Kinematic equations for uniformly accelerated motion

A car braking hard before a signal, a lift accelerating out of a ground floor, a stone dropped down a well, a sprinter exploding off the blocks — every one of these is the same physics problem wearing a different costume. In each case something moves in a straight line with constant acceleration, and you are asked some version of "how fast," "how far," or "how long." You could always answer by sketching a velocity–time graph and reading off slopes and areas. But four short algebraic formulas — the kinematic equations for uniformly accelerated motion — let you skip the sketch and go straight to the number, provided you set them up correctly. These four equations are arguably the single most exam-tested formulas in the whole of Class 11 mechanics. They connect exactly five quantities: initial velocity $u$ , final velocity $v$ , acceleration $a$ , time $t$ , and displacement $s$ . Any kinematics problem gives you three of these five and asks for a fourth (sometimes a fifth); the entire skill is recognising which equation avoids the quantity you were never given, and getting the sign convention right before a single number goes in. This article rebuilds all four equations from the definitions of velocity and acceleration you already know — first from the $v$ – $t$ graph, then with one line of calculus — and then drills them on the two situations CBSE tests hardest: a body speeding up from rest, and a body falling freely under gravity. Get comfortable here and an entire family of numerical problems, from braking distances to rocket launches, becomes routine.

Key takeaways

Four equations connect $u$ , $v$ , $a$ , $t$ , $s$ for constant acceleration only: $v = u + at$ , $s = ut + \tfrac12 at^2$ , $v^2 = u^2 + 2as$ , and $s_n = u + \tfrac{a}{2}(2n-1)$ for the $n^{\text{th}}$ second.
$v = u + at$ comes from the definition of acceleration, $a = (v-u)/t$ , rearranged — it is also the slope statement of a straight-line $v$ – $t$ graph.
$s = ut + \tfrac12 at^2$ is the area under the $v$ – $t$ graph — a rectangle of height $u$ plus a triangle for the changing part of the velocity.
$v^2 = u^2 + 2as$ is obtained by eliminating $t$ between the first two — useful whenever a problem never mentions time at all.
$s_n$ gives displacement during the $n^{\text{th}}$ second specifically, not the total displacement up to $n$ seconds — a very common point-losing confusion.
Always fix a sign convention (one direction positive) before substituting; for free fall, $a = \pm g$ depending on that choice, with $g = 9.8\ \text{m/s}^2$ .
Key idea: for straight-line motion with constant acceleration, five quantities u, v, a, t, s are linked by four algebraic equations derivable from a v–t graph, and picking the equation that omits the one unknown quantity you were not given is the entire exam skill.

The core insight

When acceleration along a straight line is constant, the five quantities initial velocity, final velocity, acceleration, time and displacement obey $v=u+at$ , $s=ut+\tfrac12 at^2$ , $v^2=u^2+2as$ and $s_n=u+\tfrac{a}{2}(2n-1)$ , all provable from the slope and area of a velocity–time graph.

Where it fits in the syllabus

This topic sits at the end of the CBSE Class 11 Physics chapter Motion in a Straight Line, right after position–time and velocity–time graphs and the definitions of average/instantaneous velocity and acceleration. The prescribed treatment is explicitly both graphical and calculus-based: you are expected to read $v=u+at$ and $s=ut+\tfrac12 at^2$ off a $v$ – $t$ graph, and also obtain them by integrating $a=dv/dt$ and $v=dx/dt$ for constant $a$ . All four listed equations — $v=u+at$ , $s=ut+\tfrac12at^2$ , $v^2=u^2+2as$ , and $s_n=u+\tfrac{a}{2}(2n-1)$ — are in scope and commonly examined together with free-fall problems. What is explicitly not in scope here: motion in two or three dimensions (that is the very next chapter, "Motion in a Plane"), and any acceleration that is not constant, which would require integrating an arbitrary $a(t)$ and lies beyond this chapter.

The physical picture

Picture a straight road, with a car's velocity plotted against time. If the acceleration is constant, that plot is a straight line — and a straight line is completely described by just two things: where it starts (the initial velocity $u$ ) and its slope (the acceleration $a$ ). Everything else — the velocity at any later instant, the distance covered, the distance in any particular second — can be read off that one straight line using nothing more than the geometry of slopes and areas.

v = u + at

where: $u$ is the velocity at time $t=0$ , $v$ is the velocity at a later time $t$ , and $a$ is the (constant) acceleration. This is simply the statement that acceleration is the graph's slope: $a = \dfrac{v-u}{t-0}$ , rearranged. Physically it says velocity changes by a fixed amount, $at$ , for every second that passes.

