In short

When acceleration is constant, three equations connect initial velocity u, final velocity v, acceleration a, displacement s, and time t: (1) v = u + at, (2) s = ut + \tfrac{1}{2}at^2, (3) v^2 = u^2 + 2as. A fourth relation, s = \tfrac{1}{2}(u + v)t, comes directly from the area of a trapezium on the v–t graph. These four equations solve every constant-acceleration problem — identify three knowns, pick the equation that contains the fourth unknown, and solve.

A car cruises at 72 km/h on NH-44 when the driver spots a cow standing in the middle of the road. The brakes slam. The car decelerates steadily — losing the same amount of speed every second — and grinds to a halt 50 metres later. How much time did the driver have? Was it enough?

This is the kind of problem where uniformly accelerated motion earns its keep. The acceleration is constant (the brakes push with a steady force), and three quantities — initial speed, final speed, and distance — are known. From those three numbers, you can extract everything else: the deceleration, the stopping time, the speed at every instant, the position at every instant. You do not need calculus for this (though you will see where the calculus hides). You just need three equations.

These are the kinematic equations — sometimes called the SUVAT equations, after the five variables they connect: s (displacement), u (initial velocity), v (final velocity), a (acceleration), and t (time). They are the single most-used set of equations in introductory physics, and they all come from one assumption: acceleration is constant.

What "uniformly accelerated" means

When you say an object has uniform acceleration, you mean the acceleration does not change with time. It is the same at t = 1 s as it is at t = 5 s or t = 100 s. Mathematically:

a = \text{constant}

This is a strong assumption. In real life, few things accelerate perfectly uniformly — a car's engine pushes harder at low revs, air resistance grows with speed, a rocket gets lighter as it burns fuel. But many situations come close enough that treating a as constant gives you the right answer to within a few percent:

For all of these, the kinematic equations give you clean, exact answers.

The velocity-time picture

Before touching algebra, look at what constant acceleration means on a v–t graph. If acceleration is constant, velocity changes at a uniform rate — and on a v–t graph, that is a straight line.

Velocity vs time for an object starting at $u = 5$ m/s with constant acceleration $a = 4$ m/s$^2$. The straight line is the signature of uniform acceleration — the slope is the acceleration, and the area under the line is the displacement.

Two facts live inside this graph:

  1. The slope of the line is the acceleration. A steeper line means faster acceleration. A horizontal line means zero acceleration (constant velocity). A line sloping downward means deceleration.

  2. The area under the line is the displacement. The region between the line and the time axis, from t = 0 to some time t, gives you how far the object has travelled. This is the geometric insight that produces all three kinematic equations.

Deriving the first equation: v = u + at

Start from the definition of acceleration. Acceleration is the rate of change of velocity:

a = \frac{dv}{dt}

Why: this is the definition. Acceleration tells you how many metres per second the velocity changes each second.

Since a is constant, you can integrate both sides with respect to time. Rearrange first:

dv = a\,dt

Why: multiplying both sides by dt separates the variables — velocity on the left, time on the right. This sets up the integration.

Integrate from the initial moment (t = 0, when v = u) to a general time t (when the velocity is v):

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

Why: the left side adds up all the tiny changes in velocity from u to v. The right side adds up all the tiny increments a\,dt over the time interval. Since a is constant, it comes out of the integral.

v - u = a \cdot t
\boxed{v = u + at} \tag{1}

Why: the integral of dv from u to v is simply v - u. The integral of a\,dt from 0 to t is at (since a is constant). Rearranging gives the first kinematic equation: the velocity at any time t equals the initial velocity plus the velocity gained through acceleration.

Read equation (1) out loud: the final velocity equals the initial velocity plus the acceleration multiplied by time. Every second, the velocity changes by a metres per second. After t seconds, the total change is at. Add that to what you started with, and you have the current velocity.

Deriving the second equation: s = ut + \tfrac{1}{2}at^2

Velocity is the rate of change of displacement:

v = \frac{ds}{dt}

Why: this is the definition. Velocity tells you how many metres the position changes each second.

