Constant acceleration and the kinematic equations

The most important special case in kinematics is constant acceleration (a = const). This case governs three crucial phenomena:

In each of these, a is a constant, and we can derive simple equations for v(t) and x(t) used throughout practical problems. These are the kinematic equations — the heart of Chapter 2.

Deriving them from the definitions

📖 We’ll use the definite integral in this section. If it isn’t yet solid, see The definite integral for physicists.

Step by step. Starting from a = dv/dt with a = const:

$$
\int_{v_0}^{v} dv’ = \int_0^t a\, dt’ = a t
$$

$$
\boxed{v(t) = v_0 + a t \quad \text{(equation 1)}}
$$

where v_0 = v(0) is the initial velocity. Velocity is linear in time.

From v = dx/dt:

$$
\int_{x_0}^{x} dx’ = \int_0^t v(t’)\, dt’ = \int_0^t (v_0 + at’)\, dt’ = v_0 t + \frac{1}{2} a t^2
$$

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

Position is quadratic in time. On an x-t plot, this is a parabola.

Derived equations — without t

Sometimes time is not in the problem (“a car stops in 100 m — what was its initial speed?”). Eliminate t from equations 1 and 2:

From equation 1: t = (v - v_0)/a. Substitute into 2 and simplify:

$$
\boxed{v^2 = v_0^2 + 2 a (x – x_0) \quad \text{(equation 3)}}
$$

Velocity as a function of position, independent of time. Very useful.

The fourth equation — without a

Sometimes acceleration is unknown. If initial velocity, final velocity, and time are known:

$$
\boxed{x – x_0 = \frac{v_0 + v}{2} \cdot t \quad \text{(equation 4)}}
$$

where (v_0 + v)/2 is the average velocity under constant acceleration (since velocity varies linearly, its average is the midpoint).

Four equations — the working table

# equation missing
1 v = v_0 + a t x
2 x - x_0 = v_0 t + ½ a t² v
3 v² = v_0² + 2a(x - x_0) t
4 x - x_0 = ½(v_0 + v) t a

Problem-solving strategy:
1. List the known quantities: of {x_0, v_0, v, a, t}, usually 3 are given
2. Identify the target
3. Pick the equation that does not contain the missing quantity — one equation with 4 variables (3 known + 1 target)
4. Solve

Complete example — a car in emergency braking

Scenario: a car moving at 72 km/h = 20 m/s. The driver hits the brakes and the car decelerates at a constant a = -5 m/s² (negative because braking).

Question 1: how long until it stops?

Given: v_0 = 20, v = 0, a = -5. Target: t. Missing: x. → equation 1.

$$
0 = 20 + (-5) t \Rightarrow t = 4\ \mathrm{s}
$$

Question 2: what is the stopping distance?

Given: v_0 = 20, v = 0, a = -5. Target: x - x_0. Missing: t. → equation 3.

$$
0 = 20^2 + 2 \cdot (-5) \cdot (x – x_0) \Rightarrow x – x_0 = \frac{400}{10} = 40\ \mathrm{m}
$$

Cross-check with equation 2: with t = 4:

$$
x – x_0 = 20 \cdot 4 + \frac{1}{2} \cdot (-5) \cdot 16 = 80 – 40 = 40\ \mathrm{m}\ ✓
$$

Question 3: at t = 2 s, what are velocity and position?

Equation 1: v = 20 - 5 \cdot 2 = 10 m/s.
Equation 2: x - x_0 = 20 \cdot 2 - ½ \cdot 5 \cdot 4 = 40 - 10 = 30 m.

So at t = 2, the car has traveled 3/4 of the stopping distance, but still has half its initial speed.

Why does stopping distance scale with velocity squared?

From equation 3 with v = 0:

$$
x_\text{stop} = \frac{v_0^2}{2|a|}
$$

Key point: stopping distance scales with the square of initial velocity, not linearly. Meaning:

Driving safety: if you can stop in 10 m at 50 km/h, at 100 km/h (just twice as fast) you need 40 m (four times). This is a fact that inexperienced drivers rarely have good intuition for.

When do the kinematic equations fail?

Only when a = const. Common cases where it breaks:

In these cases you go back to the primary definitions: a(t) = dv/dt, v(t) = dx/dt, and integrate carefully. §2.6 handles that case.

A few notes and common mistakes

1. Keep the sign of a consistent.
If the positive axis points right and you’re braking, a < 0. Don’t try to “make sense” of the sign at the same time — let the signs do their work.

2. If a = 0, the equations still hold (constant zero acceleration).
v = v_0 constant, x = x_0 + v_0 t linear. The special case of “uniform motion”.

3. Can time be negative?
In formulas, t is usually taken with t ≥ 0 (starting from time zero). But if you want to project backward to before t = 0, t < 0 works too — as long as a was really constant over that extended interval.

4. Equation 3 can give two solutions.
v² = v_0² + 2a Δx gives v = ±√(...). Pick the sign from the physics — which way is it moving?

5. Distance ≠ displacement Δx when the direction reverses.
If the object reverses before stopping (like a ball thrown up and returning), the equations give Δx — not total path length. For distance, split into segments.

What you should be able to do

After this section, you should be able to:

Preview of §2.5

The most important example of constant acceleration in physics: free fall. Any object near Earth’s surface, subject only to gravity (no air resistance), has acceleration g ≈ 9.8 m/s² downward — independent of mass, shape, or material. Everything falls with the same acceleration. This was Galileo’s discovery (see the gravity Article 1), and it becomes concrete via §2.4.

📚 See also: Halliday Vol 1, Ch 2, §2.4 — Constant Acceleration.
📖 Open reference: OpenStax University Physics Vol 1 — Chapter 3.4: Motion with Constant Acceleration.

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