Position, velocity, acceleration and higher rates

Try

A vehicle moves in the negative direction at 4 metres per second. Its acceleration is also negative. Is it slowing down because the numbers are negative, or gaining speed? Draw a number line and compare velocities −4 and −4.2. Distance traveled per second increases in the second case.

Predict

Predict whether zero acceleration implies zero velocity. Then predict whether zero velocity at one instant proves the object reverses direction. Construct examples before reviewing the formulas: a constant nonzero velocity refutes the first claim, and a velocity that touches zero without changing sign can refute the second.

Experiment

Use the synthetic position model s(t)=t³−6t²+9t metres on 0≤t≤4 seconds. Differentiating gives v(t)=3(t−1)(t−3) metres per second and a(t)=6t−12 metres per second squared. Fill a sign chart rather than relying only on a position plot.

Time interval	Velocity sign	Acceleration sign	Speed behavior
0 to 1	positive	negative	decreasing
1 to 2	negative	negative	increasing
2 to 3	negative	positive	decreasing
3 to 4	positive	positive	increasing

At t=2, velocity is −3 while acceleration is zero. At t=1 and t=3, velocity changes sign and direction reverses. Sketch aligned graphs of s, v and a, marking the same times with vertical guide lines. The sign chart provides the interpretation of their shapes.

Understand

Velocity is the derivative of position with respect to time, acceleration is the derivative of velocity, and jerk is the derivative of acceleration. If position is measured in metres and time in seconds, the units are respectively metres/second, metres/second² and metres/second³. A second derivative is not the square of the first derivative; it means differentiate again.

Speed is the nonnegative quantity |v|. Whenever v is nonzero and differentiable, its derivative is sign(v) times a. Thus speed increases when velocity and acceleration have the same sign and decreases when their signs differ. Negative acceleration by itself means velocity decreases; whether speed decreases depends on the direction of motion.

At a zero of velocity, inspect nearby signs instead of applying a formula that assumes v≠0. For s(t)=t³, velocity is 3t², positive on both sides of zero. The particle pauses instantaneously but does not reverse direction. For s(t)=t², velocity 2t changes from negative to positive at zero, so direction reverses. The speed |2t| has a corner there, demonstrating that a smooth position function need not produce a differentiable speed at every instant.

For s(t)=t⁴/12, successive derivatives are v=t³/3, a=t² and j=2t. At t=2 these values are 8/3, 4 and 4, with different units. Equal numerical acceleration and jerk at a particular time do not make them the same physical quantity.

These formulas describe a stipulated mathematical trajectory. Differentiating a fitted trajectory is not proof that a real object followed it. Measurement noise, timing errors and model mismatch can affect estimated derivatives, especially higher ones. Keep exact symbolic results for the model separate from empirical claims about motion.

Apply

For every practice motion problem, label whether the requested answer is position, signed velocity, acceleration or speed. Show sign changes to justify direction reversal. Use the signed motion investigation to revisit the distinction between displacement and distance, while keeping integral accumulation separate from this derivative analysis.

Continue

The next lesson links several changing quantities through a constraint. Related-rate problems need the same careful units and sign interpretation, plus a model equation established before values are substituted.