You already know from equation (1) that v = u + at. Substitute:

\frac{ds}{dt} = u + at

Why: replacing v with the expression from equation (1) gives you a differential equation relating displacement and time — one you can solve by integration.

Rearrange and integrate from t = 0 (when s = 0) to time t (when displacement is s):

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

Why: the left side accumulates displacement. The right side sums up v\,dt — velocity times a tiny time interval — which is the tiny distance covered in each instant.

s = \int_0^t u\,dt + \int_0^t at\,dt
s = u \cdot t + a \cdot \frac{t^2}{2}
\boxed{s = ut + \tfrac{1}{2}at^2} \tag{2}

Why: the integral of a constant u over time t is ut — the distance you would cover at the initial velocity alone. The integral of at gives \tfrac{1}{2}at^2 — the extra distance due to acceleration. The \tfrac{1}{2} appears because the velocity grows linearly, so the average contribution of the acceleration term over the interval is half the final contribution.

Here is another way to see the \tfrac{1}{2}. On the v–t graph, the displacement is the area under the velocity line. That area is a trapezium (or a triangle sitting on top of a rectangle). The rectangle has height u and width t, giving area ut. The triangle has base t and height at (the velocity gained), giving area \tfrac{1}{2} \times t \times at = \tfrac{1}{2}at^2. Add them: s = ut + \tfrac{1}{2}at^2.

Deriving the third equation: v^2 = u^2 + 2as

This equation eliminates time. It connects velocity directly to displacement, which is exactly what you need for the cow-on-the-highway problem (where you know the distance but not the time).

Start with equations (1) and (2):

From (1): t = \dfrac{v - u}{a}

Why: solve equation (1) for t so you can substitute it into equation (2) and eliminate time.

Substitute into (2):

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

Why: replacing t with \frac{v - u}{a} in equation (2) gives an equation in s, u, v, and a only — no t.

Simplify the first term:

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

Why: in the second term, one power of a in the numerator cancels one in the denominator, leaving \frac{(v-u)^2}{2a}.

Multiply both sides by 2a:

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

Why: multiplying through by 2a clears all the fractions.

Expand the right side:

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

Why: 2u(v-u) = 2uv - 2u^2 and (v-u)^2 = v^2 - 2uv + u^2.

The 2uv terms cancel, and -2u^2 + u^2 = -u^2:

2as = v^2 - u^2
\boxed{v^2 = u^2 + 2as} \tag{3}

Why: after cancellation, the result is clean and symmetric. The equation says: the square of the final velocity equals the square of the initial velocity plus twice the product of acceleration and displacement. No time variable appears — this equation connects speed directly to distance.

The fourth equation: s = \tfrac{1}{2}(u + v)t

This one comes straight from the v–t graph with no calculus at all.

On the graph, the velocity rises (or falls) linearly from u to v over time t. The displacement is the area under this line — a trapezium with parallel sides u and v, and width t.

Velocity-time trapezium showing displacement as area A v-t graph with a straight line from (0, u) to (t, v). The shaded region below is a trapezium whose area equals the displacement s = half(u + v) times t. time velocity u v t 0 Area = s ut ½(v − u)t
The displacement is the area of the shaded trapezium. The rectangle contributes $ut$ (the distance at the initial velocity), and the triangle contributes $\tfrac{1}{2}(v - u)t$ (the extra distance from acceleration). Together: $s = ut + \tfrac{1}{2}(v-u)t = \tfrac{1}{2}(u+v)t$.

The area of a trapezium is \tfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{width}:

s = \frac{1}{2}(u + v) \cdot t
\boxed{s = \tfrac{1}{2}(u + v)t} \tag{4}

Why: this is pure geometry — the area of a trapezium with parallel sides u and v and perpendicular height t. No calculus needed. The equation says: displacement equals the average velocity times time.

This fourth equation is often the fastest route to a solution when you know u, v, and t (or any two of the three) and need s.

Distance in the nth second

Sometimes a problem asks: how far does the object travel in the 5th second specifically — not the first 5 seconds, but just the 5th one?

