrm{s^{-1}}$.

What the signs mean

• $s>0$: the particle is on the positive side of the origin.
• $s<0$: the particle is on the negative side of the origin.
• $v>0$: the particle is moving in the positive direction.
• $v<0$: the particle is moving in the negative direction.
• $a>0$: velocity is increasing.
• $a<0$: velocity is decreasing.

The sign of acceleration does not by itself give the direction of motion. A particle with $v=-5$ and $a=2$ is moving in the negative direction while slowing down.

More generally:

Signs of $v$ and $a$	Behaviour
Same sign	Speed is increasing
Opposite signs	Speed is decreasing
$v=0$	Particle is instantaneously at rest

Negative acceleration is not always slowing down

If both velocity and acceleration are negative, the magnitude of velocity increases. The particle speeds up in the negative direction.

Differentiating a displacement function

When $s$ is given as a function of $t$, differentiate once for velocity and twice for acceleration.

Worked example 1: finding velocity and acceleration

A particle has displacement

$$s=t^3-6t^2+9t+4, \qquad t\geq 0,$$

where $s$ is in metres and $t$ is in seconds. Find its velocity and acceleration at time $t$, then find both at $t=2$.

Differentiate:

$$v=\frac{ds}{dt}=3t^2-12t+9.$$

Differentiate again:

$$a=\frac{dv}{dt}=6t-12.$$

At $t=2$,

$$v(2)=3(2)^2-12(2)+9=-3\ \mathrm{m\,s^{-1}},$$

and

$$a(2)=6(2)-12=0\ \mathrm{m\,s^{-2}}.$$

At this instant the particle is moving in the negative direction. Zero acceleration does not mean zero velocity.

Self-check 1

A particle has displacement $s=2t^4-5t^2+3t-7$. Find $v$ and $a$ when $t=1$.

Answer

$$v=8t^3-10t+3, \qquad a=24t^2-10.$$

Therefore

$$v(1)=1\ \mathrm{m\,s^{-1}}, \qquad a(1)=14\ \mathrm{m\,s^{-2}}.$$

Rest and change of direction

A particle is at rest when

$$v(t)=0.$$

This identifies a stationary instant, but it does not automatically prove that the particle changes direction. Direction changes only if $v$ changes sign.

Worked example 2: stationary times and direction

For the particle with

$$s=t^3-6t^2+9t+4, \qquad t\geq0,$$

find when it is at rest and determine whether it changes direction.

From Worked example 1,

$$v=3t^2-12t+9=3(t-1)(t-3).$$

Hence $v=0$ when

$$t=1 \quad\text{or}\quad t=3.$$

Test the sign of $v$ in each interval:

Time interval	Test value	Sign of $v$	Direction
$0\leq t<1$	$t=0$	positive	positive
$1<t<3$	$t=2$	negative	negative
$t>3$	$t=4$	positive	positive

The velocity changes sign at both $t=1$ and $t=3$, so the particle changes direction at both times.

There is also a quicker local check here. Since $a=6t-12$,

$$a(1)=-6\neq0, \qquad a(3)=6\neq0.$$

A nonzero acceleration at an isolated zero of velocity guarantees a sign change locally. However, a sign table remains the safest general method.

At rest does not guarantee reversal

If $v=(t-2)^2$, then $v=0$ at $t=2$, but $v$ is positive on both sides. The particle pauses instantaneously and continues in the same direction.

Displacement and total distance

Between times $t=a$ and $t=b$, the displacement is

$$s(b)-s(a)=\int_a^b v(t)\,dt.$$

This is a signed change in position. Total distance travelled counts every part of the journey positively:

$$\text{distance}=\int_a^b |v(t)|\,dt.$$

At A level, total distance is usually found by locating every time at which $v=0$, splitting the journey there, and adding the magnitudes of the separate displacements.

Worked example 3: distance after reversals

For

$$s=t^3-6t^2+9t+4,$$

find the displacement and total distance travelled from $t=0$ to $t=4$.

The particle reverses at $t=1$ and $t=3$. Calculate its position at all relevant times:

$$\begin{aligned} s(0)&=4,\\ s(1)&=1-6+9+4=8,\\ s(3)&=27-54+27+4=4,\\ s(4)&=64-96+36+4=8. \end{aligned}$$

The displacement over the whole interval is

$$s(4)-s(0)=8-4=\boxed{4\ \mathrm m}.$$

The total distance is

$$\begin{aligned} |s(1)-s(0)|+|s(3)-s(1)|+|s(4)-s(3)| &=|8-4|+|4-8|+|8-4|\\ &=4+4+4\\ &=\boxed{12\ \mathrm m}. \end{aligned}$$

Simply calculating $|s(4)-s(0)|$ would give the distance between the endpoints, not the distance travelled.

Self-check 2

A particle has velocity

$$v=3t^2-12t+9, \qquad t\geq0,$$

and $s=2$ when $t=0$. Find its total distance travelled during the first $2$ seconds.

Answer

The velocity is zero at $t=1$ and $t=3$, so only $t=1$ lies within the interval.

