Ch 4 · 14.14–14.15 Kinematics · O Level Additional Mathematics 4037
Topic 4 of 4 in Calculus — O Level Additional Mathematics 4037 — 55 lessons.
Introduction
A particle moving along a line, like a lift, a piston or a train between stations, has a displacement, a velocity and an acceleration that change with time. Calculus links all three: differentiate to go down the chain, integrate to come back up. (Syllabus 14.14–14.15.)
Key ideas — The chain s → v → a
v = ds/dt a = dv/dt = d^2s/dt^2 s = ∫ v dt v = ∫ a dt
🧑🏫 Teacher's explanation
Think of motion like a sequence of connected gears. If you know where you are (displacement), the rate at which your position changes is your velocity. If you know how fast your velocity is changing, that is your acceleration. It is a continuous chain of information: position arrow velocity arrow acceleration.
📖 Definitions
🔍 Worked example — Easy
A particle moves along a straight line such that its displacement s at time t seconds is given by s = t^3 - 6t^2 + 9t. Find the velocity of the particle when t = 4.
🔍 Worked example — Standard
The acceleration of a particle is given by a = 6t - 4. Given that the particle starts from rest at the origin (s=0 when t=0), find its displacement when t = 3.
🔍 Worked example — Exam-style
A particle moves such that its velocity is v = 2t - 5 for t ≥ 0. Find the total distance traveled by the particle between t = 0 and t = 4.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
(a) Find the acceleration of the particle when t = 3. [2] (b) Find the displacement of the particle between t = 0 and t = 3. [3] [5 marks]
Key ideas — Key phrases
🧑🏫 Teacher's explanation
Think of a particle moving along a straight line like a car driving down a long, straight road. The position of the car is its displacement s, the speedometer reading is its velocity v, and how hard you feel being pushed back into your seat is the acceleration a. These three are linked by calculus: velocity is the rate of change of displacement, and acceleration is the rate of change of velocity. \n\nImagine you are tracking a runner. If you know their position at every second, you can find their speed by seeing how much their position changes over a tiny bit of time (differentiation). If you know their speed at every second, you can find how far they ran by adding up all those tiny movements (integration). It is like a chain: s xrightarrowdiff v xrightarrowdiff a. To go backwards, you just reverse the process: a xrightarrowint v xrightarrowint s.
📖 Definitions
🔍 Worked example — Easy
A particle moves such that its displacement s (in meters) at time t seconds is given by s = t^2 + 4t + 5. Find the velocity when t = 3.
🔍 Worked example — Standard
The velocity of a particle is v = 3t^2 - 12t + 9 for t ≥ 0. Find the acceleration of the particle when it is at instantaneous rest.
🔍 Worked example — Exam-style
The acceleration of a particle is a = 6t - 4 for t ≥ 0. Given that the initial velocity is 5 m/s and the initial displacement is 2 m, find the displacement when t = 3.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Distance versus displacement
Displacement is s(t_2) - s(t_1). Distance travelled counts every metre: find when v = 0 in the interval (where the particle turns round), work out the displacement over each part, and add their sizes.
🧑🏫 Teacher's explanation
Imagine you are walking in a long hallway. If you walk 5 metres forward and then 3 metres back towards the start, your 'displacement' is only 2 metres from where you began. However, your 'distance travelled' is 8 metres because your legs still did the work of walking 8 metres. In calculus, displacement is simply the change in position, whereas distance is the total path covered.
📖 Definitions
🔍 Worked example — Standard
The velocity of a particle is v = 3t^2 - 12t + 9 for 0 ≤ t ≤ 4. Find the total distance travelled by the particle in the interval 0 ≤ t ≤ 4.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — The five graphs
Mark turning points, intercepts and end points with their values.
🧑🏫 Teacher's explanation
Imagine you are driving a car along a straight road. Kinematics is simply the study of how your position changes over time. If you look at your GPS, you see your displacement; if you look at your speedometer, you see your velocity; and if you feel a sudden push against your seat, you are experiencing acceleration.
📖 Definitions
🔍 Worked example — Easy
A particle moves such that its displacement s (in meters) at time t (in seconds) is given by s = t^2 - 4t + 3. Find the velocity and acceleration at t = 3.
🔍 Worked example — Standard
A particle's velocity is given by v = 3t^2 - 12t + 9 for t ≥ 0. Find the total displacement of the particle between t = 0 and t = 3.
🔍 Worked example — Exam-style
A particle moves with velocity v = t^2 - 4t + 3 for 0 ≤ t ≤ 4. Find the total distance traveled by the particle.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Worked examples — Example 1: polynomial velocity
v = 3t^2 - 30t + 72 m s⁻¹, with s = 0 at t = 0.
🔍 Worked example — Easy
A particle moves along a straight line such that its velocity v at time t seconds is given by v = 3t^2 - 30t + 72 m s^-1. Find the acceleration of the particle when t = 2 s.
🔍 Worked example — Exam-style
A particle moves with velocity v = 3t^2 - 30t + 72 m s^-1. Find the total distance travelled by the particle in the first 6 seconds.
⚠️ Common errors
Worked examples — Example 2: the graphs
s = 3t^3 - 10t^2 + 4t + 8 for 0 ≤ t ≤ 3.
🔍 Worked example — Easy
A particle moves such that its displacement s (in meters) at time t (in seconds) is given by s = t^2 + 4t + 3. Find the velocity and acceleration at t = 2.
🔍 Worked example — Standard
A particle's displacement is s = 3t^3 - 10t^2 + 4t + 8 for 0 ≤ t ≤ 3. Find the times when the particle is momentarily at rest.
🔍 Worked example — Exam-style
A particle moves with velocity v = 9t^2 - 20t + 4. Sketch the speed-time graph for 0 ≤ t ≤ 3.
⚠️ Common errors
Worked examples — Example 3: exponential and trigonometric motion
v = 6e^-0.5t: a = -3e^-0.5t (always decelerating); displacement in the first 2 s = ∫_0^2 6e^-0.5t dt = 12(1 - e^-1) = 7.59 m. The velocity tends to 0 but never reaches it.
🔍 Worked example — Easy
A particle moves in a straight line such that its velocity v at time t seconds is given by v = 6e^-0.5t. Find the displacement of the particle in the first 2 seconds.
🔍 Worked example — Standard
A particle moves along a straight line with velocity v = 4cos(2t) m/s. Find the acceleration a and the first time t > 0 when the particle is at rest.
🔍 Worked example — Exam-style
A particle's velocity is v = 6e^-0.5t. Show that the particle is always decelerating for t > 0.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) v = 3t^2 - 12t + 9 = 3(t - 1)(t - 3): t = 1, 3; a = 6t - 12 = -6 and 6 m s⁻². (2) v = 3t^2 - 4t; s = t^3 - 2t^2. (3) s = t^2 - 8t: displacement s(6) = -12 m; at rest at t = 4 with s(4) = -16; distance = 16 + 4 = 20 m. (4) a = 5e^-t, so 5 m s⁻² initially; v → 5 m s⁻¹.