Most students memorise four equations and spend the rest of the chapter guessing which one to use. There is only one equation. The other three are consequences. Here they are, derived once, so you never need to guess again.

Uniform Acceleration: The Four Equations Derived Once

A ball is dropped from rest. A car brakes to a stop. A rocket climbs with its engine at full thrust.

In each case, the acceleration is approximately constant — the same value at every instant of the motion. This is a special case, not the general rule. Most real accelerations vary with time. But this special case appears so frequently in problems, and in real situations approximately, that it deserves its own treatment.

The question is: if you know the acceleration is constant, what can you say about velocity and position?

The answer is: everything. Five quantities — initial velocity, final velocity, displacement, time, acceleration — are linked by a set of equations. If you know three, you can find the remaining two. But here is the thing: most students encounter these as four separate equations to memorise, and spend the rest of the chapter wondering which one to use.

There is no list to memorise. There is one starting point, and two integrations. The four equations fall out automatically.

Starting from a = constant

From K01, acceleration is the derivative of velocity:

$a = \frac{dv}{dt}$

If a is constant, this is the simplest possible differential equation. Rearrange:

$dv = a \, dt$

Integrate both sides. On the left, velocity changes from u (initial) to v (final). On the right, time runs from 0 to t:

$\int_u^v dv = \int_0^t a \, dt$

$v - u = at$

$v = u + at$

This is the first equation. It says: start with velocity u, add the accumulated change in velocity (a × t), and you get the final velocity v. Nothing more than the definition of constant acceleration, integrated once.

Integrate again

Velocity is the derivative of position:

$v = \frac{dx}{dt}$

Substitute equation (1):

$\frac{dx}{dt} = u + at$

Integrate both sides, with position going from x₀ to x and time from 0 to t. Let displacement s = x − x₀:

$\int_0^s ds = \int_0^t (u + at) \, dt$

$s = ut + \frac{1}{2}at^2$

The first term, ut, is how far the object would have gone if it never accelerated. The second term, ½at², is the extra distance accumulated by the acceleration. At t = 0, both terms are zero — correct. As t grows, the ½at² term dominates — the quadratic wins over the linear.

Eliminate time

Equations (1) and (2) both contain t. Eliminate it to get a relation between v, u, a, and s that does not involve time.

From (1): t = (v − u)/a. Substitute into (2):

$s = u \cdot \frac{v-u}{a} + \frac{1}{2}a \cdot \left(\frac{v-u}{a}\right)^2$

$2as = 2u(v-u) + (v-u)^2$

$2as = 2uv - 2u^2 + v^2 - 2uv + u^2$

$2as = v^2 - u^2$

$v^2 = u^2 + 2as$

This equation contains no t. Use it whenever time is neither given nor asked for.

The fourth equation

Average velocity under constant acceleration is exactly the mean of initial and final velocities — a fact that holds precisely only because a is constant:

$\bar{v} = \frac{u + v}{2}$

Displacement is average velocity multiplied by time:

$s = \bar{v} \cdot t = \frac{(u+v)}{2} \cdot t$

$s = \frac{(u+v)}{2} \cdot t$

This equation has no a in it. Use it when acceleration is neither given nor asked for.

The five variables, the four equations

The five quantities in uniform acceleration problems are:

u — initial velocity, v — final velocity, a — acceleration, s — displacement, t — time

Each of the four equations links four of these five. One variable is absent from each:

Equation	Missing variable
v = u + at	s
s = ut + ½at²	v
v² = u² + 2as	t
s = ½(u+v)t	a

This is the key to never guessing. Identify which variable is absent from the problem — neither given nor asked for — and that equation is the one to use. The selection is mechanical once you see the structure.

Note: 1. Assign a positive direction. Declare it at the start. Stick to it. 2. List what is given. Write u, v, a, s, t. Fill in what you know, including sign. 3. Identify the unknown. What is the problem asking for? 4. Identify the absent variable. Which of the five is neither given nor asked for? 5. Pick the equation that does not contain the absent variable. 6. Solve. One equation, one unknown — it is always algebra from here. 7. Check the sign of your answer. A negative displacement means the object moved in the negative direction. A negative velocity means it is moving opposite to your positive direction. These are results, not errors.

