Differentiate down: s → v → a: For motion in a line, velocity is the derivative of displacement (v = ds/dt) and acceleration is the derivative of velocity (a = dv/dt).
So differentiate to go from displacement to velocity to acceleration.
On a velocity–time graph the gradient is acceleration and the signed area is displacement; total distance counts area below the axis as positive.
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IB-style question — v and a
A particle has displacement s = t³ − 6t² + 9t metres.
Find its velocity and acceleration at t = 1.
Step by step
- Differentiate for v, then a.
- Substitute t = 1.
Final answer
At t = 1: velocity 0 m/s, acceleration −6 m/s².
a is the second derivative of s: Acceleration is dv/dt = d²s/dt² — differentiating displacement twice.
IB-style question — acceleration as a target
A particle moves with velocity v = t² − 4t (m s⁻¹).
Find the time at which the acceleration is 6 m s⁻².
Step by step
- Acceleration is the derivative of velocity.
- Set a = 6 and solve for t (don't just evaluate — solve).
Final answer
t = 5 s.
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Integrate up: a → v → s (with + C): Going the other way, integrate: v = ∫a dt and s = ∫v dt.
Each integration brings a + C, found from an initial condition (e.g. v at t = 0).
IB-style question — velocity from acceleration
A particle has acceleration a = 6t − 4 and velocity 2 m/s at t = 0.
Find v(t).
Step by step
- Integrate a to get v.
- Use v(0) = 2.
Final answer
v(t) = 3t² − 4t + 2.
Don't lose the + C: Each integration needs + C; an initial condition (value at t = 0) pins it down.
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v = 0 is at rest; a = 0 is a velocity extreme: The particle is at rest (and may change direction) when v = 0.
Its velocity is greatest or least when its derivative is zero, i.e. a = 0.
IB-style question — at rest & max velocity
A particle has v = 3t² − 12t + 9.
Find when it is at rest.
A second particle has v = 12 + 8t − 4t²; find its maximum velocity.
Step by step
- At rest: v = 0.
- Max velocity: a = 0 for the second particle.
Final answer
First particle at rest at t = 1 and t = 3; second particle's maximum velocity is 16 m/s (at t = 1).
Changes direction at v = 0: If v changes sign at a time when v = 0, the particle changes direction there.
Displacement = ∫v; distance = ∫|v|: Over [a, b], the displacement (net change in position) is ∫ₐᵇ v dt.
The total distance travelled is ∫ₐᵇ |v| dt — split at the times where v = 0 and add the magnitudes of each piece.
IB-style question — distance vs displacement
A particle has v = 3t² − 12t + 9 (so s = t³ − 6t² + 9t).
Find the displacement and the total distance travelled from t = 0 to t = 4.
Step by step
- Displacement = s(4) − s(0).
- Distance: v = 0 at t = 1, 3; add the magnitudes of each leg.
Final answer
Displacement = 4 m; total distance travelled = 12 m.
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IB-style question — does it return to the start?
A particle moves with velocity v = t² − 4t (m s⁻¹) for 0 ≤ t ≤ 5.
(a) Find the net displacement over the 5 seconds.
(b) Does the particle return to its starting point?
Step by step
- (a) Net displacement is the (signed) integral of velocity.
- (b) The net displacement is −25⁄3 ≠ 0, so the particle ends up 25⁄3 m on the negative side — it does NOT return to the start.
Final answer
(a) −25⁄3 ≈ −8.33 m. (b) No — net displacement ≠ 0.
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