Accelerate from rest
A particle starts from rest with acceleration 2 m/s² for 3 s. Find velocity and displacement.
- v = 0 + 2×3 = 6.
- s = 0×3 + ½×2×3² = 9.
Answer: 6 m/s; 9 m
Motion with constant acceleration
Use constant-acceleration equations with a consistent positive direction.
Before you start
You’ll need Rearranging formulae. Follow a link if you want to revise it first.
Choosing a positive direction
Velocity, acceleration and displacement are signed quantities. Choose a positive direction before substituting.
For constant acceleration, v = u + at and s = ut + ½at². These equations need the constant-acceleration assumption.
Displacement is the signed change in position; distance travelled requires accounting for any change of direction.
If a particle turns, calculate its displacement over the whole interval before deciding whether it ends to the left or right of its starting point.
Accelerating and slowing down
Decelerate without turning
A particle has u = 10 m/s and a = −2 m/s² for 3 s. Find velocity and displacement.
- v = 10 − 2×3 = 4; velocity stays positive.
- s = 10×3 + ½×(−2)×3² = 21.
Answer: 4 m/s; 21 m
Deceleration is not always a negative acceleration; its sign depends on your positive direction and the velocity.
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If the particle turns during an interval, how would you distinguish distance from displacement?