Integrating,

$$s=t^3-6t^2+9t+C.$$

Since $s(0)=2$, $C=2$. Thus

$$s(0)=2, \qquad s(1)=6, \qquad s(2)=4.$$

Hence the distance is

$$|6-2|+|4-6|=\boxed{6\ \mathrm m}.$$

Integrating acceleration

If acceleration is known as a function of time, integrate once to find velocity:

$$v=\int a\,dt.$$

An indefinite integral includes a constant. This constant represents the information lost by differentiation, usually the initial velocity.

Worked example 4: using an initial velocity

A particle has acceleration

$$a=6t-4.$$

At $t=0$, its velocity is $5\ \mathrm{m\,s^{-1}}$. Find $v$ in terms of $t$ and determine when the particle is first at rest.

Integrate:

$$v=\int(6t-4)\,dt=3t^2-4t+C.$$

Use $v(0)=5$:

$$5=C.$$

Therefore

$$v=3t^2-4t+5.$$

For rest, solve

$$3t^2-4t+5=0.$$

Its discriminant is

$$(-4)^2-4(3)(5)=-44<0.$$

There is no real solution, so the particle is never at rest. In fact, the quadratic is always positive, so it always moves in the positive direction.

Do not silently set the constant to zero

The statement $v=5$ when $t=0$ does not mean that $C$ may be omitted. It means that the initial condition must be substituted to find $C$.

Integrating twice to find displacement

Starting from acceleration, integrate once for velocity and again for displacement. Each integration introduces its own constant.

Worked example 5: two initial conditions

A particle moves with acceleration

$$a=12t-6.$$

At $t=1$, its velocity is $4\ \mathrm{m\,s^{-1}}$ and its displacement is $7\ \mathrm m$. Find $v(t)$ and $s(t)$.

First integrate acceleration:

$$v=6t^2-6t+C.$$

Use $v(1)=4$:

$$4=6-6+C,$$

so $C=4$ and

$$v=6t^2-6t+4.$$

Now integrate velocity:

$$s=2t^3-3t^2+4t+D.$$

Use $s(1)=7$:

$$7=2-3+4+D,$$

so $D=4$. Therefore

$$\boxed{v=6t^2-6t+4}, \qquad \boxed{s=2t^3-3t^2+4t+4}.$$

The two constants are unrelated. It is good practice to give them different letters.

A definite integral alternative

If $v(t_0)=v_0$, then

$$v(t)=v_0+\int_{t_0}^{t}a(u)\,du.$$

The dummy variable $u$ prevents confusion between the upper limit $t$ and the integration variable. Similarly, if $s(t_0)=s_0$,

$$s(t)=s_0+\int_{t_0}^{t}v(u)\,du.$$

This approach builds the initial condition into the formula and avoids a separate constant, but both methods are equivalent.

Variable acceleration in context

The constant acceleration formulae are not valid when $a$ depends on time. Calculus replaces them.

Worked example 6: maximum displacement

A particle starts at the origin with velocity $12\ \mathrm{m\,s^{-1}}$. Its acceleration at time $t$ is

$$a=2t-7, \qquad 0\leq t\leq6.$$

Find its greatest displacement from the origin during this interval.

Integrate acceleration:

$$v=t^2-7t+C.$$

As $v(0)=12$, $C=12$:

$$v=t^2-7t+12=(t-3)(t-4).$$

Candidates for greatest displacement occur at the endpoints or when $v=0$. Therefore test $t=0,3,4,6$.

Integrate velocity:

$$s=\frac{t^3}{3}-\frac{7t^2}{2}+12t+D.$$

Since $s(0)=0$, $D=0$. Now

$$\begin{aligned} s(0)&=0,\\ s(3)&=9-\frac{63}{2}+36=\frac{27}{2},\\ s(4)&=\frac{64}{3}-56+48=\frac{40}{3},\\ s(6)&=72-126+72=18. \end{aligned}$$

Hence the greatest displacement is

$$\boxed{18\ \mathrm m},$$

attained at $t=6$.

It would be wrong to stop after finding the stationary times. A maximum over a closed time interval can occur at an endpoint.

Worked example 7: trigonometric motion

A particle has displacement

$$s=5\sin(2t), \qquad t\geq0.$$

Find its velocity, acceleration, and the first positive time at which it is at rest.

Using the chain rule,

$$v=\frac{ds}{dt}=10\cos(2t).$$

Differentiating again,

$$a=-20\sin(2t).$$

For rest,

$$10\cos(2t)=0 \quad\Longrightarrow\quad 2t=\frac{\pi}{2}$$

at the first positive solution. Thus

$$t=\boxed{\frac{\pi}{4}\ \mathrm s}.$$

Angles are in radians because calculus formulae such as $\frac{d}{dx}\sin x=\cos x$ assume radians.

