Ch 2 · 2 Kinematics · A Level Physics 9702
Topic 2 of 3 in Physical Quantities, Kinematics and Dynamics — A Level Physics 9702 — 57 lessons.
Introduction
Kinematics describes motion without asking what causes it: where an object is, how fast it is going and how its velocity changes. Four equations, a few graphs and one key idea about projectiles cover the whole topic. (Syllabus topic 2: section 2.1.)
Key ideas — 2.1 (a) The quantities
A runner who completes one lap of a 400 m track has run a distance of 400 m but has a displacement of zero.
🧑🏫 Teacher's explanation
Imagine you are walking in a large circular park. If you walk all the way around the perimeter and end up exactly where you started, your pedometer (which tracks distance) will show a large number, but your GPS (which tracks displacement) will show zero because your position hasn't changed relative to your starting point. This is the fundamental difference between scalars and vectors.
📖 Definitions
🔍 Worked example — Easy
A cyclist travels 120 m North and then 50 m South. Calculate the total distance travelled and the final displacement.
🔍 Worked example — Standard
A particle moves along a straight line such that its displacement s is given by s(t) = 4t^2 + 2t, where t is in seconds and s is in meters. Determine the velocity at t = 3.0 s.
🔍 Worked example — Exam-style
A car accelerates from rest to a velocity of 25 m s^-1 in 5.00 s in a straight line. Calculate the average acceleration.
⚠️ Common errors
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🎯 Exam question
Key ideas — 2.1 (b) Graphs
Displacement–time (s–t) graph: the gradient is the velocity. A straight line means constant velocity; a curve means changing velocity (draw a tangent to find the instantaneous velocity).
🧑🏫 Teacher's explanation
Think of graphs as a 'movie' of an object's journey. A displacement-time graph tells you where the object is at any moment. If the line is steep, the object is moving fast; if it is flat, the object has stopped. The 'steepness' or gradient tells you how fast the position is changing, which is velocity.\n\nA velocity-time graph is like a speedometer reading over time. If the line is straight and sloping upwards, the object is speeding up at a steady rate (constant acceleration). If the line is flat, the speed isn't changing (zero acceleration). The area under this graph represents the total distance covered, because you are essentially multiplying 'how fast' by 'how long'.\n\nImagine a car driving. A displacement-time graph shows its position on a map. A velocity-time graph shows its speedometer reading. An acceleration-time graph shows how hard the driver is pressing the gas pedal.
📖 Definitions
🔍 Worked example — Easy
A particle moves such that its displacement s (m) at time t (s) is given by s = 5t^2. Calculate the velocity of the particle at t = 3 s.
🔍 Worked example — Standard
A car accelerates from rest to 25 m/s in 5.0 s with constant acceleration. Sketch the velocity-time graph and find the displacement covered.
🔍 Worked example — Exam-style
A ball is thrown upwards. Its velocity v (m/s) is given by v = 20 - 9.81t for the first 2.04 s. Find the total displacement after 2.04 s using the area under the v-t graph.
⚠️ Common errors
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🎯 Exam question
Key ideas — 2.1 (c) Deriving the equations of uniformly accelerated motion
For constant acceleration a, initial velocity u, final velocity v, time t and displacement s:
🧑🏫 Teacher's explanation
Imagine you are driving a car. If you press the accelerator steadily, your speed doesn't just jump; it increases by the same amount every second. This is 'uniform acceleration'. Because the change is steady, your 'average speed' during the trip is simply the halfway point between where you started and where you ended.
📖 Definitions
🔍 Worked example — Easy
A car starts from rest and accelerates at a constant rate of 3.0 m s^-2 for 5.0 s. Calculate the final velocity.
🔍 Worked example — Standard
A ball is thrown vertically upwards with an initial velocity of 12 m s^-1. Calculate the maximum height reached. Assume g = 9.81 m s^-2 and ignore air resistance.
🔍 Worked example — Exam-style
A stone is dropped from the top of a cliff and hits the ground after 3.5 s. Calculate the height of the cliff. Assume g = 9.81 m s^-2.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — 2.1 (d) Free fall
An object falling with no air resistance has a constant acceleration g = 9.81 m s⁻² downwards (near the Earth's surface), whatever its mass. If you take upwards as positive, a = -9.81 m s⁻².
🧑🏫 Teacher's explanation
Imagine you drop a heavy bowling ball and a light tennis ball at the same time in a vacuum. Without air pushing back, they fall at exactly the same rate. This is because gravity pulls on everything with the same acceleration, regardless of how much it weighs.
📖 Definitions
🔍 Worked example — Easy
A stone is dropped from the top of a cliff. Calculate its velocity after 3.0 s has elapsed. Assume g = 9.81 m s^-2 and ignore air resistance.
🔍 Worked example — Standard
A ball is thrown upwards from the ground with an initial velocity of 15 m s^-1. Calculate the maximum height reached by the ball. Use g = 9.81 m s^-2.
🔍 Worked example — Exam-style
An object is released from rest at a height of 50 m above the ground. Calculate the time it takes to hit the ground. Assume g = 9.81 m s^-2.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — 2.1 (e) Projectile motion: perpendicular components
When an object is thrown, the horizontal and vertical motions are independent (with no air resistance):
🧑🏫 Teacher's explanation
Imagine you are standing on a cliff and you throw a ball straight out horizontally, while simultaneously dropping another ball from the same height. Even though one is moving sideways and the other is falling straight down, they will both hit the ground at the exact same time. This is because gravity only pulls downwards; it doesn't care how fast you are moving sideways.
📖 Definitions
🔍 Worked example — Standard
A projectile is launched from a height of 40 m with a horizontal velocity of 15 m s^-1. Calculate the time it takes to hit the ground (assume g = 9.81 m s^-2 and no air resistance).
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Practical notes
Alternatives: a light gate timing a card or "picket fence" falling through it, or video analysis. A small dense ball minimises air resistance. A non-zero intercept on the graph shows a systematic error (e.g. a delay in the magnet releasing the ball).
Worked example
Example 1. A car accelerates uniformly from 12 m s⁻¹ to 30 m s⁻¹ in 6.0 s. Find its acceleration and the distance travelled.
🔍 Worked example — Easy
A car accelerates uniformly from u = 12 m s^-1 to v = 30 m s^-1 in t = 6.0 s. Calculate the acceleration a and the total distance s travelled.
🔍 Worked example — Exam-style
A ball rolls off a horizontal table at a velocity of 3.0 m s^-1. The table is 1.2 m high. Calculate the horizontal distance (range) from the edge of the table to where the ball lands.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) From the first, t = (v-u)/a; substitute: s = tfrac12(u+v)(v-u)/a = (v^2 - u^2)/2a, so v^2 = u^2 + 2as. (2) s = tfrac12 × 9.81 × 2.5^2 = 30.7 m. (3) 0 = 25^2 + 2a(500) ⇒ a = -0.625 m s⁻²; t = 25/0.625 = 40 s. (4) u_y = 10 m s⁻¹, u_x = 17.3 m s⁻¹; (a) t = 2u_y/g = 20/9.81 = 2.04 s; (b) range = 17.3 × 2.04 = 35 m. (5) (a) acceleration; (b) displacement.