Kinematics in a Straight Line | A-Maths Study Guide

Secondary 4 Additional Mathematics · Study guide

Use calculus to describe straight-line motion, find reversals and calculate the whole distance travelled.

Concept → worked examples → practice22 min read · practice linked below

One straight line, three connected quantities

Choose a positive direction and a fixed origin $O$ . Displacement $s(t)$ is signed position relative to $O$ ; velocity $v(t)$ is its rate of change; acceleration $a(t)$ is the rate of change of velocity.

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

In the reverse direction, integrate and use initial conditions. If displacement is in metres and time in seconds, velocity is in m/s and acceleration in m/s². Speed is $|v|$ , and distance from the origin is $|s|$ . Neither is necessarily the total distance travelled.

Translate the wording before calculating
Wording	Condition
Starts from rest	$v(0)=0$ .
Starts from the origin	$s(0)=0$ .
Returns to the origin	$s(t)=0$ at a later allowed time.
Returns to its initial position	$s(t)=s(0)$ , which need not be zero.
Instantaneously at rest	$v=0$ ; check signs to decide whether direction changes.
Extreme velocity	Find candidates where $a=0$ , then justify their nature and check the interval.
Speed increasing	$v$ and $a$ have the same sign, away from rest.

The direction of the calculation matters. Starting at the origin fixes displacement, while starting from rest fixes velocity.

Read motion from the displacement function

Example 1: rest times and changes of direction

A particle has $s=t^3-6t^2+9t$ metres for $0\le t\le4$ seconds. Find its velocity, acceleration and times of rest.

v=3t^2-12t+9=3(t-1)(t-3),\qquad a=6t-12.

It is at rest at $t=1$ s and $t=3$ s. Velocity is positive for $0\le t<1$ , negative for $1<t<3$ , and positive for $3<t\le4$ . Therefore it reverses direction at both rest times.

At $t=2$ , acceleration is zero but velocity is $-3\text{ m/s}$ , so the particle is moving. This is its minimum velocity on the stated interval; the quadratic $v=3(t-2)^2-3$ confirms it.

Total distance: split at direction changes

Displacement change over $[a,b]$ is $s(b)-s(a)=\int_a^b v\,dt$ . Total distance is $\int_a^b|v|\,dt$ . Find zeros of velocity, retain only those in the time interval, check signs, and add the lengths of each one-direction leg. Average velocity is displacement change divided by elapsed time; average speed is total distance divided by elapsed time.

Example 2: distance, displacement and two different averages

For the particle in Example 1, find total distance and average speed over the first 4 seconds.

The relevant positions are $s(0)=0$ , $s(1)=4$ , $s(3)=0$ , $s(4)=4$ , all in metres. The three legs therefore have lengths 4, 4 and 4 m:

\text{Total distance}=|4-0|+|0-4|+|4-0|=12\text{ m}.

The displacement change is only $4-0=4$ m. Thus average speed is $12/4=3\text{ m/s}$ , whereas average velocity is $4/4=1\text{ m/s}$ . The 4 m distance from the origin at the end does not include the journey back and forth.

Recover motion from acceleration or velocity

Example 3: two integrations with two initial conditions

A particle has $a=6t-4\text{ m/s}^2$ , with $v(0)=3\text{ m/s}$ and $s(0)=-2\text{ m}$ , for $t\ge0$ .

v=3t^2-4t+C_1,\quad C_1=3.

s=t^3-2t^2+3t+C_2,\quad C_2=-2.

Hence $v=3t^2-4t+3$ and $s=t^3-2t^2+3t-2$ . The velocity quadratic has discriminant $16-36=-20<0$ and positive leading coefficient, so velocity is always positive. It never reverses direction.

In the first second, $s(1)=0$ , so the particle travels 2 m from its initial position $-2$ m to the origin. Its final displacement from the origin is zero, while its displacement change is 2 m.

Example 4: a moment of rest without a reversal

For $t\ge0$ , a particle starts at the origin with velocity $v=(t-2)^2\text{ m/s}$ . Does it turn at $t=2$ ?

Velocity is positive on both sides of 2 and zero only at 2. It pauses instantaneously but does not reverse. Integrating gives

s=\frac{t^3}{3}-2t^2+4t+C,\qquad C=0.