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

where: $s$ is the displacement covered in time $t$ . Physically this is the area under the $v$ – $t$ graph between $0$ and $t$ — a rectangle of height $u$ and width $t$ (the distance you'd cover at constant velocity $u$ ) plus a triangle of base $t$ and height $at$ (the "extra" distance from speeding up), whose area is $\tfrac12\, t\,(at) = \tfrac12 at^2$ .

v^2 = u^2 + 2as

where: this equation has no time in it at all — get it by substituting $t=(v-u)/a$ from the first equation into the second and simplifying. It is the equation of choice whenever a problem tells you $u$ , $a$ , and $s$ (or asks for $s$ given $u$ , $v$ , $a$ ) and never mentions how long the motion took.

s_n = u + \frac{a}{2}(2n-1)

where: $s_n$ is the displacement during the $n^{\text{th}}$ second — that is, between $t=n-1$ and $t=n$ seconds — not the total displacement up to $n$ seconds. It follows from subtracting the displacement up to $(n-1)$ seconds from the displacement up to $n$ seconds, both computed with the second equation above.

Figure

A velocity time graph with a straight sloping line starting at height u on the vertical axis, rising to height v at time t, the area under the line split into a lower rectangle labelled u times t and an upper triangle labelled one half a t squared, with the slope of the line labelled acceleration a

The same equations fall out of two short integrations, and CBSE expects you to know this path too. Start from the definition of acceleration and integrate with respect to time, treating $a$ as a constant that can be pulled outside the integral:

\int_u^v dv' = \int_0^t a\, dt' \quad\Rightarrow\quad v - u = at \quad\Rightarrow\quad v = u + at

Now integrate velocity, $v(t) = u + at$ , with respect to time to get displacement:

s = \int_0^t (u + at')\, dt' = ut + \frac{1}{2}at^2

Both routes — graph and calculus — must give the same two equations, and CBSE examiners sometimes ask you to derive them either way, so it is worth being fluent in both.

Derivation ADeriving v squared equals u squared plus two a s

The goal of this derivation is to produce a kinematic equation that connects initial velocity, final velocity, acceleration and displacement while containing no time variable at all — because many real problems (a car skidding to a stop over a known distance, a bullet penetrating a known depth) tell you everything except how long the event took, and forcing time into the calculation would be unnecessary extra work. We build it by eliminating $t$ between the two equations already derived, $v = u + at$ and $s = ut + \tfrac12 at^2$ , using nothing beyond algebra.

Step 1. Start from the first kinematic equation and isolate time. Since $v = u + at$ , rearranging gives $t = \dfrac{v - u}{a}$ , valid whenever $a \neq 0$ (there is genuinely constant, nonzero acceleration).

Step 2. Substitute this expression for $t$ into the displacement equation $s = ut + \tfrac12 at^2$ . This gives $s = u\left(\dfrac{v-u}{a}\right) + \dfrac12 a\left(\dfrac{v-u}{a}\right)^2$ .

Step 3. Expand the second term carefully. $\dfrac12 a\left(\dfrac{v-u}{a}\right)^2 = \dfrac12 a \cdot \dfrac{(v-u)^2}{a^2} = \dfrac{(v-u)^2}{2a}$ . So the equation becomes $s = \dfrac{u(v-u)}{a} + \dfrac{(v-u)^2}{2a}$ .

Step 4. Put both terms over the common denominator $2a$ : $s = \dfrac{2u(v-u) + (v-u)^2}{2a}$ . Factor out $(v-u)$ from the numerator: $s = \dfrac{(v-u)\big[2u + (v-u)\big]}{2a} = \dfrac{(v-u)(v+u)}{2a}$ .

Step 5. Recognise $(v-u)(v+u) = v^2 - u^2$ (difference of squares), so $s = \dfrac{v^2 - u^2}{2a}$ . Multiplying both sides by $2a$ and rearranging gives the final result, $v^2 = u^2 + 2as$ — a purely algebraic consequence of the first two equations, containing exactly the four quantities $u$ , $v$ , $a$ , $s$ and no time at all.

