Displacement
d = xf − xi
Change in position, including direction.
metres · m · vectorReturn to your starting point: displacement = 0, even though distance travelled is not zero.
AS91171 MECHANICSYear 12 Physics · Interactive learning
Linear Motion / Chapter activities
Linear Motion Lab
Year 12 Physics · Observe, predict, simulate, calculate, explain
Created for Alex Sun · University of Auckland
01 / The whole picture
One dimension. Three connected quantities.
Motion tells a story: where an object is, how quickly its position changes, and how its velocity changes.
Choose a positive direction first. On the track: right → positive, left ← negative. For vertical motion: up ↑ positive.
Velocity
v̄ = ΔdΔt
Average rate of change of displacement.
metres per second · m/s · vectorInstantaneous velocity is the gradient of the tangent to a displacement–time graph.
Acceleration
ɑ̄ = ΔvΔt
Average rate of change of velocity.
m/s² · vectorThe sign tells you the direction of acceleration. It does not, on its own, tell you whether speed increases.
Constant-acceleration equations
Use these only when acceleration is constant.
(1)vf = vi + ɑt
(2)d = vi + vf2t
(3)d = vit + 12ɑt2
(4)vf2 = vi2 + 2ɑd
(5)d = vft − 12ɑt2
d: displacement · vi: initial velocity · vf: final velocity · a: acceleration · t: elapsed time
Read motion from a graph
d–t gradient → velocity
A steeper slope means a greater speed.
v–t gradient → acceleration
Signed v–t area → displacement
Area below v = 0 is negative. Add absolute areas to find total distance.
05 / Choose before calculating
Which equation connects what you know?
1. Identify the quantities
2. Choose an equation
3. Show substitution & interpretation
Question-type guide
Choose the relationship before calculating. Equations below assume constant acceleration unless the graph method is used.
vi: initial velocity · vf: final velocity · d: displacement · a: acceleration · t: elapsed time. The live v readout is velocity at the selected time, so it is vf for the interval from launch to that instant.
Average acceleration
Clue: Initial and final velocity, and elapsed time
Find: Acceleration a
Think: Choose a positive direction before subtracting velocities
Use: ɑ = vf − vit
Watch out: Negative acceleration does not necessarily mean slowing down.
Final / initial velocity
Clue: Acceleration, elapsed time, and one velocity
Find: vf or vi
Think: Velocity changes by at under constant acceleration
Use: vf = vi + ɑt; vi = vf − at
Watch out: Keep the signs of velocity and acceleration; speed is the magnitude of velocity.
Displacement from initial velocity
Clue: Initial velocity, acceleration and elapsed time
Find: Displacement d
Think: Use an equation that does not require final velocity
Use: d = vit + 12ɑt2
Watch out: Displacement is signed. If the object reverses, distance travelled is not simply |d|.
Displacement from two velocities
Clue: Initial and final velocity, and elapsed time
Find: Displacement d
Think: Under constant acceleration, average velocity is the mean of the endpoints
Use: d = vi + vf2t
Watch out: This average-velocity shortcut requires constant acceleration.
Displacement from final velocity
Clue: Final velocity, acceleration and elapsed time
Find: Displacement d
Think: Use an equation that does not require initial velocity
Use: d = vft − 12ɑt2
Watch out: The acceleration term is subtracted; retain the sign of a when substituting.
Stopping time / stopping distance
Clue: Initial velocity and a constant braking acceleration
Find: Time and displacement until the object stops
Think: Set vf = 0 at the stopping instant
Use: t = 0 − viɑ; 0 = vi² + 2ad
Watch out: Stop the braking model at v = 0 unless continued acceleration and reversal are intended. Reaction distance is a separate stage.
Motion without time
Clue: Three of d, vi, vf and a
Find: The remaining quantity
Think: Eliminate time from the calculation
Use: vf2 = vi2 + 2ɑd
Watch out: Taking a square root gives a magnitude. Choose the velocity sign from the physical direction.
Find elapsed time
Clue: Velocities and acceleration, or displacement and initial conditions
Find: Elapsed time t
Think: Choose a linear equation when possible; otherwise solve a quadratic
Use: t = vf − viɑ; d = vit + 12ɑt2
Watch out: The first form requires a ≠ 0. For a quadratic, retain only times that fit the situation; a position may be reached twice.
Free fall / vertical throw
Clue: Initial vertical velocity and a time or height change
Find: Vertical velocity, displacement or time
Think: Use the same equations with up positive and a = −g
Use: vf = vi − gt; d = vit − 12gt²; vf² = vi² − 2gd
Watch out: At the highest point v = 0, but a = −g. Displacement from launch differs from height above ground.
Two vehicles / catch-up
Clue: Starting positions and each vehicle’s motion
Find: Time and position of a meeting
Think: Use a shared clock and set the two positions equal
Use: xA = xA,0 + vi,At + 12ɑAt²; xB = xB,0 + vi,Bt + 12ɑBt²; solve xA = xB
Watch out: Include the starting gap and any start delay. Equal position does not imply equal velocity.
Displacement–time graph
Clue: A graph of displacement against time
Find: Average or instantaneous velocity
Think: Read the slope: secant for an interval, tangent for an instant
Use: Average velocity = ΔdΔt; instantaneous velocity = tangent gradient
Watch out: A negative gradient means negative velocity. The graph’s height is displacement, not speed.
Velocity–time graph
Clue: A graph of velocity against time
Find: Acceleration, displacement or distance
Think: Read the gradient and the signed area
Use: ɑ = ΔvΔt; d = signed area; distance = sum of absolute areas
Watch out: Area below the time axis is negative displacement. Split at v = 0 when finding total distance.
Track auto-fits the full motion. Objects are not to scale; velocity and acceleration arrows use different scales.