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Physics · Motion in a Straight Line

Kinematic Equations for Uniform Acceleration

Three short equations cover almost every NEET problem on one-dimensional motion under constant acceleration. They are not three independent laws — they are three algebraic rearrangements of the same content, each missing one of the five quantities (u, v, a, s, t). This deep-dive derives them two ways, introduces the nth-second formula s$_n$ = u + a(2n − 1)/2 that NEET 2022 and 2021 tested back-to-back, fixes the sign convention, and walks through three abstract-motion examples — one for each equation. Free-fall lives in the next subtopic; here every example is horizontal.

The three equations — what they say

For straight-line motion with constant acceleration, NCERT §2.4 records three relations between the five quantities: initial velocity $u$ (NCERT writes $v_0$), final velocity $v$, acceleration $a$, displacement $s$ (NCERT writes $x$ or $x - x_0$), and time $t$. The three equations are:

$$\boxed{\;v = u + at\;} \qquad \boxed{\;s = ut + \tfrac{1}{2}at^2\;} \qquad \boxed{\;v^2 = u^2 + 2as\;}$$

Read them in order. The first connects $u$, $v$, $a$, $t$ — no displacement. The second connects $u$, $a$, $t$, $s$ — no final velocity. The third connects $u$, $v$, $a$, $s$ — no time. There is also a fourth, often-useful form built from the average-velocity definition,

$$s = \tfrac{1}{2}(u + v) \, t,$$

which NCERT writes as Eq. (2.7a). Under constant acceleration, the average velocity is the arithmetic mean of initial and final — a fact already met in average vs. instantaneous velocity.

The single assumption powering all of these is that $a$ is a constant of motion — same magnitude, same direction, throughout the interval. If $a$ varies with time or position, the equations fail; you must return to $a = dv/dt$ and $v = dx/dt$ and integrate directly.

"The kinematic equations are true only for motion in which the magnitude and the direction of acceleration are constant during the course of motion."

NCERT Class 11 Physics, §2.4, Points to Ponder 6

Derivation 1 — from the definition of acceleration

The cleanest route uses calculus and works because uniform acceleration means $a$ is constant under the integral sign. Start from the definition of instantaneous acceleration,

$$a = \frac{dv}{dt} \quad \Longrightarrow \quad dv = a\,dt.$$

Integrate both sides from $t = 0$, when the velocity is $u$, to a later instant $t$, when the velocity is $v$:

$$\int_{u}^{v} dv = \int_{0}^{t} a\,dt = a \int_{0}^{t} dt \quad \Longrightarrow \quad v - u = at.$$

Rearranging gives the first equation: $v = u + at$. Notice that $a$ came outside the integral only because it is constant — that is where the uniform-acceleration assumption enters mathematically.

For the second equation, start from $v = dx/dt$, write $dx = v\,dt$, and substitute the just-derived expression for $v$:

$$dx = (u + at)\,dt.$$

Integrate from $x = 0$ at $t = 0$ to $x = s$ at time $t$:

$$s = \int_{0}^{t}(u + at)\,dt = ut + \tfrac{1}{2}at^2.$$

For the third equation, use the chain rule trick NCERT shows in Example 2.2:

$$a = \frac{dv}{dt} = \frac{dv}{dx}\cdot\frac{dx}{dt} = v\,\frac{dv}{dx} \quad \Longrightarrow \quad v\,dv = a\,dx.$$

Integrate from initial $(x=0, v=u)$ to final $(x=s, v=v)$:

$$\int_{u}^{v} v\,dv = a \int_{0}^{s} dx \quad \Longrightarrow \quad \tfrac{1}{2}(v^2 - u^2) = as \quad \Longrightarrow \quad v^2 = u^2 + 2as.$$

NCERT notes the advantage: each step uses only the definitions of $v$ and $a$, so the same method extends to non-uniform acceleration once you stop pulling $a$ outside the integral.

Derivation 2 — from the v–t graph

The graphical derivation is the route most students remember on the day of the exam, because it is purely geometric. Plot velocity against time for uniformly accelerated motion: the curve is the straight line shown in NCERT Fig. 2.5, starting at $v = u$ at $t = 0$ and rising linearly with slope $a$ to reach $v = u + at$ at time $t$.