Worked examples

Board level - Car accelerating from rest

A car starts from rest and accelerates uniformly at $2.5\ \text{m/s}^2$ for $8$ seconds. Find (a) its final velocity, (b) the distance it covers in this time, and (c) the distance covered in the 8th second alone.

Step 1. Set up the sign convention and list the givens. Take the direction of motion as positive throughout. "Starts from rest" means $u = 0$ . Given also: $a = 2.5\ \text{m/s}^2$ , $t = 8\ \text{s}$ .

Step 2. Find the final velocity using $v = u + at$ , since $u$ , $a$ , $t$ are all known and $v$ is asked. $v = 0 + (2.5)(8) = 20\ \text{m/s}$ .

Step 3. Find the total distance using $s = ut + \tfrac12 at^2$ , again because $u$ , $a$ , $t$ are all known. $s = (0)(8) + \tfrac12(2.5)(8)^2 = \tfrac12(2.5)(64) = 80\ \text{m}$ .

Step 4. Find the distance in the 8th second using $s_n = u + \dfrac{a}{2}(2n-1)$ with $n = 8$ — not the total distance formula, because the question asks for distance during the 8th second only. $s_8 = 0 + \dfrac{2.5}{2}(2\times 8 - 1) = 1.25 \times 15 = 18.75\ \text{m}$ .

(a) Final velocity = 20 m/s. (b) Total distance in 8 s = 80 m. (c) Distance in the 8th second = 18.75 m (part of the 80 m total, not extra distance beyond it).

Competitive level - Ball dropped down a well, cross-checked three ways

A ball is dropped (not thrown) from a height of $44.1\ \text{m}$ above the ground, with $g = 9.8\ \text{m/s}^2$ and air resistance ignored. Find (a) the time taken to hit the ground, and (b) the impact speed — and verify the answer using two different kinematic equations.

Step 1. Fix the sign convention: take downward as positive since the ball only moves downward. "Dropped" means released from rest, so $u = 0$ ; also $a = g = 9.8\ \text{m/s}^2$ and $s = 44.1\ \text{m}$ .

Step 2. Find time using $s = ut + \tfrac12 at^2$ , the only equation linking $s$ , $u$ , $a$ , $t$ without needing $v$ first. $44.1 = 0 + \tfrac12(9.8)t^2 = 4.9\,t^2$ , so $t^2 = 44.1/4.9 = 9$ , giving $t = 3\ \text{s}$ .

Step 3. Find impact speed using $v = u + at$ . $v = 0 + (9.8)(3) = 29.4\ \text{m/s}$ .

Step 4. Cross-check independently using $v^2 = u^2 + 2as$ , which never used the time computed in Step 2 — a genuinely independent path. $v^2 = 0 + 2(9.8)(44.1) = 864.36$ , so $v = \sqrt{864.36} \approx 29.4\ \text{m/s}$ , matching Step 3 exactly, which confirms the value of $t$ found in Step 2 was consistent.

(a) Time to hit the ground = 3 s. (b) Impact speed = 29.4 m/s, confirmed by two independent equations. Timing note: in an exam, do not compute $t$ and then $v$ and then separately verify with $v^2=u^2+2as$ unless the question explicitly asks for a check — that triples your working for no extra marks. Pick the single fastest route: if a question only asks for impact speed and gives $s$ , $u$ , $a$ , go directly to $v^2 = u^2+2as$ and skip finding $t$ altogether; save the cross-check technique for when you have spare time or the question explicitly asks you to verify.