Warning: Using v = u + at when v is not the final velocity at time t. These equations describe the object's state at one moment (t = 0) and another moment (t). If the problem gives you velocity at some intermediate moment, re-define your initial conditions.

> > **Forgetting that s is displacement, not distance.** If the object reverses direction during the interval, s can be zero or negative even though the object covered real ground. For total distance, split the interval at the point of reversal (where v = 0) and add separately. > > **Treating u as always zero.** "Starts from rest" means u = 0. "Starts moving" does not. Read carefully. > > **Sign errors on a.** Gravity is approximately −10 m/s² if upward is positive. If you set up the problem with upward positive and then write a = +10, every sign in the answer will be wrong. > > **Using these equations when acceleration is not constant.** These four equations are valid only for uniform (constant) acceleration. For variable acceleration, return to calculus — the equations of K01.

A worked example

A car is moving at 20 m/s when the driver applies brakes, giving a constant deceleration of 4 m/s². Find: (a) the velocity after 3 seconds, (b) the distance covered before stopping, (c) the time to stop.

Setup. Positive direction: forward (direction of initial motion). u = +20 m/s, a = −4 m/s² (braking opposes motion).

Part (a) — velocity after 3 s.

Absent variable: s. Use v = u + at.

$v = 20 + (-4)(3) = 20 - 12 = +8 \text{ m/s}$

After 3 seconds, the car is still moving forward at 8 m/s.

Part (b) — distance to stop.

At the moment of stopping, v = 0. Absent variable: t. Use v² = u² + 2as.

$0 = 20^2 + 2(-4)s \implies 8s = 400 \implies s = 50 \text{ m}$

The car covers 50 m before stopping.

Part (c) — time to stop.

v = 0, u = 20, a = −4. Absent variable: s. Use v = u + at.

$0 = 20 + (-4)t \implies t = 5 \text{ s}$

Twist variant

Same car, same braking. The driver applies brakes 10 m before a wall. Does the car stop in time?

From part (b), the stopping distance is 50 m. The wall is at 10 m. The car does not stop — it hits the wall.

How fast does it hit? Use v² = u² + 2as with s = 10 m:

$v^2 = 400 + 2(-4)(10) = 400 - 80 = 320$

$v = \sqrt{320} \approx 17.9 \text{ m/s}$

The car hits the wall at nearly 18 m/s — having barely slowed. This is why stopping distance matters: the relationship between speed and stopping distance is quadratic, not linear. Double the speed, quadruple the stopping distance.

The nth second

The displacement covered in the nth second alone — not in the first n seconds, in the nth second only.

Displacement in first n seconds: $s_n = un + \frac{1}{2}an^2$

Displacement in first (n−1) seconds: $s_{n-1} = u(n-1) + \frac{1}{2}a(n-1)^2$

Subtract:

$s_{\text{nth}} = u + \frac{a}{2}(2n-1)$

This is not a fifth independent equation — it is derived from (2). But it appears often enough in JEE problems that you should recognise it immediately. The displacement in successive seconds increases by a fixed amount a — an arithmetic progression with common difference a. This is a direct consequence of constant acceleration.

What to remember

Constant acceleration means one integration gives v = u + at, a second gives s = ut + ½at². The other two equations are algebraic rearrangements of these two.

Five variables. Four equations. Each equation is missing one variable. Identify the absent variable, pick the matching equation.

s is displacement — signed, direction-dependent. Total distance may differ if the object reverses.

These equations break down the moment acceleration varies. For variable acceleration: return to derivatives and integrals.

K03 takes the framework built here and applies it to two simultaneous constant-acceleration motions at right angles — projectile motion. The key idea there is that horizontal and vertical motions are completely independent of each other, and can be solved separately using these same equations.