Kinematics — Motion in a Straight Line

CIE IGCSE Additional Mathematics· 14.14, 14.15· 25 min read

1. Core Kinematic Relationships & Sign Conventions★★☆☆☆⏱ 7 min

📘 Definition

Kinematic Calculus Relationships

$v = \frac{d s}{d t}$ , $a = \frac{d v}{d t} = \frac{d ^{2} s}{d t ^{2}}$ , $s = \int v d t$ , $v = \int a d t$

For straight line motion, the three core quantities are linked by calculus: velocity is the first derivative of displacement, acceleration is the first derivative of velocity (or second derivative of displacement). The reverse relationships use integration, with constants of integration representing initial conditions.

📐 Worked Example

A particle moves in a straight line with displacement $s = 2 t^{3} - 5 t^{2} + 3 t$ meters from the origin at time $t$ seconds. Find the velocity of the particle at $t = 2$ s.

1
Differentiate the displacement function to get velocity:
$v = \frac{ds}{dt} = 6t^2 - 10t + 3$
2
Substitute $t = 2$ into the velocity function:
$v = 6(2)^2 - 10(2) + 3 = 24 - 20 + 3 = 7$
3
Final velocity is $7$ m/s in the positive direction.

Exam tip:

Always state your chosen positive direction in written answers to avoid sign error mark deductions.

2. Constant vs Variable Acceleration Problems★★★☆☆⏱ 8 min

The 0606 syllabus tests both constant acceleration and variable acceleration, and in 0606 both are handled with calculus. Differentiate to move from displacement to velocity to acceleration ( $v = \frac{d s}{d t}$ , $a = \frac{d v}{d t}$ ), and integrate to move back the other way ( $v = \int a d t$ , $s = \int v d t$ ), using the initial conditions to fix each constant of integration. When the acceleration is constant, integrating a constant simply gives a velocity that is linear in $t$ ; the List of formulas provides no separate constant-acceleration formulas, so you integrate in every case.

📐 Worked Example

A particle has constant acceleration $a = 4$ m/s², initial velocity $u = 2$ m/s, and initial displacement $s = 0$ at $t = 0$ . Find the displacement at $t = 3$ s.

1
Integrate acceleration to get velocity, add constant of integration $C$ :
$v = \int 4 \, dt = 4t + C$
2
Use initial condition $v = 2$ at $t = 0$ to find $C = 2$ , so $v = 4 t + 2$
3
Integrate velocity to get displacement, add constant $D$ :
$s = \int (4t + 2) \, dt = 2t^2 + 2t + D$
4
Use initial condition $s = 0$ at $t = 0$ to find $D = 0$ , substitute $t = 3$ :
$s = 2(3)^2 + 2(3) = 18 + 6 = 24$
5
Final displacement is $24$ m.

📐 Worked Example

A particle has variable acceleration $a = 2 t + 1$ m/s², with initial velocity $0$ m/s. Find velocity at $t = 2$ s.

1
Integrate acceleration to get velocity:
$v = \int (2t + 1) \, dt = t^2 + t + C$
2
Use initial condition $v = 0$ at $t = 0$ to find $C = 0$ , substitute $t = 2$ :
$v = (2)^2 + 2 = 6$
3
Final velocity is $6$ m/s.

Exam tip:

Whether the acceleration is constant or variable, use differentiation and integration throughout — 0606 provides no ready-made kinematics formulas, so find every velocity or displacement by integrating and applying the initial conditions.

3. Motion Graph Interpretation & Sketching★★★☆☆⏱ 6 min

Graph Type	Gradient Represents	Area Under Graph Represents
Displacement-Time ( $s$ - $t$ )	Velocity ( $v$ )	No physical meaning
Velocity-Time ( $v$ - $t$ )	Acceleration ( $a$ )	Change in displacement (signed area)
Speed-Time (speed- $t$ )	Magnitude of acceleration	Total distance travelled (sum of positive areas)
Acceleration-Time ( $a$ - $t$ )	Not required in 0606	Change in velocity

📐 Worked Example

A $v$ - $t$ graph has a straight line from $(0, 0)$ to $(2, 4)$ , a horizontal line to $(5, 4)$ , then a straight line down to $(7, 0)$ . Find the total distance travelled over 7 seconds.

