01 Check the conditions
Use these equations only when acceleration is constant throughout the chosen interval, along one fixed axis.
Choose +x first. Keep signed velocities, acceleration and displacement consistent with it.
u = initial velocity; v = final velocity (m/s)
a = acceleration (m/s²); t = elapsed time (s)
Δx = x - x₀ = displacement (m)
Revision aid: first learn position, velocity and acceleration. Scope excludes projectiles and variable acceleration.
02 Choose what you need
| v = u + at | displacement not needed |
| Δx = ut + ½at² | final velocity not needed |
| v² = u² + 2aΔx | time not needed |
| Δx = ½(u + v)t | acceleration not needed explicitly |
All four still require constant acceleration.
v² does not tell you the sign of v. Use direction and time information.
Δx is signed displacement. If motion reverses, find distance by adding the lengths travelled on each side of the turn.
03 Contrast: displacement or distance?
Take right as positive. A particle has u = +4 m/s and constant a = -2 m/s² for 4 s. Find final velocity, displacement and distance.
Pause here and try before reading the check.
v = 4 - 2 × 4 = -4 m/s
Δx = 4 × 4 + ½ × (-2) × 4² = 0 m
Turn: 0 = 4 - 2t, so t = 2 s. It travels 4 m right, then 4 m left. Distance = 8 m.
Same starting and ending position: zero displacement, but nonzero distance. The particle reverses because the stated acceleration continues after it stops.