The area under this line between $t = 0$ and $t = t$ equals the displacement $s$ (NCERT records this in §2.3 — "the area under a velocity–time curve represents the displacement"). The region is a trapezium with parallel sides of length $u$ and $v$, separated by the time interval $t$. The standard trapezium-area formula gives:

$$s = \tfrac{1}{2}(u + v)\,t. \qquad (\star)$$

This is the average-velocity equation. To get the three named equations from it:

  • Eliminate $v$ using $v = u + at$: substitute into ($\star$) and simplify, $s = \tfrac{1}{2}(u + u + at)\,t = ut + \tfrac{1}{2}at^2$. Second equation done.
  • Eliminate $t$ using $t = (v - u)/a$: substitute into ($\star$), $s = \tfrac{1}{2}(u + v) \cdot \tfrac{v - u}{a} = \tfrac{v^2 - u^2}{2a}$, which rearranges to $v^2 = u^2 + 2as$. Third equation done.
  • The first equation, $v = u + at$, is just the linear slope of the v–t line: rise / run $=$ $(v - u)/t = a$.

The same trapezium splits into a rectangle of height $u$ and base $t$ (area $ut$) plus a triangle of height $(v - u) = at$ (area $\tfrac{1}{2}at^2$). The two pieces literally are the two terms of $s = ut + \tfrac{1}{2}at^2$ — the rectangle is the distance you would have covered at the initial velocity; the triangle is the extra distance because you sped up.

Displacement in the nth second — Galileo's law and its trap

A separate but related formula gives the displacement during only the nth second of the motion — that is, between $t = n - 1$ and $t = n$. Call this $s_n$. It is simply the total displacement up to time $n$ minus the total displacement up to time $n - 1$:

$$s_n = s(n) - s(n - 1).$$

Substitute $s(t) = ut + \tfrac{1}{2}at^2$ in both terms:

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

Expand $(n-1)^2 = n^2 - 2n + 1$ and collect terms:

$$s_n = u + \tfrac{1}{2}a(2n - 1) \quad \text{or equivalently} \quad \boxed{\;s_n = u + \frac{a(2n - 1)}{2}.\;}$$

For an object starting from rest ($u = 0$), this collapses to $s_n = \tfrac{a}{2}(2n - 1)$, which is in the ratio $1 : 3 : 5 : 7 : \ldots$ for successive seconds. This is exactly Galileo's law of odd numbers recorded in NCERT Example 2.5: a body falling from rest covers distances in the simple ratio of consecutive odd numbers in equal time intervals. NEET 2022 lifted this directly into a 4-second freely-falling question whose answer was the 1 : 3 : 5 : 7 ratio.

Sign convention — the most common loss of marks

The kinematic equations are algebraic: all five quantities carry signs (NCERT Points to Ponder 5). Time is naturally positive, but the other four can be negative. The discipline is simple:

  1. Pick a positive direction at the start and write it down. Once made, do not change it mid-solution.
  2. Assign signs consistently. A velocity along the positive direction is positive; against it, negative. An acceleration opposing the motion (braking) takes the opposite sign to the velocity.
  3. Substitute with signs included. Do not strip a minus because the answer "should" be positive — the equation produces the correct sign automatically.

Consider a car moving in the $+x$ direction at $u = +20$ m/s, braking at $a = -4$ m/s². Asked for displacement until it stops ($v = 0$), use $v^2 = u^2 + 2as$:

$$0 = (20)^2 + 2(-4)\,s \quad \Longrightarrow \quad s = \frac{400}{8} = +50\,\text{m}.$$

The displacement comes out positive — the car is still moving forward while it decelerates. If you had plugged $|a| = 4$ and a separate minus sign, you would have double-counted the negative and got $s = -50$ m, which is physically wrong: the car never travels backward during braking.

Which equation to use — a single decision rule

Every constant-acceleration problem gives you three of the five quantities $\{u, v, a, s, t\}$ and asks for a fourth. The table below identifies, for each "missing-and-not-asked" quantity, the equation that does not contain it. Use the equation that omits the quantity you neither know nor want.

You are given You want to find Quantity not in play Use this equation
$u$, $a$, $t$ $v$ $s$ (displacement) $v = u + at$
$u$, $v$, $a$ $t$ $s$ (displacement) $v = u + at$, solved for $t$
$u$, $a$, $t$ $s$ $v$ (final velocity) $s = ut + \tfrac{1}{2}at^2$
$u$, $v$, $t$ $s$ $a$ (acceleration) $s = \tfrac{1}{2}(u + v)\,t$
$u$, $a$, $s$ $v$ $t$ (time) $v^2 = u^2 + 2as$
$u$, $v$, $s$ $a$ $t$ (time) $v^2 = u^2 + 2as$, solved for $a$
$u$, $a$, $n$ (an integer) $s_n$ (nth-second) $v$ and final $s$ $s_n = u + a(2n - 1)/2$

Two operational tips. First, scan the question for the words "time", "displacement", "velocity" and circle the one that is neither given nor asked — that is the quantity absent from your chosen equation. Second, in two-step problems (find $t$, then use it to find $s$), treat the second step as a fresh substitution. Mixing five symbols across two equations in one block is where arithmetic errors live.