During the first 4 seconds, distance equals displacement change because $v\ge0$ : $s(4)-s(0)=64/3-32+16=16/3\text{ m}$ . A zero of velocity is a candidate split point, not automatic evidence of a direction change.

Trigonometric motion: keep the time interval and radians

Example 5: two reversals in a stated interval

A particle starts from $O$ with $v=4+8\cos2t\text{ m/s}$ , for $0\le t\le\pi$ seconds. Find its total distance.

At rest, $\cos2t=-1/2$ . Since $0\le2t\le2\pi$ , the solutions are $2t=2\pi/3,4\pi/3$ , giving $t=\pi/3,2\pi/3$ .

s=4t+4\sin2t,\qquad s(0)=0.

Velocity has signs $+, -, +$ across these times. The positions at the two turns are $4\pi/3+2\sqrt3$ and $8\pi/3-2\sqrt3$ . Finally $s(\pi)=4\pi$ . Thus

\begin{aligned}\text{Distance}&=s(\pi/3)+[s(\pi/3)-s(2\pi/3)]\\&\quad+[s(\pi)-s(2\pi/3)]\\&=\frac{4\pi}{3}+8\sqrt3\text{ m}.\end{aligned}

The acceleration is $-16\sin2t\text{ m/s}^2$ . Its least value on this interval is $-16\text{ m/s}^2$ , attained at $t=\pi/4$ . That is a different question from finding rest times.

Returning and approaching a limiting position

Example 6: return to the origin is not a rest condition

A particle starts from $O$ with $v=t^2-5t+4\text{ m/s}$ , for $t\ge0$ . Find when it first returns to $O$ .

s=\frac{t^3}{3}-\frac{5t^2}{2}+4t.

Returning means $s=0$ , so $t(2t^2-15t+24)=0$ . Reject $t=0$ as the starting instant. The two later solutions are

t=\frac{15\pm\sqrt{33}}4.

The first return is at $(15-\sqrt{33})/4$ seconds. At that instant $a=2t-5=(5-\sqrt{33})/2\text{ m/s}^2$ . The rest times, obtained from $v=(t-1)(t-4)=0$ , are 1 and 4 seconds; neither is the first return time.

Example 7: a limit need not be attained

A particle starts at $O$ with $v=4e^{-t}-2e^{-2t}\text{ m/s}$ , for $t\ge0$ . Describe its direction and eventual position.

Since $v=2e^{-t}(2-e^{-t})>0$ , it always moves in the positive direction. Integrating and using $s(0)=0$ gives

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

As $t$ grows, both exponentials approach zero, so $s$ approaches 3 m. For every finite $t\ge0$ , $3-s=e^{-t}(4-e^{-t})>0$ , so it never reaches 3 m. Its distance from $O$ has limiting value 3 m, with no attained maximum over all finite times.

Its acceleration is $a=-4e^{-t}+4e^{-2t}<0$ for $t>0$ . Because velocity is positive, the speed decreases, approaching zero without a finite rest time.

Explore speed, direction and distance together

Take $v=2t-4$ , $a=2$ , and $s(0)=0$ , so $s=t^2-4t$ . Before 2 seconds, velocity and acceleration have opposite signs and the particle slows down. After 2 seconds, both are positive and it speeds up.

Elapsed time t (seconds):4

The fixed graph shows velocity for 0 ≤ t ≤ 4. The moving marker selects the elapsed time; the output totals the journey only up to that time.

At t = 4 s: velocity 4 m/s; displacement 0 m; total distance 8 m. Moving in the positive direction and speeding up.

Common mistakes and useful checks

Using $s=0$ for rest: rest means $v=0$ ; $s=0$ means being at the chosen origin.
Assuming every rest is a reversal: check the sign of velocity on both sides.
Calling negative acceleration deceleration: speed decreases only when velocity and acceleration have opposite signs.
Using displacement for total distance: include every leg after direction changes.
Forgetting the time domain: discard negative or out-of-interval solutions.
Mixing units: minimum velocity has velocity units, not acceleration units.
Assuming a limiting position is reached: distinguish finite times from an eventual limit.

Continue with topic practice

Use the Secondary 4 kinematics practice bank for exam questions and their worked solutions. Attempt a question on paper before comparing methods. This guide supports Secondary 4 learning and revision of the full Secondary Additional Mathematics course.