2.1 Equations of motion
- Syllabus
- 9702–2028–2029
- Topic
- 2.1
- Level
- AS
Learning objectives
2.1.1Distance, displacement, speed, velocity and acceleration• define and use distance, displacement, speed, velocity and acceleration2.1.2Distance, displacement, speed, velocity and acceleration• use graphical methods to represent distance, displacement, speed, velocity and acceleration2.1.3Displacement from the area under a velocity–time graph• determine displacement from the area under a velocity–time graph2.1.4Velocity using the gradient of a displacement–time graph• determine velocity using the gradient of a displacement–time graph2.1.5Acceleration using the gradient of a velocity–time graph• determine acceleration using the gradient of a velocity–time graph2.1.6Derive, from the definitions of velocity and acceleration, equations that• derive, from the definitions of velocity and acceleration, equations that represent uniformly accelerated motion in a straight line2.1.7Equations that represent uniformly accelerated motion in a straight line• solve problems using equations that represent uniformly accelerated motion in a straight line, including the motion of bodies falling in a uniform gravitational field without air resistance2.1.8The acceleration of free fall using a falling object• describe an experiment to determine the acceleration of free fall using a falling object2.1.9Motion due to a uniform velocity in one direction and a uniform acceleration• describe and explain motion due to a uniform velocity in one direction and a uniform acceleration in a perpendicular direction
Five quantities describe how position changes with time
Distance is total path length; displacement is directed change in position. Speed is rate of change of distance; velocity is rate of change of displacement. Acceleration is rate of change of velocity.
average speed = total distance / total time; average velocity = displacement / time; average acceleration = change in velocity / time = (v − u)/t.
After one 400 m lap back to the start, distance is 400 m and displacement is zero, so average speed is non-zero but average velocity is zero.
Distance and speed are scalars and non-negative. Displacement, velocity and acceleration are vectors; a negative component indicates direction along the chosen axis.
Motion graphs encode values, slopes and direction
A distance-time graph is non-decreasing and its gradient is speed. A displacement-time graph may rise or fall and its gradient is velocity. Speed-time values are non-negative; velocity-time values may be positive, zero or negative. An acceleration-time graph shows acceleration directly.
A straight sloping line on a position-time graph means constant speed or velocity. A changing gradient means acceleration. On a velocity-time graph, a horizontal line means constant velocity and a sloping line means acceleration.
On a displacement-time graph, a downward slope means negative velocity; the displacement itself need not be negative. On a velocity-time graph, crossing the time axis means reversal of direction.
Graph height, gradient and area are different information. Always read the axis labels before interpreting the shape.
Signed area under a velocity-time graph gives displacement
Over a time interval, displacement equals the signed area between the velocity-time graph and the time axis. Area above the axis is positive and area below it is negative for the chosen direction.
Split the region at zero crossings and changes of shape, calculate rectangles, triangles or trapezia, then add their signed areas. Units are (m s⁻¹) × s = m.
A triangular region from 0 to 4 s with peak velocity 6 m s⁻¹ has displacement ½ × 4 × 6 = 12 m when it lies above the axis.
Total distance is the sum of the magnitudes of all areas; displacement allows positive and negative areas to cancel.
Displacement-time gradient gives velocity
Velocity is the gradient of a displacement-time graph. A secant through two points gives average velocity over an interval; a tangent at one point gives instantaneous velocity.
Choose two well-separated points on the tangent and calculate Δs/Δt with units m s⁻¹. Positive, zero and negative slopes mean positive velocity, rest at that instant and negative velocity.
For s = t² metres, instantaneous velocity at t = 3 s is the tangent gradient 2t = 6 m s⁻¹; average velocity from 2 s to 3 s is (9 − 4)/(3 − 2) = 5 m s⁻¹.
The graph height is displacement, not velocity. A horizontal tangent means zero velocity at that instant, not necessarily that motion never resumes.
Velocity-time gradient gives acceleration
Acceleration is the gradient of a velocity-time graph. A secant gives average acceleration over an interval; a tangent gives instantaneous acceleration at a point.
Calculate Δv/Δt with units m s⁻². A straight sloping segment has constant acceleration; a horizontal segment has zero acceleration.
Velocity changing from +10 m s⁻¹ to +4 m s⁻¹ in 3 s gives average acceleration (4 − 10)/3 = −2 m s⁻² and the object slows during that interval.
Negative acceleration means velocity changes in the negative direction. It means slowing only when velocity is positive; if velocity is negative, speed may increase.
Derive constant-acceleration equations from definitions
For constant acceleration, a = (v − u)/t, so v = u + at.
Velocity changes linearly, so average velocity is (u + v)/2 and displacement is s = ½(u + v)t.
Substitute v = u + at into the average-velocity equation to get s = ut + ½at². Eliminate t using t = (v − u)/a to obtain v² = u² + 2as.
These derivations require straight-line motion with constant acceleration and consistent signed u, v, a and s.
Apply constant-acceleration equations with one sign convention
Use v = u + at, s = ut + ½at², v² = u² + 2as and s = ½(u + v)t only for straight-line motion with constant acceleration.
Choose a positive direction, assign signs to u, v, a and s, list known quantities and select the equation containing the required value and no extra unknown.
A car changes from 5.0 to 17 m s⁻¹ in 4.0 s at constant acceleration: a = 3.0 m s⁻² and s = ½(5.0 + 17) × 4.0 = 44 m.
For ideal free fall, acceleration is constant downward with magnitude g. If upward is positive, a = −g. Split the motion if acceleration changes; air resistance violates the constant-g-only model.
Determine g from a timed falling object
Hold a small dense ball with an electromagnet above a trapdoor or light gate. Measure vertical distance s from the same release reference point to the timing endpoint. Releasing the ball starts the electronic timer; detection at the endpoint stops it.
Record fall time t for several distances s, repeat each timing and average. Keep the release mechanism and distance reference fixed; use a sufficiently large fall distance to reduce percentage timing uncertainty.
Released from rest with negligible air resistance, s = ½gt². Plot s against t²: a straight line through the origin supports the model and gradient = g/2, so g = 2 × gradient.
Electronic release/detection reduces reaction-time effects but does not remove distance calibration, alignment or air-resistance errors. A non-zero intercept is evidence to investigate, not something to force through the origin.
Projectile components share time but not acceleration
With negligible air resistance, horizontal acceleration is zero and horizontal velocity is constant; vertical acceleration is constant downward with magnitude g. The two components describe one motion and therefore use the same time t.
Taking upward as positive: x = uₓt and vₓ = uₓ; y = uᵧt − ½gt² and vᵧ = uᵧ − gt.
For a horizontal launch from height h, uᵧ = 0. Setting vertical displacement to −h gives flight time √(2h/g), so horizontal range is uₓ√(2h/g).
Gravity does not create horizontal acceleration in the ideal model. Air resistance would couple the components and make horizontal velocity decrease.