Acceleration

From §2.2, we saw that instantaneous velocity can change over time. The rate of change of velocity is itself a fundamental quantity — we call it acceleration. It matters especially because, by Newton’s second law (F = ma), acceleration ties directly to force. So when we ask “what is changing the motion?”, the answer lives in acceleration.

Average acceleration

If an object’s velocity is v_1 at time t_1 and v_2 at t_2, the average acceleration over [t_1, t_2]:

$$
a_\text{avg} = \frac{\Delta v}{\Delta t} = \frac{v_2 – v_1}{t_2 – t_1}
$$

• Vector (signed). The sign gives the direction of velocity change
• SI unit: m/s²
• In 1D: a signed scalar (positive or negative)

Example: a train starts at v(0) = 0 and reaches v = 24 m/s at t = 40 s:

$$
a_\text{avg} = \frac{24 – 0}{40 – 0} = 0.6\ \mathrm{m/s^2}
$$

Meaning: on average, its speed increased by 0.6 m/s every second.

Instantaneous acceleration

Just as in §2.2 we passed from average velocity to instantaneous by shrinking Δt, here:

$$
a(t) = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt}
$$

Instantaneous acceleration is the derivative of velocity with respect to time. And since v = dx/dt:

$$
\boxed{a(t) = \frac{dv}{dt} = \frac{d^2 x}{dt^2}}
$$

Instantaneous acceleration is the second derivative of position with respect to time.

Example: for x(t) = 5t + 3t² − 0.1 t³ (from §2.2):

$$
v(t) = \frac{dx}{dt} = 5 + 6t – 0.3 t^2
$$

$$
a(t) = \frac{dv}{dt} = 6 – 0.6 t\ \mathrm{m/s^2}
$$

• at t = 0: a = +6 m/s² (positive acceleration — speeding up)
• at t = 10: a = 0 (instantaneous zero — velocity at maximum)
• at t = 20: a = −6 m/s² (negative acceleration — slowing down or reversing)

Four sign cases — driving at a traffic light

Let’s walk through a scenario that covers all sign combinations of v and a: a car at a traffic light. Positive axis points forward.

Stage 1 — start from rest (0 → 20 m/s in 10 s):
– v > 0 (moving forward), a > 0 (accelerating)
– Speed is increasing — v and a have the same sign

Stage 2 — brake to stop (20 → 0 m/s in 5 s):
– v > 0 (still moving forward), a < 0 (a = -4 m/s²)
– Speed is decreasing — v and a have opposite signs

Stage 3 — reverse gear (0 → −15 m/s in 6 s):
– v < 0 (moving backward), a < 0 (a = -2.5 m/s²)
– Speed is increasing — same signs (both negative)

Stage 4 — brake from reverse (−15 → 0 in 3 s):
– v < 0 (still moving backward), a > 0 (a = +5 m/s²)
– Speed is decreasing — opposite signs

Golden rule

It’s not the sign of a but whether v and a have the same sign that determines how speed changes:

sign of v	sign of a	what happens?
+	+	speeding up (positive direction)
+	−	slowing down (still positive direction)
−	+	slowing down (still negative direction)
−	−	speeding up (negative direction)

Common mistake: thinking “negative acceleration” means “braking”. No — negative acceleration just means “acceleration in the negative direction of the axis”. If the object is moving in the negative direction, that acceleration is speeding it up, not slowing it down.

Graphical interpretation

On a v-t plot:
– Average acceleration = slope of the chord between two points
– Instantaneous acceleration = slope of the tangent at that point

On an x-t plot:
– Instantaneous acceleration = curvature (concavity) at that point
– Positive acceleration → curve is concave up
– Negative acceleration → curve is concave down
– Zero acceleration → straight line (constant slope = constant velocity)

“Deceleration”?

Sometimes you hear “deceleration”. In physics this word is ambiguous and best avoided. Colloquially it means “slowing down”, but that is not the same as negative acceleration:

• Slowing down means: v and a have opposite signs (regardless of which is which)
• Negative acceleration means: a < 0 (regardless of v)

In the car scenario above: stage 3 (reversing) has negative acceleration but is speeding up — so it isn’t “deceleration” despite a < 0.

Just stick with the sign of a and the same-sign rule for v — it’s clearer.

Acceleration and mass: preview of Newton’s second law

Why did we pick acceleration as the fundamental kinematic quantity — not, say, the third derivative of position (called “jerk”)?

The answer: Newton. His second law says:

$$
\vec F = m \vec a
$$

Force ties directly to acceleration. So whenever you know any force (gravity, friction, spring, …), you immediately have acceleration; integrate for velocity; integrate again for position. All of dynamics starts from this equation.

So a is not just an abstract quantity — it is the bridge between kinematics and dynamics. Whenever anything changes an object’s acceleration, the next question is always: “what force did that?”

A few notes and common mistakes

1. a = 0 means constant velocity, not zero velocity.
If a = 0, the car could be moving at 50 m/s — its speed just isn’t changing. v = 0 and a = 0 are two different things.

2. Instantaneous acceleration can jump unexpectedly.
If x(t) has a jerk (like a sudden kick), v(t) can be discontinuous — and a(t) can be infinite. In the real world this can’t happen (since F = ma requires infinite force), but in idealized problems it’s sometimes modeled as a delta function.

3. mgh and F = ma must give consistent results.
Free fall of an object near Earth: force F = -mg (negative, downward), so a = -g = -9.8 m/s². Independent of mass! This is a direct consequence of F = ma — when F ∝ m (as in gravity), m cancels. §2.4 and §2.5 build on this.

4. Units.
a in SI: m/s². In everyday life: g (surface gravity, 1 g ≈ 9.8 m/s²). A sports car reaching 0-100 km/h in 3 s:

$$
a = \frac{100/3.6 – 0}{3} \approx 9.26\ \mathrm{m/s^2} \approx 0.94\ g
$$

That is about 0.9× gravity. Military manned rides sometimes hit 9g or more.

5. Second derivative of a polynomial, drop two degrees.
If x(t) = t^n, then v = n t^(n-1), a = n(n-1) t^(n-2). For n = 2: a = 2 (constant). This special case of “constant acceleration” is central in §2.4.

What you should be able to do

After this section, you should be able to:

• Compute average acceleration as Δv/Δt from two velocity values
• Take instantaneous acceleration as dv/dt (or d²x/dt²) from x(t) or v(t)
• Interpret the 4 sign combinations (v±, a±) — decide whether speed is increasing or decreasing
• Read slope of v-t as acceleration and curvature of x-t as acceleration
• Explain why the word “deceleration” is ambiguous and best avoided
• Recognize acceleration’s role in Newton’s second law (F = ma) and know why acceleration is the bridge between kinematics and dynamics
• Convert between m/s² and g (gravity)

Preview of §2.4

Most important physics problems involve constant acceleration — free fall, motion on a frictionless incline, horizontal projectile launch, a car with uniform braking. For that case, a = const, and we can integrate to get v(t) and x(t):

$$
v(t) = v_0 + a t
$$

$$
x(t) = x_0 + v_0 t + \frac{1}{2} a t^2
$$

From these come the kinematic equations used in every practical problem. That is §2.4.

📚 See also: Halliday Vol 1, Ch 2, §2.3 — Acceleration.
📖 Open reference: OpenStax University Physics Vol 1 — Chapter 3.3: Average and Instantaneous Acceleration.
📖 Feynman Lectures Vol I — Ch 8: Motion §8-3 (a beautiful discussion of “why acceleration, not velocity, is what matters”).