Worked examples — one per equation

Example 1 — find final velocity from $u$, $a$, $t$

Worked example · v = u + at

A trolley on a frictionless horizontal track starts at $u = 3$ m/s and accelerates uniformly at $a = 5$ m/s² along the direction of motion. What is its velocity after 4 s? (Setup adapted from NIOS Example 2.10.)

Setup. Take the direction of motion as positive. Given $u = +3$ m/s, $a = +5$ m/s², $t = 4$ s. Displacement is neither given nor asked, so use $v = u + at$.

Substitute. $v = 3 + (5)(4) = 3 + 20 = 23$ m/s.

Distractor 1 — forgot $u$: $v = at = 20$ m/s. Wrong; the trolley was not at rest. Distractor 2 — used $s = ut + \tfrac{1}{2}at^2$: $s = 12 + 40 = 52$ m. That is the displacement, not the velocity — the wrong equation for the question asked.

Example 2 — find displacement from $u$, $v$, $a$ (no time)

Worked example · v² = u² + 2as

A bullet enters a wooden plank moving horizontally at $u = 200$ m/s and exits at $v = 100$ m/s. The wood applies a uniform retarding acceleration of magnitude $7500$ m/s² on the bullet. How thick is the plank?

Setup. Take the direction of motion as positive. The bullet slows down, so $a$ opposes velocity: $a = -7500$ m/s². Time is neither given nor asked — use $v^2 = u^2 + 2as$.

Substitute. $(100)^2 = (200)^2 + 2(-7500)\,s$ $\Longrightarrow$ $10000 = 40000 - 15000\,s$ $\Longrightarrow$ $s = 30000/15000 = 2$ m.

The plank is 2 m thick.

Distractor — plugged $|a|$ with no sign: $10000 = 40000 + 15000\,s$ $\Longrightarrow$ $s = -2$ m. The minus tells you the sign was wrong; physically the bullet cannot travel backward through the plank. Always carry $a$'s sign explicitly.

Example 3 — nth-second displacement

Worked example · s_n = u + a(2n − 1)/2

A particle moving along the x-axis starts at $u = 2$ m/s and accelerates uniformly at $a = 4$ m/s² in the direction of motion. Find the displacement during the 5th second of its motion.

Setup. Take the direction of motion as positive. The "5th second" means the interval from $t = 4$ s to $t = 5$ s — one second long. Use the nth-second formula with $n = 5$.

Substitute. $s_5 = u + \dfrac{a(2n - 1)}{2} = 2 + \dfrac{4 \cdot 9}{2} = 2 + 18 = 20$ m.

Cross-check by subtraction. $s(5) = (2)(5) + \tfrac{1}{2}(4)(5)^2 = 10 + 50 = 60$ m. $s(4) = (2)(4) + \tfrac{1}{2}(4)(4)^2 = 8 + 32 = 40$ m. $s_5 = s(5) - s(4) = 60 - 40 = 20$ m. Same answer — and this is the safest method on test day if you forget the closed-form.

Distractor 1 — confused $s_n$ with $s$ in $n$ seconds: answered 60 m. That is total displacement after 5 s, not the 5th-second slice. Distractor 2 — set $n = 4$: $s_4 = 2 + 4(7)/2 = 16$ m, the 4th-second displacement; off-by-one from misreading "during the 5th second".

Quick recap

What this subtopic locked in

  • Three equations, one assumption. $v = u + at$; $s = ut + \tfrac{1}{2}at^2$; $v^2 = u^2 + 2as$. Valid only for constant $a$ along the line of motion.
  • Two derivations. Calculus (integrate $a = dv/dt$ and $v = dx/dt$); v–t graph (trapezium area equals displacement; eliminate one variable each time).
  • Average-velocity form. $s = \tfrac{1}{2}(u + v)\,t$ — the bridge between the three equations.
  • nth-second formula. $s_n = u + a(2n - 1)/2$, with units of length even though the expression looks like velocity. Galileo's odd-numbers law follows when $u = 0$.
  • Sign discipline. Pick a positive direction; assign signs to $u$, $v$, $a$, $s$; substitute with signs included. There is one $v^2 = u^2 + 2as$, not two.
  • Equation choice. Use the equation that omits the quantity neither given nor asked.