The displacement in the first n seconds is s_n = un + \tfrac{1}{2}an^2 (from equation 2). The displacement in the first (n-1) seconds is s_{n-1} = u(n-1) + \tfrac{1}{2}a(n-1)^2. The distance covered in the nth second alone is:

S_n = s_n - s_{n-1}

Why: subtract the total distance up to (n-1) seconds from the total distance up to n seconds. What remains is the distance covered during just that one second.

S_n = \left[un + \tfrac{1}{2}an^2\right] - \left[u(n-1) + \tfrac{1}{2}a(n-1)^2\right]

Expand:

S_n = un + \tfrac{1}{2}an^2 - u(n-1) - \tfrac{1}{2}a(n^2 - 2n + 1)
S_n = un + \tfrac{1}{2}an^2 - un + u - \tfrac{1}{2}an^2 + an - \tfrac{1}{2}a

Why: expanding (n-1)^2 = n^2 - 2n + 1 and distributing the \tfrac{1}{2}a and u through the brackets.

The un terms cancel. The \tfrac{1}{2}an^2 terms cancel. What remains:

\boxed{S_n = u + \frac{a(2n - 1)}{2}} \tag{5}

Why: after cancellation, the distance in the nth second depends on the initial velocity, the acceleration, and n itself. The factor (2n - 1) gives the sequence 1, 3, 5, 7, ... — the distances covered in successive seconds form an arithmetic progression (when u = 0, they go as \tfrac{a}{2}, \tfrac{3a}{2}, \tfrac{5a}{2}, ...). This is a beautiful result that Galileo observed by rolling balls down inclined planes.

Seeing acceleration: dots that spread apart

What does constant acceleration look like? Imagine marking the position of an object every 0.5 seconds. If the object moves at constant velocity, the dots are evenly spaced. If it accelerates, each successive gap is larger than the last — the object covers more distance in each interval because it is moving faster.

Animated: uniformly accelerated motion showing position dots getting farther apart A ball starts from rest and accelerates at 5 m/s squared. Its position dots are marked every 0.5 seconds, showing increasing gaps — the signature of acceleration.
A ball starts from rest with acceleration $a = 5$ m/s$^2$. Position: $x = \tfrac{1}{2}(5)t^2 = 2.5t^2$. The ghost markers at 0.5 s intervals show the gaps growing — that is the visual fingerprint of acceleration. Click replay to watch again.

Notice the pattern in the gaps. In the first 0.5 s, the ball travels 2.5 \times 0.25 = 0.625 m. In the next 0.5 s, it travels from 0.625 m to 2.5 m — a gap of 1.875 m. Each interval covers three times more distance than the previous one. That ratio of 1 : 3 : 5 : 7 is exactly the (2n-1) pattern from equation (5).

Sign conventions: choosing the positive direction

Before solving any problem, you must decide which direction is positive. This is not physics — it is bookkeeping. But getting it wrong is the most common source of sign errors.

The rule: pick one direction as positive and stick with it for the entire problem. All velocities, accelerations, and displacements in that direction are positive; all in the opposite direction are negative.

Sign convention: choosing the positive direction Two examples of sign conventions. Top: a car braking, with rightward positive. Bottom: a ball in free fall, with upward positive. Car braking (horizontal) +x v > 0 a < 0 Ball in free fall (vertical) +y (up) g ↓ so a = −g
Top: for a car moving right and braking, take rightward as positive. Velocity is positive, but acceleration (braking force) is negative. Bottom: for a ball thrown upward, take upward as positive. Then $g$ acts downward, so $a = -g = -9.8$ m/s$^2$.

Common convention choices:

Scenario Typical positive direction Then a is...
Car braking on a road Direction of motion (forward) Negative (deceleration)
Ball thrown upward Upward a = -g (gravity pulls down)
Ball dropped from a height Downward a = +g (gravity is in the positive direction)
Train accelerating Direction of motion Positive

The physics does not care which direction you call positive. The signs take care of the direction for you — as long as you are consistent. If you choose upward as positive and the ball ends up below where it started, the displacement s will be negative. That is not an error; it is the sign convention working correctly.

