Motion Along a Line

You’ll be able to

Relate position, velocity, and acceleration through differentiation
Determine when a particle is speeding up or slowing down
Find total distance traveled versus displacement

Position, velocity, acceleration

For a particle moving along a line with position x(t), the velocity is the derivative v(t) = x′(t) and the acceleration is a(t) = v′(t) = x″(t). Velocity is a signed quantity: its sign tells direction and its magnitude is speed. The particle is momentarily at rest when v(t) = 0, and it changes direction exactly when v(t) changes sign (crosses zero, not merely touches it).

Speeding up vs. slowing down

A particle speeds up when velocity and acceleration have the same sign, and slows down when they have opposite signs. The intuition: acceleration in the direction of motion adds speed; acceleration against the motion removes it. Speed itself is |v(t)|. So do not confuse "negative acceleration" with "slowing down" — a particle moving in the negative direction (v < 0) with a < 0 is actually speeding up.

The motion chain and distance

v(t) = x′(t) · a(t) = v′(t) · total distance = ∫ₐᵇ |v(t)| dt

Displacement is ∫v dt (signed); total distance is the integral of |v| (unsigned), splitting at direction changes.

Displacement vs. total distance

Displacement over [a, b] is the net change in position, x(b) − x(a), equal to ∫ₐᵇ v(t) dt — it can be zero even after long travel. Total distance is how far the particle actually moves, ∫ₐᵇ |v(t)| dt. To compute it you find where v = 0, split the interval at those times, integrate v on each piece, and add the absolute values. These two questions are distinct, and the exam tests whether you know which is being asked.

Worked example

A particle has velocity v(t) = t² − 4t + 3 for t ≥ 0. At t = 4, is it speeding up or slowing down?

1.Find velocity at t = 4: v(4) = 16 − 16 + 3 = 3, so v is positive (moving in the positive direction).
2.Find acceleration a(t) = v′(t) = 2t − 4, so a(4) = 8 − 4 = 4, which is positive.
3.Compare the signs: velocity is positive and acceleration is positive — the same sign.
4.Same-sign velocity and acceleration means the particle is gaining speed.

Answer: At t = 4 the particle is speeding up, since v(4) = 3 and a(4) = 4 share the same (positive) sign, so its speed is increasing.

Watch out

A particle changes direction only where v(t) changes sign, not merely where v = 0. If v touches zero and keeps the same sign afterward (a double root), the particle pauses but does not turn around.

Checkpoint

At a certain instant a particle has velocity v = −3 and acceleration a = −2. Is it speeding up or slowing down?

Checkpoint

A particle moves with v(t) = 2t − 6. Over the interval [0, 5], the displacement and total distance are:

On the exam

Read motion questions carefully: "how far from start" (displacement, signed) and "total distance traveled" (unsigned) are different. For total distance, always locate where v = 0, split the interval, and sum the absolute values of the pieces.

Answer the 2 checkpoints as you read.