Particle Motion: Speed, Acceleration, and Direction
- Distinguish velocity, speed, and acceleration and their sign conditions
- Determine when a particle changes direction and when it is speeding up
- Compute total distance traveled versus displacement
Three quantities, and the one that is not a derivative
Position x(t), velocity v(t) = x′(t), and acceleration a(t) = v′(t) = x″(t) are a chain of derivatives. Speed = |v(t)| is not a derivative of anything — it is the magnitude of velocity, and that distinction is the source of most errors in this topic. Velocity carries direction through its sign: positive means moving in the positive direction, negative means moving in the negative direction. Speed is always non-negative. A particle with velocity −7 has speed 7, and a particle whose velocity changes from −7 to −9 is slowing down in no sense at all — it is speeding up while moving backward.
Changing direction versus speeding up
These are different questions with different tests, and the exam asks both in the same problem to see whether you know it. A particle changes direction where v(t) changes sign — so find where v = 0 and check that the sign actually flips there, since v can touch zero without crossing. A particle is speeding up where v and a have the same sign, and slowing down where they have opposite signs. The reasoning is that acceleration in the direction of motion increases the magnitude of velocity regardless of which direction that is. So v < 0 with a < 0 means speeding up, which is the case students most often get wrong because both numbers are negative and the word "negative acceleration" sounds like slowing.
Displacement, total distance, and position
Three integrals that look similar and are not. Displacement over [a, b] is ∫v(t)dt — signed, and it can be zero for a particle that has moved a great deal. Total distance traveled is ∫|v(t)|dt, which requires splitting the interval at every zero of v and adding the absolute values of the pieces. Position at time b is x(a) + ∫v(t)dt, which needs an initial condition. If a question asks how far the particle has traveled and you report displacement, you have answered a different question — and on a problem where the particle reverses, the two answers differ.
A particle moves with velocity v(t) = t² − 6t + 8 for t ≥ 0. Find when it changes direction, and find its total distance traveled on [0, 3].
- 1.Factor: v(t) = (t − 2)(t − 4), so v = 0 at t = 2 and t = 4.
- 2.Sign analysis: for t < 2 both factors are negative so v > 0; for 2 < t < 4 the first is positive and the second negative so v < 0; for t > 4 both are positive so v > 0. Direction changes at t = 2 and t = 4.
- 3.On [0, 3], v changes sign only at t = 2, so split there.
- 4.On [0, 2], v ≥ 0. ∫₀² (t² − 6t + 8)dt = [t³/3 − 3t² + 8t]₀² = (8/3 − 12 + 16) − 0 = 8/3 + 4 = 20/3.
- 5.On [2, 3], v ≤ 0. ∫₂³ (t² − 6t + 8)dt = [t³/3 − 3t² + 8t]₂³ = (9 − 27 + 24) − (8/3 − 12 + 16) = 6 − 20/3 = −2/3.
- 6.Total distance = |20/3| + |−2/3| = 20/3 + 2/3 = 22/3.
v(t) = 0 does not by itself mean a change of direction. For v(t) = (t − 3)², the velocity is zero at t = 3 and positive on both sides, so the particle pauses and continues the same way. Always check that the sign actually flips.
At time t = 4 a particle has v(4) = −5 and a(4) = −2. The particle is
A particle has velocity v(t) with v > 0 on [0, 2] and v < 0 on [2, 5]. Total distance traveled on [0, 5] is
Answer the 2 checkpoints as you read.
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