Acceleration and Its Sign defined acceleration as the rate at which velocity changes. With Velocity as a Derivative in hand, that rate is one more derivative.
Key idea
$$a(t) = \dfrac{dv}{dt} = \dfrac{d^2x}{dt^2}.$$ Acceleration is the slope of the $v$–$t$ graph and the concavity of the $x$–$t$ graph. Its units are $(\text{m/s})/\text{s} = \text{m/s}^2$. The velocity at an instant and the acceleration at that instant are separate facts: either can be zero while the other is not.
Why it works
The derivative at an instant depends on how a quantity is changing there, not on its value there. A velocity can pass through zero while it is changing quickly, like a ball at the top of its flight: just before the top its velocity is a little upward, just after it is a little downward. The change from $+$ to $-$ happens at a steady rate of $9.8\ \text{m/s}$ every second, so $a = -9.8\ \text{m/s}^2$ (up positive) right through the top.
Worked example: Differentiate twice
A particle has $x(t) = t^3 - 6t^2 + 5t$ (m, s). Find its velocity and acceleration at $t = 1\ \text{s}$.
- $v(t) = 3t^2 - 12t + 5$, so $v(1) = 3 - 12 + 5 = -4\ \text{m/s}$.
- $a(t) = \dfrac{dv}{dt} = 6t - 12$, so $a(1) = -6\ \text{m/s}^2$.
- Both are negative: the particle moves in $-x$ and its velocity is becoming more negative, so it is speeding up.
Answer: $v = -4\ \text{m/s}$, $a = -6\ \text{m/s}^2$ (speeding up in the $-x$ direction).
Worked example: At rest, but accelerating
For $x(t) = 2t^3 - 9t^2 + 12t$ (m, s), the particle is at rest at $t = 1\ \text{s}$. What is its acceleration then?
- $v(t) = 6t^2 - 18t + 12 = 6(t - 1)(t - 2)$, which is zero at $t = 1$.
- $a(t) = 12t - 18$, so $a(1) = -6\ \text{m/s}^2$.
- The velocity is passing from positive to negative: the particle is turning around, and the nonzero acceleration is what turns it.
Answer: $a(1) = -6\ \text{m/s}^2$.
Common mistake
Thinking that zero velocity at an instant means zero acceleration. If the acceleration were also zero, the velocity would stay at zero and the object would sit still. A ball at the top of its throw doesn't hover. Check: look at the velocity a moment before and a moment after. If they differ, the acceleration is not zero.
| $v$ | $a$ | What happens |
|---|---|---|
| $+$ | $+$ | moving in $+x$, speeding up |
| $+$ | $-$ | moving in $+x$, slowing down |
| $-$ | $-$ | moving in $-x$, speeding up |
| $-$ | $+$ | moving in $-x$, slowing down |
| $0$ | $\neq 0$ | turning around (or starting from rest) |