Describing Motion
Speed and Velocity
Speed tells you how fast an object moves, but not which direction. It is a scalar quantity — it has magnitude but no direction. The equation is:
$$s = v \times t$$
Where \(s\) is distance (m), \(v\) is speed (m/s), and \(t\) is time (s). Rearranged: \(v = \frac{s}{t}\) or \(t = \frac{s}{v}\). Typical speeds to be familiar with: walking ~1.5 m/s, running ~3 m/s, cycling ~6 m/s, a car in a town ~13 m/s, and the speed of sound in air ~340 m/s.
Velocity is speed in a stated direction. It is a vector quantity. An object travelling around a circular track at constant speed has a changing velocity, because direction is continuously changing — so it is always accelerating.
Distance (scalar) measures the total path length travelled. Displacement (vector) measures how far — and in what direction — an object is from its starting point. A runner who completes one full lap of a 400 m track has covered a distance of 400 m but has a displacement of zero.
Scalars have magnitude only: speed, distance, mass, energy. Vectors have magnitude AND direction: velocity, displacement, force, acceleration. An object moving in a circle has a constantly changing velocity (direction changes) even if speed is constant.
Acceleration
Acceleration is how quickly velocity changes. It can mean speeding up, slowing down (deceleration), or changing direction. The equation is:
$$a = \frac{v - u}{t}$$
Where \(a\) is acceleration (m/s2), \(v\) is final velocity (m/s), \(u\) is initial velocity (m/s), and \(t\) is time (s). A negative acceleration (deceleration) means the object is slowing down. For example, a car braking from 20 m/s to rest in 4 s has \(a = \frac{0 - 20}{4} = -5 \text{ m/s}^2\).
Question: A cyclist accelerates from 4 m/s to 12 m/s in 5 seconds. Calculate the acceleration.
Answer: \(a = \frac{v - u}{t} = \frac{12 - 4}{5} = \frac{8}{5} = 1.6 \text{ m/s}^2\). The cyclist is accelerating at 1.6 m/s2 in the direction of travel.
Distance-Time Graphs
On a distance-time graph, the gradient (slope) of the line equals the speed. Interpreting the shapes:
• Horizontal line — the object is stationary (speed = 0).
• Straight diagonal line — constant speed; steeper gradient means faster.
• Curve getting steeper — the object is accelerating.
• Curve getting shallower — the object is decelerating.
To find speed at a specific moment on a curved distance-time graph, draw a tangent to the curve at that point and calculate the gradient of the tangent. The tangent is a straight line that just touches the curve at one point and has the same slope as the curve there.
On a distance-time graph: gradient = speed. Horizontal = stopped. Straight line = constant speed. Curve getting steeper = accelerating. To find instantaneous speed on a curve, draw a tangent and calculate its gradient.
Velocity-Time Graphs
On a velocity-time graph, two pieces of information can be extracted:
• The gradient equals the acceleration. Steep upward slope = large acceleration. Horizontal line = constant velocity (zero acceleration). Downward slope = deceleration.
• A line below the x-axis means the object is moving in the opposite direction to the positive direction defined in the question.
The area under the line on a velocity-time graph equals the distance travelled. For a triangular shape: area = \(\frac{1}{2} \times \text{base} \times \text{height}\). For a rectangular shape: area = base × height. Complex shapes are split into triangles and rectangles and summed.
Scenario: A train accelerates from 0 to 30 m/s in 10 s, then travels at constant 30 m/s for 20 s, then decelerates to rest in 5 s.
Acceleration phase: gradient = \(\frac{30 - 0}{10} = 3 \text{ m/s}^2\).
Constant phase: gradient = 0, so acceleration = 0 m/s2, constant velocity.
Deceleration phase: gradient = \(\frac{0 - 30}{5} = -6 \text{ m/s}^2\) (deceleration of 6 m/s2).
Uniform and Non-Uniform Acceleration
Uniform (constant) acceleration produces a straight line on a velocity-time graph. Non-uniform acceleration (changing acceleration) produces a curve. For uniform acceleration, you can also use:
$$v^2 = u^2 + 2as$$
Where \(s\) is distance (m). This equation is useful when time is not given — for example, calculating braking distance given initial speed and deceleration. Real-world journeys almost always involve non-uniform acceleration — average speed is found by dividing total distance by total time.
Motion and Road Safety
The physics of motion has direct consequences for road safety. Stopping distance is the total distance a vehicle travels from when a driver notices a hazard to when the vehicle comes to rest. It has two parts:
• Thinking distance — the distance covered at constant speed during the driver’s reaction time (typically 0.2–0.9 s depending on alertness). Increased by tiredness, alcohol, or drug use.
• Braking distance — the distance covered while decelerating to rest. Increased by higher speed (because kinetic energy \(E_k = \frac{1}{2}mv^2\) grows with the square of speed), wet or icy roads (reduced friction), or worn tyres and brakes.
Because braking distance scales with \(v^2\), doubling speed more than doubles stopping distance. A car travelling at 30 mph on a dry road requires roughly 23 m to stop; at 60 mph, the stopping distance is approximately 73 m — more than three times greater for twice the speed. This is why speed limits in residential areas are low — the consequences of a collision at lower speed are far less severe.