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O-Level Physics · Secondary 3 and 4

Kinematics

Kinematics is the study of how objects move, without yet asking why they move. It is one of the first topics in O-Level Physics and gives you the language of distance, displacement, speed, velocity and acceleration, together with the motion graphs used to describe them. Almost every later mechanics topic, from Newton's laws to work and energy, assumes you can read a speed-time graph and calculate an acceleration quickly and correctly, so it is worth making this topic secure early.

What the syllabus expects

  • State what is meant by speed and velocity and calculate average speed as total distance travelled divided by total time taken
  • Distinguish between scalar and vector quantities, such as distance and displacement
  • Define acceleration and calculate it using a = (v − u)/t
  • Plot and interpret distance-time and speed-time graphs
  • Find acceleration from the gradient of a speed-time graph and distance travelled from the area under it
  • State that the acceleration of free fall near the Earth's surface is constant and approximately 10 m/s²
  • Describe the motion of falling objects with and without air resistance, including terminal velocity

Distance, displacement, speed and velocity

Distance is the total length of path travelled. It is a scalar, so it has size only. Displacement is the distance moved in a stated direction from the starting point, so it is a vector. Speed is distance travelled per unit time, and velocity is the rate of change of displacement, which is speed in a stated direction. Both are measured in metres per second (m/s). Average speed is total distance divided by total time. A runner who completes one lap of a 400 m track in 80 s has an average speed of 400 ÷ 80 = 5 m/s, but because she ends where she started, her displacement and average velocity are both zero. To convert km/h to m/s, divide by 3.6, so 72 km/h is 20 m/s.

Acceleration

Acceleration is the rate of change of velocity: a = (v − u)/t, where u is the initial velocity, v is the final velocity and t is the time taken. Its unit is m/s², which you can read as metres per second, every second. A car that speeds up from 5 m/s to 25 m/s in 4 s has an acceleration of (25 − 5) ÷ 4 = 5 m/s², so its speed rises by 5 m/s every second. If the object slows down, v is smaller than u and the acceleration comes out negative. This is called a deceleration. Uniform acceleration means the velocity changes by equal amounts in equal intervals of time, which shows up as a straight sloping line on a speed-time graph.

Reading motion graphs

On a distance-time graph, the gradient gives the speed. A straight sloping line means constant speed, a horizontal line means the object is at rest, and a curve that gets steeper means the object is speeding up. On a speed-time graph, the gradient gives the acceleration and the area under the graph gives the distance travelled. A horizontal line means constant speed, a straight line sloping upwards means uniform acceleration, and a line sloping down means uniform deceleration. For example, a trolley that accelerates uniformly from rest to 6 m/s in 3 s has an acceleration of 6 ÷ 3 = 2 m/s², and it travels ½ × 3 × 6 = 9 m in that time, the area of the triangle under the line. Split complicated graphs into triangles and rectangles and add the areas.

Free fall and terminal velocity

Without air resistance, every object near the Earth falls with the same constant acceleration of about 10 m/s², whatever its mass. A hammer and a feather dropped in a vacuum land together. With air resistance the story changes. At the moment of release the speed is zero, so the only force is the weight and the acceleration is g. As the object speeds up, air resistance increases, so the resultant downward force decreases and the acceleration decreases. The speed-time graph becomes less steep. Eventually air resistance grows until it equals the weight. The resultant force is then zero, the acceleration is zero and the object falls at a constant speed called its terminal velocity. When a skydiver opens a parachute, air resistance suddenly exceeds the weight, so she decelerates until she reaches a new, lower terminal velocity.

Worked example

A car starts from rest and accelerates uniformly to 20 m/s in 8.0 s. It then travels at a constant 20 m/s for 12 s before braking uniformly to rest in 4.0 s. (a) Calculate the acceleration in the first 8.0 s. (b) Calculate the total distance travelled. (c) Calculate the average speed for the whole journey.

  1. Acceleration in the first stage: a = (v − u)/t = (20 − 0) ÷ 8.0 = 2.5 m/s².
  2. Distance in the first stage is the area of the triangle under the speed-time graph: ½ × 8.0 × 20 = 80 m.
  3. Distance at constant speed is the area of the rectangle: 12 × 20 = 240 m.
  4. Distance while braking is the area of the second triangle: ½ × 4.0 × 20 = 40 m.
  5. Total distance = 80 + 240 + 40 = 360 m.
  6. Total time = 8.0 + 12 + 4.0 = 24 s, so average speed = total distance ÷ total time = 360 ÷ 24 = 15 m/s.

Answer: (a) 2.5 m/s² (b) 360 m (c) 15 m/s

Common mistakes

  • Reading the distance straight off the vertical axis of a speed-time graph. That axis shows speed. Distance is the area under the line.
  • Finding average speed by averaging the starting and finishing speeds. For a journey with several stages or stops, average speed is always total distance divided by total time.
  • Writing a = (u − v)/t, which flips the sign. Always take final velocity minus initial velocity, so a slowing object gives a negative acceleration.
  • Believing heavier objects fall faster. Without air resistance all objects have the same acceleration of free fall. Any difference in real life comes from air resistance.
  • Saying no forces act on an object at terminal velocity. The weight and air resistance both still act, they are simply equal and opposite, so the resultant force is zero.
  • Mixing units, such as a speed in km/h with a time in seconds, or a graph with time in minutes. Convert to m and s before calculating.

Exam tips

  • Before using a graph, check whether it is distance-time or speed-time and note the units on each axis. The same shape means different things on the two graphs.
  • When finding distance from a speed-time graph, label each triangle and rectangle, write each area separately, then add them. Method marks are easier to award when each part is visible.
  • For terminal velocity explanations, give the forces in order: weight constant, air resistance increases with speed, resultant force decreases, acceleration decreases, forces become equal, constant velocity.

Practice questions

1. Easy: A cyclist travels 4.5 km in 15 minutes. Calculate her average speed in m/s.

Show answer

5.0 m/s. Convert first: 4.5 km = 4500 m and 15 min = 15 × 60 = 900 s. Average speed = 4500 ÷ 900 = 5.0 m/s.

2. Medium: A ball is dropped from rest. Ignoring air resistance and taking the acceleration of free fall as 10 m/s², find its speed after 1.5 s and the distance it has fallen in that time.

Show answer

15 m/s and 11.25 m (about 11 m). Speed: v = u + at = 0 + 10 × 1.5 = 15 m/s. The speed-time graph is a straight line from 0 to 15 m/s over 1.5 s, so distance = area of triangle = ½ × 1.5 × 15 = 11.25 m.

3. Harder: A train moving at 30 m/s brakes with uniform deceleration and stops in 25 s. Find (a) the deceleration, (b) the braking distance and (c) its speed 10 s after the brakes are applied.

Show answer

(a) a = (0 − 30) ÷ 25 = −1.2 m/s², so the deceleration is 1.2 m/s². (b) Braking distance = area of triangle = ½ × 25 × 30 = 375 m. (c) The speed falls by 1.2 m/s every second, so after 10 s it is 30 − 1.2 × 10 = 18 m/s.

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