1
Total distance is the total area under the $v$ - $t$ graph (all values are positive, so no need for absolute values).
2
Area of first triangle: $0.5 \times 2 \times 4 = 4$ m
3
Area of rectangle: $3 \times 4 = 12$ m
4
Area of second triangle: $0.5 \times 2 \times 4 = 4$ m
5
Total distance: $4 + 12 + 4 = 20$ m

Exam tip:

For displacement from a $v$ - $t$ graph, keep signs of areas below the $t$ -axis. For distance, take absolute values of all areas.

4. Scalar vs Vector Quantity Distinctions★★★★☆⏱ 5 min

A common exam trap is mixing scalar and vector quantities. Displacement and velocity are vectors (can be negative), while distance and speed are scalars (always non-negative). When calculating total distance over an interval, you must account for direction reversals.

📐 Worked Example

A particle's displacement is given by $s = t^{2} - 4 t$ meters, $t$ in seconds. Find the total distance travelled between $t = 0$ and $t = 3$ s.

1
Find when velocity is zero (direction reversal point):
$v = \frac{ds}{dt} = 2t - 4 = 0 \implies t = 2$
2
Calculate displacement at key points: $s (0) = 0$ , $s (2) = 4 - 8 = - 4$ m, $s (3) = 9 - 12 = - 3$ m
3
Calculate distance for each segment: $∣ s (2) - s (0) ∣ = 4$ m, $∣ s (3) - s (2) ∣ = 1$ m
4
Total distance: $4 + 1 = 5$ m

Exam tip:

Never calculate total distance as the difference between final and initial displacement, as this ignores direction reversals.

5. Common Pitfalls

Wrong move:

Reaching for a ready-made constant-acceleration formula instead of integrating

Why:

0606 lists no kinematics formulas, and such a shortcut fails as soon as the acceleration varies with time.

Correct move:

Always use differentiation and integration ( $a = \frac{d v}{d t}$ , $v = \int a d t$ , $s = \int v d t$ ), fixing the constant of integration from the initial conditions — this works for constant and variable acceleration alike.

Wrong move:

Calculating displacement from a speed-time graph

Why:

Speed is a scalar quantity, so the area under a speed-time graph gives total distance, not signed displacement.

Correct move:

Use a velocity-time graph (with sign values) to calculate displacement.

Wrong move:

Omitting the constant of integration when solving kinematics integrals

Why:

The constant represents initial velocity or displacement, so omitting it leads to incorrect final values.

Correct move:

Use given initial conditions (e.g. $s = 0$ at $t = 0$ ) to solve for the constant every time you integrate.

Wrong move:

Calculating total distance as $s_{f ina l} - s_{ini t ia l}$

Why:

If the particle reverses direction, the net displacement is smaller than the total path length travelled.

Correct move:

Find all points where $v = 0$ , calculate displacement for each segment, and sum the absolute values of each segment's displacement.

6. Quick Reference Cheatsheet

Quantity	Type	Calculus Relationship	Graph Property
Displacement ( $s$ )	Vector	$s = \int v d t$	Gradient of $s$ - $t$ graph = $v$
Velocity ( $v$ )	Vector	$v = \frac{d s}{d t} = \int a d t$	Gradient of $v$ - $t$ graph = $a$ , Area under $v$ - $t$ = $Δ s$
Acceleration ( $a$ )	Vector	$a = \frac{d v}{d t} = \frac{d ^{2} s}{d t ^{2}}$	Area under $a$ - $t$ graph = $Δ v$
Distance	Scalar	Sum of absolute displacement segments	Area under speed-time graph
Speed	Scalar	Magnitude of velocity	Absolute value of $v$ on $v$ - $t$ graph

7. Frequently Asked

When do I use differentiation vs integration in kinematics?

Use differentiation if you have a displacement function and need velocity/acceleration, or a velocity function and need acceleration. Use integration if you have an acceleration function and need velocity/displacement, or a velocity function and need displacement.

What is the difference between distance and displacement?

Displacement is a vector that measures net change in position from the origin, so it can be negative. Distance is a scalar that measures total path travelled, so it is always positive, even if the particle reverses direction.

Going deeper

study_guideDifferentiation Rules for 0606
study_guideIntegration Basics for 0606

What's Next

Now that you have mastered straight line kinematics for CIE IGCSE Additional Mathematics 0606, you can move on to applying calculus to more advanced problems involving exponential and trigonometric motion functions, which are frequently tested in Paper 2. This topic is often combined with other calculus skills in longer structured questions, so make sure you are confident with differentiation and integration rules for all function types in the syllabus. Practice past paper kinematics questions to familiarize yourself with common exam phrasing and trap scenarios.