Picture: a drone on a vertical line: Position, velocity and acceleration are linked: each one is the derivative of the one before.
Picture: a drone on a vertical line
- Imagine a drone rising and falling on a single vertical line.
- Its position s(t) tells you where it is.
- How fast that position is changing is its velocity.
- How fast the velocity is changing is its acceleration.
Differentiate down the chain: Each step is just a derivative — differentiating moves you DOWN the chain:
position → (differentiate) → velocity → (differentiate) → acceleration.
Velocity vs speed
- Velocity has a sign: positive means moving in the positive direction (up).
- Negative means moving the other way (down).
- Speed is |velocity| — the size only.
Two key moments to look for: Instantaneous rest (the particle is momentarily stopped) happens when v = 0.
Acceleration is zero when a = 0 — this is where the velocity itself is largest or smallest (a maximum/minimum of v), so it is how you find the greatest velocity in the interior of an interval.
IB-style question — a drone's velocity and acceleration
A drone moves on a vertical line so that its velocity is v(t) = 3t² − 12t + 9 (m s⁻¹) for 0 ≤ t ≤ 4 seconds, where up is positive.
(a) Find the acceleration a(t).
(b) Find the times when the drone is instantaneously at rest.
(c) Find the acceleration at t = 2.
Step by step
- (a) Acceleration is the derivative of velocity, so differentiate v(t).
- (b) The drone is at rest when v = 0, so solve the quadratic (a GDC solver does this instantly).
- Read off the roots, both inside 0 ≤ t ≤ 4.
- (c) Substitute t = 2 into a(t).
Final answer
(a) a(t) = 6t − 12 m s⁻². (b) The drone is momentarily at rest at t = 1 s and t = 3 s (where it changes direction). (c) a(2) = 0, so at t = 2 s the velocity is neither increasing nor decreasing — that is the midpoint where v is at its minimum.
Direction changes need a SIGN change, not just v = 0: The particle changes direction only where v actually switches sign. Almost always v = 0 is such a point, but check: if v just touches zero and stays the same sign (a repeated root), the particle does not turn around — it only pauses.
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Picture: running the chain backwards: Integrating moves UP the chain, from acceleration back to velocity and position.
The chain, backwards: If you only know the acceleration, you can rebuild the velocity, and from the velocity rebuild the position — by integrating:
acceleration → (integrate) → velocity → (integrate) → position.
Every integration adds + C
- Every integration adds a + C.
- That constant is the unknown starting value.
- Pin it down with an initial condition the question gives you — usually the velocity or position at t = 0.
IB-style question — a probe's velocity and position from its acceleration
A probe travels along a straight track. Its acceleration is a(t) = 6t − 4 (m s⁻²). At t = 0 its velocity is 5 m s⁻¹ and its displacement from the start is 0 m.
(a) Find the velocity v(t).
(b) Find the displacement s(t), and hence the displacement at t = 3 s.
Step by step
- (a) Velocity is the integral of acceleration — add a + C.
- Use the initial condition v(0) = 5 to find C.
- Write the velocity.
- (b) Position is the integral of velocity — add a new + C.
- Use s(0) = 0 to find this C, then substitute t = 3.
Final answer
(a) v(t) = 3t² − 4t + 5 m s⁻¹. (b) s(t) = t³ − 2t² + 5t, so at t = 3 s the probe is 24 m from the start. Each + C came from an initial condition — without them you would only know the shape, not the actual values.
One condition per integration: You integrate twice to get from acceleration to position, so you need two pieces of information — typically v(0) and s(0). Find the first C from the velocity condition BEFORE integrating again, so the second integration is of the complete velocity.