Home / IB DP Maths 2026, 2027 & 2028 / Application and Interpretation HL / IB Mathematics AI AHL Kinematic problems involving displacement MAI Study Notes

IB Mathematics AI AHL Kinematic problems involving displacement MAI Study Notes - New Syllabus

IB Mathematics AI AHL Kinematic problems involving displacement MAI Study Notes

LEARNING OBJECTIVE

  • Kinematic problems involving displacement s, velocity v and acceleration a

Key Concepts: 

  •  Kinematics in calculus 

MAI HL and SL Notes – All topics

 KINEMATICS IN CALCULUS (MOTION IN ONE DIMENSION)

Displacement, Velocity, and Acceleration – Core Definitions

In one-dimensional motion, displacement s(t) describes the position of an object at time t. 

The velocity v(t) is the rate of change of displacement with respect to time.

Acceleration a(t) is the rate of change of velocity with respect to time.

Key formulas:

  • \( v(t) = \dfrac{ds}{dt} \) – Velocity is the first derivative of displacement.

  • \( a(t) = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2} \) – Acceleration is the second derivative of displacement.

  • \( a = \dfrac{dv}{ds} \cdot \dfrac{ds}{dt} = v \cdot \dfrac{dv}{ds} \) – Alternate form using chain rule.

Example

A particle moves with velocity \( v(t) = 3t^2 – 12t + 9 \).Find:

  • Displacement from \( t = 0 \) to \( t = 3 \)
  • Total distance travelled from \( t = 0 \) to \( t = 3 \)
  • Acceleration at \( t = 2 \)
▶️Answer/Explanation

Displacement:

$ s = \int_0^3 (3t^2 – 12t + 9)\,dt = [t^3 – 6t^2 + 9t]_0^3 = (27 – 54 + 27) – 0 = 0 $

Total distance:
We find roots: \( v(t) = 0 \) when \( 3t^2 – 12t + 9 = 0 \Rightarrow t = 1, 3 \)

Break into intervals: [0,1], [1,3]

$ \int_0^1 |v(t)|dt = \int_0^1 (3t^2 – 12t + 9)dt = [t^3 – 6t^2 + 9t]_0^1 = (1 – 6 + 9) = 4 $ $ \int_1^3 |v(t)|dt = -\int_1^3 (3t^2 – 12t + 9)dt = -[t^3 – 6t^2 + 9t]_1^3 = -[(27 – 54 + 27) – (1 – 6 + 9)] = -[0 – 4] = 4 $

Total distance = 4 + 4 = 8

Acceleration:

$ a(t) = \frac{dv}{dt} = 6t – 12, \quad a(2) = 6(2) – 12 = 0 $

DISPLACEMENT AND DISTANCE

Displacement from Velocity – Integration

If velocity \( v(t) \) is known, the net change in position (displacement) over a time interval \([t_1, t_2]\) can be calculated using:

$ \text{Displacement} = \int_{t_1}^{t_2} v(t)\, dt $

This gives the signed area under the velocity-time curve, which may be negative if motion is in the opposite direction.

Example

A particle’s velocity is given by $v(t) = 3t^2 – 6t$ for $0 \le t \le 4$. Find the displacement over this interval.

▶️Answer/Explanation

Solution:

$\int_0^4 (3t^2 – 6t)\, dt = [t^3 – 3t^2]_0^4 = (64 – 48) – (0) = 16$ meters

Total Distance Travelled

The total distance travelled takes into account all movement, regardless of direction. It is found by integrating the magnitude of velocity over time:

$ \text{Total distance} = \int_{t_1}^{t_2} |v(t)|\, dt $

This is always a positive quantity and represents the full path length covered.

Example 

A particle moves with velocity $v(t) = t^2 – 4t + 3$ on the interval $0 \le t \le 4$. Find the total distance travelled.

▶️Answer/Explanation

Factor: $v(t) = (t – 1)(t – 3)$

Sign changes at $t = 1$ and $t = 3$. Split integrals:

$\int_0^1 v(t)\,dt + \int_1^3 -v(t)\,dt + \int_3^4 v(t)\,dt$

Compute each to get total distance: $1 + 4 + 3 = 8$ units

 SPEED VS VELOCITY

Velocity is a vector quantity – it has both magnitude and direction.

Speed is the magnitude of velocity:

$ \text{Speed} = |v(t)| $

Speed is always non-negative, whereas velocity can be negative depending on direction.

Example – Difference Between Speed and Velocity

A car’s velocity is $v(t) = -5$ m/s for $0 \le t \le 10$. What is its speed?

▶️Answer/Explanation

Speed is the magnitude of velocity:

$|v(t)| = |-5| = 5$ m/s

NEWTON NOTATION

Notation Using Dots 

In physics, derivatives with respect to time are often written using dots above the variable:

  • \( \dot{x} = \dfrac{dx}{dt} \) – First derivative (velocity)

  • \( \ddot{x} = \dfrac{d^2x}{dt^2} \) – Second derivative (acceleration)

This notation is especially common in mechanics and differential equations.

Example 

Given $x(t) = 2t^3 – 3t^2 + 4$, find $\dot{x}$ and $\ddot{x}$ using dot notation.

▶️Answer/Explanation

$\dot{x} = \frac{dx}{dt} = 6t^2 – 6t$

$\ddot{x} = \frac{d^2x}{dt^2} = 12t – 6$

Scroll to Top