Solving problems: the recipe

Every uniformly accelerated motion problem gives you three of the five variables (s, u, v, a, t) and asks you to find a fourth. The strategy is always the same:

  1. List what you know. Write down s, u, v, a, t and fill in the three known values.
  2. Choose your sign convention. Decide which direction is positive. Assign signs to all your knowns accordingly.
  3. Pick the right equation. Each equation connects four of the five variables. Choose the one that contains the unknown you want and the three values you know.
Equation Contains Does NOT contain
v = u + at v, u, a, t s
s = ut + \tfrac{1}{2}at^2 s, u, a, t v
v^2 = u^2 + 2as v, u, a, s t
s = \tfrac{1}{2}(u + v)t s, u, v, t a
  1. Substitute and solve. Plug in the three knowns, solve for the unknown. Check the sign of your answer — it tells you the direction.

Worked examples

Example 1: The cow on NH-44

A car is travelling at 72 km/h on a national highway when the driver spots a cow and brakes. The car decelerates uniformly and stops in 50 m. Find the deceleration and the stopping time.

Before and after diagram for car braking Left: car moving right at 20 m/s. Right: car stopped after 50 m, with a cow standing ahead. 72 km/h 50 m stopped 🐄
The car starts at 72 km/h (20 m/s), brakes uniformly, and comes to rest after 50 m.

Step 1. Convert and list the knowns.

u = 72 km/h = \dfrac{72 \times 1000}{3600} = 20 m/s

Why: convert to SI. Divide by 3.6 to go from km/h to m/s. Here, 72/3.6 = 20.

v = 0 (the car stops), s = 50 m, a = ?, t = ?

Step 2. Choose sign convention: take the direction of motion (forward) as positive.

u = +20 m/s, v = 0, s = +50 m. The acceleration will come out negative (deceleration).

Step 3. Find a using equation (3), since you know v, u, and s but not t.

v^2 = u^2 + 2as
0 = (20)^2 + 2a(50)
0 = 400 + 100a
a = \frac{-400}{100} = -4 \text{ m/s}^2

Why: the negative sign confirms deceleration — the acceleration opposes the direction of motion, as expected for braking.

Step 4. Find t using equation (1).

v = u + at
0 = 20 + (-4)t
4t = 20
t = 5 \text{ s}

Why: the car loses 4 m/s every second. Starting at 20 m/s, it takes 20/4 = 5 seconds to stop.

Result: The deceleration is 4 m/s^2. The stopping time is 5 seconds.

The $v$–$t$ graph for the braking car. The straight line drops from 20 m/s to 0 over 5 seconds. The area under this line (a triangle) equals the displacement: $\tfrac{1}{2} \times 20 \times 5 = 50$ m — matching the given stopping distance.

What this shows: When you know three of the five SUVAT variables, the equations give you the other two cleanly. The v–t graph confirms the algebra — the triangle's area is exactly 50 m.

Example 2: Stone dropped from a tower

A stone is released from rest at the top of a 100 m tall building. Take g = 10 m/s^2. Neglect air resistance. Find (a) the time to reach the ground, and (b) the velocity at impact.

Animated: stone falling from a 100 m tower A stone is released from rest at the top of a 100 m tower. It accelerates downward at 10 m/s squared, covering more distance each second: positions at 5 m, 20 m, 45 m, 80 m. It hits the ground at about 4.47 seconds. 100 m 0 m
The stone starts from rest at the top and accelerates at $g = 10$ m/s$^2$ downward. Its position is $y = \tfrac{1}{2}(10)t^2 = 5t^2$ m below the top. Ghost markers at $t = 1, 2, 3, 4$ s show how the gaps grow — 5 m, 15 m, 25 m, 35 m — following the 1 : 3 : 5 : 7 pattern. Click replay to watch again.

Step 1. List the knowns.

u = 0 (released from rest), s = 100 m, a = g = 10 m/s^2

Why: take downward as positive (the direction of motion). Then a = +g = +10 m/s^2 and s = +100 m.

(a) Find time t. Use equation (2), since you know s, u, and a.

s = ut + \tfrac{1}{2}at^2
100 = 0 + \tfrac{1}{2}(10)t^2
100 = 5t^2
t^2 = 20
t = \sqrt{20} = 2\sqrt{5} \approx 4.47 \text{ s}

Why: with u = 0, the equation simplifies to s = \tfrac{1}{2}at^2. Solving for t gives t = \sqrt{2s/a}.

