Straight-Line Motion: Position, Velocity, and Acceleration
Position, velocity, and acceleration are linked by differentiation: velocity is the derivative of position, and acceleration is the derivative of velocity (second derivative of position).
The derivative chain of motion
If s(t) is position, then v(t) = s'(t) is velocity, and a(t) = v'(t) = s''(t) is acceleration. Speed is |v(t)|, the magnitude of velocity (always non-negative).
An object speeds up when velocity and acceleration have the SAME sign, and slows down when they have OPPOSITE signs — this is a common exam distinction from just looking at the sign of acceleration alone.
The object is momentarily at rest when v(t) = 0. It changes direction at a time where v(t) = 0 AND v changes sign around that time (not just touches zero).
Worked Example
s(t) = t³ - 6t² + 9t. Find when the object is at rest and determine if it's speeding up or slowing down at t=1.
- v(t) = s'(t) = 3t² - 12t + 9 = 3(t²-4t+3) = 3(t-1)(t-3). At rest when v(t)=0: t=1 or t=3.
- a(t) = v'(t) = 6t - 12.
- At t=1: v(1)=0 (at rest, not moving — so 'speeding up/slowing down' is really about the instant just after), a(1) = 6(1)-12 = -6 (negative).
- Just after t=1, check sign of v just past 1: at t=1.1, v=3(0.1)(-1.9)<0, so velocity becomes negative while acceleration is negative — same sign, so the object speeds up (in the negative direction) just after t=1.
Answer: At rest at t=1 and t=3; just after t=1 the object speeds up (v and a share the same sign).
Key terms (3)
- Velocity
- v(t) = s'(t), the rate of change of position; can be positive or negative depending on direction.
- Speed
- |v(t)|, the magnitude of velocity — never negative.
- Speeding up
- Occurs when velocity and acceleration have the same sign.
Practice Quiz
If v(t) = -4 and a(t) = -2 at some time t, is the object speeding up or slowing down?