Common traps

Mixing sign conventions mid-problem. Once "upward positive" or "downward positive" is chosen, $u$ , $v$ , $a$ , $s$ must all follow it for the rest of that problem; switching partway through is the single most common source of a wrong final answer.
Confusing $s_n$ with total displacement. $s_n = u + \tfrac{a}{2}(2n-1)$ gives displacement during the $n^{\text{th}}$ second only; total displacement up to $n$ seconds is the separate quantity $s = un + \tfrac12 an^2$ . Students frequently use the wrong one under time pressure.
Reaching for time when it was never needed. If a problem gives $u$ , $a$ , $s$ and asks for $v$ (or vice versa), using $v = u+at$ first (which requires computing $t$ via another equation) wastes a step; $v^2 = u^2 + 2as$ gets there directly.
Assuming "dropped" and "thrown" mean the same $u$ . "Dropped from rest" fixes $u = 0$ ; "thrown downward" or "thrown upward" both give a nonzero initial speed that must be read from the problem, never assumed to be zero.
Applying these four equations to non-uniform acceleration. All four are derived assuming $a$ is strictly constant; if a problem states or implies acceleration changes with time, none of these equations may be used as they stand.

Common trapThese equations apply only to constant acceleration along one straight line

It is tempting to reach for $v=u+at$ or $v^2=u^2+2as$ for any motion problem, but both derivations above used $a$ = constant at every step — pulling $a$ outside an integral, treating the $v$ – $t$ graph as a straight line, and so on. If acceleration changes with time (a rocket burning fuel and getting lighter, air resistance growing with speed), these four equations simply do not apply, and you would need calculus with the actual, varying $a(t)$ — a treatment kept out of this chapter. Equally, all four equations describe motion along one straight line only; the moment a problem involves two directions at once (a projectile, a car turning a corner), you have left this chapter and entered "Motion in a Plane," where velocity and acceleration must be treated as vectors with components — that is deliberately not covered here.

You should be able to…

State all four kinematic equations for uniformly accelerated motion and explain what each of the five symbols $u$ , $v$ , $a$ , $t$ , $s$ represents.
Derive $v = u+at$ and $s = ut + \tfrac12 at^2$ both from the slope/area of a $v$ – $t$ graph and by integrating $a = dv/dt$ and $v = dx/dt$ .
Derive $v^2 = u^2 + 2as$ algebraically by eliminating $t$ , and explain why it is the natural choice when a problem gives no time.
Choose and consistently apply one sign convention through an entire free-fall or vertical-throw problem.
Correctly distinguish $s_n$ (displacement during the $n^{\text{th}}$ second) from total displacement up to $n$ seconds, and compute either on demand.
Recognise when these equations do NOT apply — non-uniform (time-varying) acceleration, or motion that is not confined to a single straight line.

Ask the tutor: Want a set of mixed kinematics problems where you have to pick the right equation each time, or targeted practice keeping sign conventions straight in free-fall and vertical-throw questions — ask the AI tutor to drill you on the kinematic equations for uniformly accelerated motion.

If I drop a ball and throw another one downward from the same height, do they hit the ground at the same time?

No, and this is a favourite CBSE trick. A dropped ball has $u = 0$ ; a ball thrown downward has some nonzero $u > 0$ from the start. Using $s = ut + \tfrac12 at^2$ with the same $s$ and $a$ but a larger $u$ gives a smaller $t$ — the thrown ball reaches the ground sooner. They would only arrive together if both truly start with $u = 0$ , i.e. both are simply dropped, or if you are comparing a dropped ball against a ball thrown upward from the same point, which is a different, longer-time scenario entirely because it must first rise and come back down.

Can acceleration be negative even if the object is speeding up?

Yes — this trips up almost everyone at first, because "negative" gets read as "slowing down." Acceleration is negative only relative to whatever direction you called positive. If you take upward as positive and drop a ball, it speeds up as it falls, yet its acceleration is $-g$ , because it is speeding up in the negative (downward) direction. A car braking while moving in the positive direction, by contrast, is slowing down with negative acceleration. The sign tells you the direction of $a$ relative to your chosen positive axis, not whether the object is speeding up or slowing down — you must check that separately by comparing the signs of $v$ and $a$ .

Do these equations work for a lift moving up and then suddenly stopping?

Only within each phase of constant acceleration, treated separately. If a lift accelerates uniformly for the first two seconds and then moves at constant velocity, or decelerates to a stop, you cannot use one single set of $u$ , $v$ , $a$ , $t$ , $s$ across the whole journey, because $a$ is not the same constant throughout. Instead, split the motion into stages where $a$ is genuinely constant in each stage — apply the kinematic equations separately to each stage, using the final velocity of one stage as the initial velocity $u$ of the next. This "stitching" trick is common in CBSE numericals involving lifts, trains, and multi-stage rocket problems.