(b) Find velocity at impact. Use equation (3), since you know u, a, and s.

v^2 = u^2 + 2as = 0 + 2(10)(100) = 2000
v = \sqrt{2000} = 20\sqrt{5} \approx 44.7 \text{ m/s}

Why: the velocity grows as the square root of the distance fallen. After 100 m, the stone is moving at nearly 45 m/s — about 161 km/h.

Quick check using equation (1):

v = u + at = 0 + 10 \times 4.47 = 44.7 \text{ m/s} \checkmark

Why: the answer from equation (3) matches equation (1) with the time from part (a). Both routes give the same velocity — a good sanity check.

Result: The stone takes approximately 4.47 seconds to reach the ground and hits at about 44.7 m/s (\approx 161 km/h).

What this shows: A 100 m fall under gravity — roughly the height of a 30-storey building — produces a terrifying impact speed. This is why construction sites require safety nets and harnesses. The equations give you both the time and the speed from just the height and the value of g.

Common confusions

If you came here to learn the kinematic equations and use them, you have what you need. The rest of this section explores the equations from a graphical and calculus perspective that appears in JEE-level problems.

The v–t area method for multi-stage motion

Many JEE problems involve an object that accelerates for a while, then moves at constant velocity, then decelerates. The v–t graph for such motion is a trapezoid or a sequence of line segments. The total displacement is the total area under the graph — and you can compute that area as a sum of rectangles and triangles without using any kinematic equation at all.

For example, a Rajdhani Express accelerates uniformly from rest to 160 km/h (about 44.4 m/s) in 120 seconds, cruises for 600 seconds, then brakes uniformly to a stop in 90 seconds. The total displacement is:

  • Acceleration phase: \tfrac{1}{2} \times 44.4 \times 120 = 2664 m
  • Cruising phase: 44.4 \times 600 = 26640 m
  • Braking phase: \tfrac{1}{2} \times 44.4 \times 90 = 1998 m
  • Total: 31302 m \approx 31.3 km

The area method generalises to any piecewise-linear v–t graph — you do not need separate equations for each phase.

Deriving the equations using calculus directly

If you are comfortable with calculus, the three equations are really just one equation integrated twice:

a = \text{const} \implies \frac{dv}{dt} = a \implies v = u + at
v = \frac{ds}{dt} = u + at \implies s = \int_0^t (u + at)\,dt = ut + \tfrac{1}{2}at^2

The third equation, v^2 = u^2 + 2as, comes from a slick trick. Multiply both sides of a = dv/dt by v = ds/dt:

a \cdot v = v \cdot \frac{dv}{dt} = \frac{ds}{dt} \cdot \frac{dv}{dt}

But a \cdot v = a \cdot \frac{ds}{dt}, so:

a\,\frac{ds}{dt} = v\,\frac{dv}{dt}
a\,ds = v\,dv

Why: this chain-rule manipulation eliminates dt, linking v directly to s without time.

Integrate:

\int_0^s a\,ds = \int_u^v v\,dv
as = \frac{v^2}{2} - \frac{u^2}{2}
v^2 = u^2 + 2as

This v\,dv = a\,ds technique is powerful and shows up repeatedly in JEE problems involving variable acceleration as well — though in that case a is a function of s or v, and the integral is more complex.

When acceleration is not constant

The kinematic equations derived above work only when a is constant. When acceleration varies with time, position, or velocity, you must go back to the definitions and integrate directly:

v(t) = u + \int_0^t a(t')\,dt'
s(t) = \int_0^t v(t')\,dt'

For example, if a(t) = 6t (acceleration increasing linearly with time), then:

v = u + \int_0^t 6t'\,dt' = u + 3t^2
s = \int_0^t (u + 3t'^2)\,dt' = ut + t^3

Neither of these is a standard kinematic equation — they are specific to this particular acceleration function. The kinematic equations are the special case where a(t) is a constant, and the integrals simplify to the clean forms you derived above.

Where this leads next