Motion Along a Line

You’ll be able to

Relate position, velocity, and acceleration through derivatives
Determine when a particle moves right, left, or is at rest, and when it speeds up or slows down
Distinguish displacement from total distance traveled

Position, velocity, acceleration

For a particle on a line with position s(t), velocity is v(t) = s′(t) and acceleration is a(t) = v′(t) = s″(t). The sign of velocity gives direction: v > 0 moves in the positive direction, v < 0 in the negative direction, and v = 0 means momentarily at rest — the candidates for turning around. Speed is |v(t)|, the magnitude regardless of direction.

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. This is subtle: a negative acceleration does not mean slowing down — if the particle is already moving in the negative direction (v < 0), a negative acceleration speeds it up. Compare the two signs, not the sign of acceleration alone.

Displacement vs. total distance

Displacement over [a, b] is the net change in position, s(b) − s(a) — it can be zero even after lots of movement. Total distance adds up how far the particle actually traveled, treating each direction as positive. To get total distance you must split the interval at every time the particle changes direction (where v = 0 and switches sign) and sum the absolute changes in position over each piece.

Motion relationships

v(t) = s′(t) · a(t) = v′(t) · speed = |v(t)|

Speeding up ⇔ v and a share a sign. Direction changes where v = 0 and switches sign.

Worked example

A particle has velocity v(t) = t² − 4t + 3 for t ≥ 0. When is it moving left, and is it speeding up or slowing down at t = 1?

1.Factor the velocity: v(t) = (t − 1)(t − 3), which is zero at t = 1 and t = 3.
2.Test signs: v < 0 between the roots (1 < t < 3), so the particle moves left on (1, 3).
3.Find acceleration: a(t) = v′(t) = 2t − 4, so a(1) = 2(1) − 4 = −2.
4.At t = 1, v is transitioning through 0; just after t = 1 velocity is negative and a(1) = −2 is negative — same sign — so it is speeding up.

Answer: The particle moves left on 1 < t < 3. At t = 1 it is momentarily at rest and beginning to speed up in the negative direction, since v and a are both heading negative.

Watch out

Displacement and total distance are different questions. If a particle goes out and comes back, its displacement can be 0 while its total distance is large. Only split at direction changes when the problem asks for total distance traveled.

Checkpoint

A particle has velocity v(t) = 2t − 6. At what time is it momentarily at rest?

Checkpoint

At an instant a particle has velocity v = −4 and acceleration a = −2. Is it speeding up or slowing down?

On the exam

For "speeding up or slowing down", always compare the signs of v and a at that instant. Same sign ⇒ speeding up; opposite signs ⇒ slowing down. The sign of acceleration by itself tells you nothing about speed.

Answer the 2 checkpoints as you read.