Calculus was invented to describe motion. With Higher Derivatives and the rate language of Leibniz Notation and Units, position, velocity and acceleration become one function and its first two derivatives.
Key idea
For an object moving along a line with position $s(t)$: velocity $v(t) = s'(t)$, acceleration $a(t) = v'(t) = s''(t)$, and speed $= \lvert v(t) \rvert$. The object is at rest when $v = 0$, moves in the positive direction when $v > 0$, and speeds up when $v$ and $a$ have the same sign.
Why it works
Velocity is the rate of change of position, so it carries a sign: positive for forward, negative for backward. Speed is how fast, without direction. Acceleration is the rate of change of velocity. If $v$ and $a$ are both negative, the velocity is becoming more negative, so the object moves backward faster and faster: its speed grows. Speeding up means the size of $v$ grows, and that happens exactly when $a$ pushes in the same direction as $v$.
Worked example: A full motion analysis
A particle has position $s(t) = t^3 - 6t^2 + 9t$ (meters, $t \ge 0$ in seconds). When is it at rest, when does it move backward, and when is it speeding up?
- $v(t) = 3t^2 - 12t + 9 = 3(t - 1)(t - 3)$. At rest when $t = 1$ or $t = 3$.
- Sign of $v$: positive on $[0, 1)$, negative on $(1, 3)$, positive after 3. So it moves backward on $(1, 3)$.
- $a(t) = 6t - 12$: negative before $t = 2$, positive after.
- Same signs: on $(1, 2)$ both are negative, and on $(3, \infty)$ both are positive.
Answer: At rest at $t = 1$ and $t = 3$; backward on $(1, 3)$; speeding up on $(1, 2)$ and $(3, \infty)$.
Worked example: Thrown straight up
A ball's height is $s(t) = -16t^2 + 64t + 5$ feet. When is it highest, and how fast is it going when it passes 53 feet on the way down?
- $v(t) = -32t + 64$, which is 0 at $t = 2$: the top of the flight, at $s(2) = 69$ ft.
- $-16t^2 + 64t + 5 = 53$ gives $t^2 - 4t + 3 = 0$, so $t = 1$ (going up) or $t = 3$ (coming down).
- $v(3) = -96 + 64 = -32$ ft/s.
Answer: Highest at $t = 2$ s; at $t = 3$ s its velocity is $-32$ ft/s (speed 32 ft/s, falling).
Common mistake
Thinking negative acceleration always means slowing down. In the first example, on $(1, 2)$ the acceleration is negative and the particle speeds up, because it is already moving in the negative direction. Check by comparing signs: same signs, speeding up; opposite signs, slowing down.
Total distance traveled is not always $\lvert s(\text{end}) - s(\text{start}) \rvert$: the object may turn around. Split the trip at the times when $v = 0$. We return to this in Net Change and Total Distance.