AP Calculus AB and BC

Velocity vs Acceleration

Q: Is acceleration just the second derivative?

Yes, when the function is position. Since a(t) = s''(t), the sign of acceleration is the concavity of the position graph.

Velocity is the first derivative of position and acceleration is the second, so acceleration measures how the velocity is changing, not how fast the particle is going. When velocity and acceleration share a sign the particle is speeding up; when their signs differ it is slowing down.

Velocity

Use when: The question asks for the signed rate of change of position, which direction the particle is moving, when it is at rest, or when it changes direction.

Acceleration

Use when: The question asks how the velocity itself is changing, or asks whether the particle is speeding up, which you settle by comparing the sign of $a(t)$ with the sign of $v(t)$ .

Side by side

	Velocity	Acceleration
What it measures	The rate of change of position	The rate of change of velocity
Definition	$v(t) = s'(t)$	$a(t) = v'(t) = s''(t)$
When it is zero	The particle is momentarily at rest	The velocity is momentarily not changing
Its sign alone tells you	The direction of motion	Nothing about speeding up or slowing down
Its definite integral gives	Displacement, $s(b) - s(a)$	Change in velocity, $v(b) - v(a)$
Common trap	Reading a negative value as slow rather than backward	Reading a negative value as slowing down

Position, velocity, and acceleration are one chain of derivatives. Differentiating $s(t)$ gives $v(t)$ , and differentiating again gives $a(t)$ . Running the chain backwards, $\int a(t)\,dt$ recovers velocity and $\int v(t)\,dt$ recovers position, each up to a constant that an initial condition supplies.

Speeding up is a question about agreement of signs, not about the sign of acceleration. A particle with $v(3) = -6$ and $a(3) = -2$ is speeding up, because it is moving in the negative direction and gaining speed. The same particle with $a(3) = 2$ is slowing down even though the acceleration is positive.

At rest does not mean no acceleration

Solving $v(t) = 0$ finds the moments the particle is at rest, and the acceleration at those moments is usually not zero. A ball at the top of its flight has velocity $0$ and acceleration $-32$ feet per second squared, which is exactly why it does not stay there.

Frequently asked questions

Does negative acceleration mean the particle is slowing down?

Only when the velocity is positive. If the velocity is also negative, the particle is speeding up in the negative direction.

What does zero acceleration mean?

The velocity is not changing at that instant. The particle can still be moving quickly; constant velocity gives zero acceleration at every time.

Is acceleration just the second derivative?

Yes, when the function is position. Since $a(t) = s''(t)$ , the sign of acceleration is the concavity of the position graph.

In the CED: Unit 4: Contextual Applications