Derivatives of Motion
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.
Velocity is the derivative of position, and acceleration is the derivative of velocity: v = dx/dt, a = dv/dt = d²x/dt². Given x(t), you can find everything about the motion at any instant just by differentiating.
1. Three graphs, one motion
2. Reading the motion
At rest means v = 0. Solve v(t) = 0, and those are the turning points of x(t).
Maximum or minimum position happens where v = 0 (and changes sign).
Speeding up or slowing down? Compare the signs of v and a at that instant. A negative acceleration doesn’t mean slowing down if v is also negative.
Speed is |v|, never negative.
3. Example, then try it
A particle’s position is x(t) = 4t³ − 2t² + 5 (SI units). Find its velocity and acceleration at t = 2 s.
v = dx/dt = 12t² − 4t, so v(2) = 48 − 8 = 40 m/s.
a = dv/dt = 24t − 4, so a(2) = 44 m/s².
For x(t) = t³ − 3t² (graphed below), find when the particle is at rest and its minimum position. Is it speeding up or slowing down at t = 1.5 s? When is its acceleration zero?
Velocity: v = 3t² − 6t = 3t(t − 2), which is zero at t = 0 and t = 2 s.
Minimum: x(2) = 8 − 12 = −4 m.
At 1.5 s: v = 6.75 − 9 = −2.25 m/s and a = 6t − 6 = +3 m/s². Opposite signs, so it’s slowing down.
a = 0: 6t − 6 = 0 at t = 1 s, the inflection point of x(t).
Answer: at rest at 0 and 2 s; min −4 m; slowing down at 1.5 s; a = 0 at 1 s.
4. Common mistakes
A particle moves with v(t) = t² − 4t (m/s). At t = 3 s, is it speeding up or slowing down?
v(3) = 9 − 12 = −3 m/s and a = 2t − 4 = +2 m/s². They have opposite signs, so the particle is slowing down.
Ready to practice?
5 questions on Derivatives of Motion, each with a hint and an explanation for every wrong answer.