Kinematic Equations of Motion

The three kinematic equations — v = v₀ + at, x = x₀ + v₀t + ½at², and v² = v₀² + 2a(x − x₀) — are the most-used formulas in school mechanics, and the most misused. Both problems have the same cure: derive them once from first principles and note exactly when they are allowed. NTA tests the derivations indirectly but relentlessly — through validity-based Assertion–Reason items, "which equation needs no time" shortcuts, the nth-second formula, and braking/acceleration numericals worth 4 marks each. This page derives all three by calculus from the definitions of the previous pages, verifies them graphically on the v–t plot, adds the average-velocity form and the distance-in-nth-second bonus formula, and closes with a selection strategy that tells you which equation to reach for before you start substituting numbers.

What are the Kinematic Equations? — Complete Theory

1. The setup and its one big condition

Take motion along a fixed axis with constant acceleration a. Let v₀ be the velocity at t = 0 and x₀ the position at t = 0. Everything below follows from just two definitions — a = dv/dt and v = dx/dt — plus the constancy of a. That constancy is not a technicality: every equation on this page is false when a varies with time (drag, springs, oscillations), and NTA checks this understanding explicitly. If a is constant, the v–t graph is a straight line, and — as the figure below shows — the displacement is its trapezoidal area, which is where the ½ and the t² come from.

2. Equation 1: v = v₀ + at (velocity–time)

Start from a = dv/dt, so dv = a dt. Integrate both sides, left over the velocity change, right over time:

∫_v₀^v dv = ∫₀^t a dt → v − v₀ = at

Dimensional check: [v₀] = [L T⁻¹]; [a][t] = [L T⁻²][T] = [L T⁻¹] ✔ — both terms are velocities, so adding them is legal. Validity: constant a. Use when the displacement is neither asked nor given.

3. Equation 2: x = x₀ + v₀t + ½at² (position–time)

Now feed Equation 1 into v = dx/dt: dx = (v₀ + at) dt. Integrate from x₀ to x and from 0 to t:

∫_x₀^x dx = ∫₀^t (v₀ + at) dt → x − x₀ = v₀t + ½at²

Dimensional check: [v₀t] = [L]; [a][t²] = [L T⁻²][T²] = [L] ✔ — every term is a length. Note it gives position, not distance: if the particle reverses mid-journey, the t² bookkeeping still tracks net position, while distance must be gathered leg by leg (the distinction built on the distance and displacement page).

4. Equation 3: v² = v₀² + 2a(x − x₀) (time-free)

Solve Equation 1 for t = (v − v₀)/a and substitute into Equation 2:

x − x₀ = v₀(v − v₀)/a + ½a(v − v₀)²/a² = (v − v₀)(v₀ + v)/2a = (v² − v₀²)/2a

Dimensional check: [v²] = [L² T⁻²]; [a][Δx] = [L T⁻²][L] = [L² T⁻²] ✔. Use when time is unknown and unwanted — the braking-distance workhorse, since stopping problems rarely quote a stopping time.

5. Bonus forms: average velocity and the nth second

Because v grows linearly when a is constant, the average velocity over any interval is the midpoint of the endpoints: v̄ = (v₀ + v)/2 — this fails the moment a varies. Hence a fourth relation:

And a favourite NTA integer-question gadget: the distance in the nth second is the position difference between t = n and t = n − 1:

s_n = x(n) − x(n−1) = v₀ + ½a(n² − (n−1)²) → s_n = v₀ + a(2n − 1)/2

Sanity check with v₀ = 0, a = 2: s₁ = 1 m, s₂ = 3 m, s₃ = 5 m — the 1 : 3 : 5 pattern, summing to ½a n² as they must. Keep the two ideas apart: distance in the nth second (this formula) versus distance in n seconds (½an² for a rest start).

Visualising the Derivation: Area under v–t

Solved Examples on the Kinematic Equations (Step-by-Step)

Solved Example 1 — Braking: pick the time-free equation

Q. A car moving at 72 km h⁻¹ brakes uniformly at 5 m s⁻². Find the time to stop and the distance covered while stopping.

Step 1. Convert: 72 km h⁻¹ = 72 × 5/18 = 20 m s⁻¹. Take the direction of motion positive, so a = −5 m s⁻², v = 0.

Step 2. Time: v = v₀ + at → 0 = 20 − 5t → t = 4 s.

Step 3. Distance (no time needed): v² = v₀² + 2aΔx → 0 = 400 + 2(−5)Δx → Δx = 400/10 = 40 m.

Step 4. Cross-check: x = v₀t + ½at² = 20(4) − ½(5)(16) = 80 − 40 = 40 m ✔ — both equations agree.

Solved Example 2 — nth second vs n seconds

Q. A body starts from rest with a = 2 m s⁻². Find the distance covered in the 3rd second, and the total distance in the first 3 seconds.

Step 1. Third second: s₃ = v₀ + a(2n−1)/2 = 0 + 2(5)/2 = 5 m.

Step 2. First three seconds: x = ½at² = ½(2)(9) = 9 m.

Step 3. Consistency: s₁ + s₂ + s₃ = 1 + 3 + 5 = 9 m ✔ — the nth-second values are the rungs of the ladder whose top is the total.

Practice Questions on the Kinematic Equations (With Solutions)

Practice Q1 (Numerical-value type)

Q. A train moving at 20 m s⁻¹ accelerates uniformly at 1 m s⁻² for 10 s. Its final speed is ______ m s⁻¹.

Solution. v = v₀ + at = 20 + 1(10) = 30 m s⁻¹. (Equation 1 suffices — no displacement data involved.)

Practice Q2 (Single-correct MCQ)

Q. Which kinematic relation lets you solve a problem without knowing the time? (a) v = v₀ + at (b) x = x₀ + v₀t + ½at² (c) v² = v₀² + 2a(x − x₀) (d) none of these

Solution. Equation 3 contains no t — it links velocities, acceleration and displacement directly, which is why stopping-distance problems use it first.

Practice Q3 (Assertion–Reason)

Q. Assertion (A): The kinematic equations may be applied to a freely falling stone. Reason (R): Free fall near the Earth's surface has constant acceleration g = 9.8 m s⁻² (air resistance neglected).
Options: (a) both true, R explains A · (b) both true, R does not explain A · (c) A true, R false · (d) A false, R true

Solution. The equations demand constant a; gravity supplies exactly that near the surface. R is the condition that licenses A — the same logic runs the free fall page.

Key Formulas & Takeaways

FAQs on the Kinematic Equations

What are the three kinematic equations?

For constant acceleration: v = v₀ + at, x = x₀ + v₀t + ½at², and v² = v₀² + 2a(x − x₀). A useful fourth form is x − x₀ = ½(v₀ + v)t.

When are the kinematic equations valid?

Only when acceleration is constant in magnitude and direction along one line. With time-varying a — air drag, springs — they fail and you must integrate a = dv/dt directly.

What is the distance travelled in the nth second?

s_n = v₀ + a(2n − 1)/2, the difference between positions at t = n and t = n − 1 seconds. It is not the total distance in n seconds, which for v₀ = 0 is ½an².

How do you choose which kinematic equation to use?

List what is known and unknown: time not asked → v² = v₀² + 2aΔx; final velocity not asked → x = x₀ + v₀t + ½at²; displacement not asked → v = v₀ + at; acceleration not asked → x − x₀ = ½(v₀ + v)t.