A useful extension: acceleration as a function of displacement

Sometimes acceleration is given in terms of $s$, not $t$. The chain rule gives

$$a=\frac{dv}{dt} =\frac{dv}{ds}\frac{ds}{dt} =v\frac{dv}{ds}.$$

Therefore

$$\boxed{a=v\frac{dv}{ds}}.$$

This relation is especially useful when time does not appear in the data.

Worked example 8: eliminating time

A particle moves in a straight line with acceleration $a=6s$. When $s=1$, its speed is $4\ \mathrm{m\,s^{-1}}$. Find its possible speeds when $s=3$.

Use $a=v\frac{dv}{ds}$:

$$v\frac{dv}{ds}=6s.$$

Separate and integrate:

$$\int v\,dv=\int6s\,ds,$$

so

$$\frac12v^2=3s^2+C.$$

When $s=1$, the speed is $4$, so $v^2=16$ regardless of direction:

$$8=3+C,$$

giving $C=5$. Hence

$$\frac12v^2=3s^2+5.$$

At $s=3$,

$$\frac12v^2=27+5=32,$$

so $v^2=64$. Thus the possible velocities are $v=\pm8\ \mathrm{m\,s^{-1}}$, and the requested speed is

$$\boxed{8\ \mathrm{m\,s^{-1}}}.$$

The integration determines $v^2$, so additional information would be needed to choose the direction.

Choosing the correct method

Given	Required	Operation
$s(t)$	$v(t)$	Differentiate once
$s(t)$	$a(t)$	Differentiate twice
$v(t)$	$a(t)$	Differentiate once
$v(t)$ and one position	$s(t)$	Integrate once, then use the position
$a(t)$ and one velocity	$v(t)$	Integrate once, then use the velocity
$a(t)$, one velocity and one position	$s(t)$	Integrate twice, using both conditions
$a(s)$	relation between $v$ and $s$	Use $a=v\frac{dv}{ds}$

Common misconceptions

Confusing position with distance travelled

$s=3$ means the particle is $3$ units in the positive direction from the origin. It says nothing about the length of its previous journey.

Treating speed as signed

Velocity can be negative. Speed is $|v|$ and is never negative.

Using $v=0$ as proof of reversal

Check the sign of $v$ on both sides of the stationary time. An even repeated factor may touch zero without changing sign.

Using constant acceleration formulae for variable acceleration

Formulae such as $v=u+at$ assume that $a$ is constant. If $a=3t^2$, integrate instead.

Discarding solutions without checking the model

Time usually satisfies $t\geq0$, and a question may restrict it further. Reject solutions because they lie outside the stated time interval, not merely because there are several of them.

Mixed self-check

Question 1

A particle has velocity

$$v=t^2-5t+4, \qquad t\geq0.$$

Find its acceleration at $t=3$, and determine the times when it changes direction.

Answer

$$a=2t-5,$$

so $a(3)=1\ \mathrm{m\,s^{-2}}$.

Also

$$v=(t-1)(t-4),$$

so the particle is at rest at $t=1$ and $t=4$. The factors are simple, so $v$ changes sign at each time. The particle changes direction at both.

Question 2

A particle has acceleration $a=8-6t$. At $t=1$, $v=7$, and at $t=0$, $s=-2$. Find $s$ in terms of $t$.

Answer

Integrate acceleration:

$$v=8t-3t^2+C.$$

Using $v(1)=7$ gives $7=8-3+C$, so $C=2$. Then

$$s=4t^2-t^3+2t+D.$$

Using $s(0)=-2$ gives $D=-2$. Therefore

$$\boxed{s=4t^2-t^3+2t-2}.$$

Question 3

A particle has displacement $s=t^3-3t^2-9t+5$ for $0\leq t\leq5$. Find the total distance travelled.

Answer

$$v=3t^2-6t-9=3(t-3)(t+1).$$

Within $0\leq t\leq5$, the only stationary time is $t=3$. Calculate positions:

$$s(0)=5, \qquad s(3)=27-27-27+5=-22, \qquad s(5)=125-75-45+5=10.$$

Therefore the total distance is

$$|-22-5|+|10-(-22)|=27+32=\boxed{59\ \mathrm m}.$$

Question 4

Explain why a particle with $v=-2$ and $a=-3$ is speeding up.

Answer

The velocity is negative, so the particle moves in the negative direction. The acceleration is also negative, so the velocity becomes more negative. Its magnitude $|v|$, which is its speed, therefore increases.

Exam checklist

Before finishing a calculus kinematics problem, check that you have:

• differentiated or integrated in the correct direction;
• included and evaluated every integration constant;
• kept velocity and acceleration signed;
• used $v=0$ to locate possible reversals;
• checked signs when a change of direction matters;
• split total distance at every reversal;
• tested interval endpoints in maximum or minimum problems;
• rejected times outside the modelled interval;
• included appropriate units and given exact answers unless told otherwise.

Next steps

• Connect derivatives and integrals to gradients and signed areas in kinematics graphs.
• Compare variable acceleration with the special case of constant acceleration.
• Extend motion into two dimensions in projectiles and two dimensional motion.
• Apply forces to changing motion through Newton’s laws.