Free-fall under gravity is the canonical application of these equations with $a = -g$, but it has enough variants — vertical throws, drops, two-step round trips — to deserve its own subtopic. We pick that up in Motion under gravity (free fall) next.

NEET PYQ Snapshot — Kinematic Equations

Three NEET PYQs from 2023, 2022 and 2021 — each testing a different equation. The fourth is a direct NCERT exercise that turns up regularly in chapter mocks.

NEET 2023

A bullet from a gun is fired on a rectangular wooden block with velocity $u$. When the bullet travels 24 cm through the block along its length horizontally, the velocity of the bullet becomes $u/3$. It then further penetrates into the block in the same direction before coming to rest exactly at the other end of the block. The total length of the block is

  1. 30 cm
  2. 27 cm
  3. 24 cm
  4. 28 cm
Answer: (2) 27 cm

Why: Use $v^2 = u^2 + 2as$ twice. Segment 1 ($u \to u/3$ over 24 cm): $(u/3)^2 = u^2 + 2(-a)(24)$ gives $48a = u^2 - u^2/9 = 8u^2/9$, so $2a \cdot 24 = 8u^2/9$. Segment 2 ($u/3 \to 0$ over $s$ cm): $0 = (u/3)^2 + 2(-a)\,s$ gives $2a \cdot s = u^2/9$. Dividing: $s/24 = (u^2/9)/(8u^2/9) = 1/8$, so $s = 3$ cm. Total length $= 24 + 3 = \textbf{27 cm}$. Distractor (3) 24 cm stops after segment 1; (1) 30 cm mis-divides the ratio; (4) 28 cm uses arithmetic average instead of squared-velocity ratio.

NEET 2021

A small block slides down on a smooth inclined plane, starting from rest at time $t = 0$. Let $S_n$ be the distance travelled by the block in the interval $t = n - 1$ to $t = n$. The ratio $\dfrac{S_n}{S_{n+1}}$ is

  1. $\dfrac{2n}{2n - 1}$
  2. $\dfrac{2n - 1}{2n}$
  3. $\dfrac{2n - 1}{2n + 1}$
  4. $\dfrac{2n + 1}{2n - 1}$
Answer: (3)

Why: The acceleration along the incline is constant ($a = g\sin\theta$). The block starts from rest ($u = 0$), so the nth-second formula gives $S_n = \tfrac{a}{2}(2n - 1)$ and $S_{n+1} = \tfrac{a}{2}(2(n+1) - 1) = \tfrac{a}{2}(2n + 1)$. Dividing: $\dfrac{S_n}{S_{n+1}} = \dfrac{2n - 1}{2n + 1}$. Distractor (4) flips numerator and denominator; (1) and (2) drop the $(2n - 1)$ pattern. The acceleration $g\sin\theta$ cancels and never appears in the answer — a useful sanity check.

NEET 2018

A toy car with charge $q$ moves on a frictionless horizontal plane under the influence of a uniform electric field $\vec{E}$. Due to the force $q\vec{E}$, its velocity increases from 0 to 6 m/s in one second. At that instant the field is reversed. The car continues to move for two more seconds under the reversed field. The average velocity and the average speed of the car between $t = 0$ and $t = 3$ s respectively are

  1. 2 m/s, 4 m/s
  2. 1 m/s, 3 m/s
  3. 1 m/s, 3.5 m/s
  4. 1.5 m/s, 3 m/s
Answer: (2) 1 m/s, 3 m/s

Why: Acceleration in segment 1: $a = (6 - 0)/1 = 6$ m/s². After reversal, $a = -6$ m/s²; using $v = u + at$ with $u = 6$, $a = -6$, the car reaches $v = 0$ at $t = 2$ s (i.e., after 1 more second), then moves backward and reaches $v = -6$ m/s at $t = 3$ s. Net displacement: forward triangle $\tfrac{1}{2}(1)(6) = 3$ m, then forward decelerating triangle $\tfrac{1}{2}(1)(6) = 3$ m, then backward triangle $\tfrac{1}{2}(1)(6) = -3$ m. Net $= +3$ m. Average velocity $= 3/3 = \textbf{1 m/s}$. Path length $= 3 + 3 + 3 = 9$ m; average speed $= 9/3 = \textbf{3 m/s}$. Distractor (4) 1.5 m/s halves the path-length sum incorrectly. This problem stitches together $v = u + at$ and the v–t-area interpretation of displacement.

NCERT Exercise

A car moving along a straight highway with a speed of $126$ km/h is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop? (NCERT Exercise 2.5)

  1. $a = -3.0625$ m/s²; $t = 11.4$ s
  2. $a = -3.0625$ m/s²; $t = 9.8$ s
  3. $a = -4.9$ m/s²; $t = 7.1$ s
  4. $a = -1.75$ m/s²; $t = 20$ s
Answer: (1)

Why: Convert $u = 126$ km/h $= 126 \times 5/18 = 35$ m/s. Use $v^2 = u^2 + 2as$ with $v = 0$, $s = 200$ m: $0 = (35)^2 + 2a(200)$ gives $a = -1225/400 = \textbf{-3.0625 m/s}^2$. Then $v = u + at$ with $v = 0$: $t = -u/a = 35/3.0625 \approx \textbf{11.43 s}$. Two equations, single-unknown discipline at each step. Distractor (3) doubles the deceleration (forgot $2as$, used $as$); (2) uses $u = 30$ m/s by sloppy conversion; (4) halves the deceleration.

FAQs — Kinematic Equations

Short answers to the questions NEET aspirants ask most about the three equations.

When can the three kinematic equations be applied?
Only when the acceleration is constant in both magnitude and direction along the line of motion. If $a$ varies with time or position the three equations are invalid; you must integrate $dv/dt$ or $v\,dv/dx$ directly. NCERT records this restriction explicitly in §2.4 — the equations follow from assuming $a$ is a constant of motion.
How do I choose which equation to use?
Identify which quantity is missing in the problem. If time $t$ is not given and not asked, use $v^2 = u^2 + 2as$. If displacement $s$ is not given and not asked, use $v = u + at$. If final velocity $v$ is not given and not asked, use $s = ut + \tfrac{1}{2}at^2$. The equation that does not contain the missing quantity is the right one — that is what the "Which equation to use" table in this article encodes.
Why does $s_n = u + a(2n - 1)/2$ have units of length even though it looks like velocity?
Because $s_n$ is the displacement during a one-second interval, the implicit time factor of 1 s is hidden. Strictly $s_n = u\cdot(1\,\text{s}) + a\cdot(2n - 1)/2\cdot(1\,\text{s})$. When you substitute $u$ in m/s and $a$ in m/s², you must multiply by $\Delta t = 1$ s to recover units of metres. NEET examiners exploit this by changing the time interval — the formula assumes 1 s.
Is $v^2 = u^2 + 2as$ the same as $v^2 = u^2 - 2as$ for deceleration?
Yes and no. The equation $v^2 = u^2 + 2as$ is the only correct form; the minus sign in $v^2 = u^2 - 2as$ is built into $a$ being negative. If a vehicle decelerates from $u = 20$ m/s at $a = -4$ m/s² over $s = 50$ m, $v^2 = 400 + 2(-4)(50) = 0$. Plugging $|a| = 4$ with a separate minus sign double-counts. Always carry $a$ with its sign.
Can I use the equations if the object reverses direction?
Yes, provided the acceleration stays the same throughout — which it does, for example, for an object thrown up that comes back down (acceleration is $g$ downward the whole time). The equations are algebraic in $s$, $u$, $v$ and $a$, so direction reversals are handled automatically by signs. Keep one positive axis throughout the problem.
What is the difference between $s_n$ and $s$ in $n$ seconds?
Total displacement after $n$ seconds is $s = un + \tfrac{1}{2}an^2$. The nth-second displacement $s_n = u + a(2n - 1)/2$ is the displacement during only the nth second alone, that is, between $t = n - 1$ and $t = n$. They are related by $s_n = s(n) - s(n - 1)$. NEET 2022 and NEET 2021 both tested this directly — a common error is to answer the wrong one.
Related Topics
Motion in a Straight Line — Parent chapter Full chapter overview: kinematics in 1D, graphs, equations, relative velocity, free fall. Position, Path Length & Displacement The starting concepts — origin, sign convention, vector vs scalar distinction. Average vs Instantaneous Velocity From $\Delta x/\Delta t$ to $dx/dt$; speed vs velocity; graphical slopes. Acceleration Average and instantaneous acceleration; v–t graphs; positive vs negative acceleration. Relative Velocity in One Dimension $v_{BA} = v_B - v_A$; head-on, same-direction and overtaking problems. Motion Under Gravity (Free Fall) The kinematic equations with $a = -g$; vertical throws and Galileo's law of odd numbers.