Kinematics — vidhyapath 2027

Concept Core

Uniform acceleration, projectiles, and relative motion — the three pillars.

Equations of Motion — Uniform Acceleration

These three equations apply when acceleration a is constant (uniform). Initial velocity = u, final = v, time = t, displacement = s.

v = u + at [velocity-time] s = ut + ½at² [displacement-time] v² = u² + 2as [velocity-displacement]

Fourth equation (less common): s = ½(u+v)t — useful when a is not given.

Free Fall & Vertical Motion

Under gravity, a = g = 9.8 ≈ 10 m/s² (downward). Apply equations of motion with a = g (if taking down as positive) or a = −g (if up is positive).

Time to reach maximum height: t = u/g (at peak, v = 0)

Maximum height: H = u²/2g

Time of flight (thrown up): T = 2u/g

Projectile Motion

Launched at angle θ with speed u. Horizontal and vertical are independent.

Horizontal: uₓ = ucosθ (constant, no acceleration)

Vertical: uᵧ = usinθ, a = −g

Range R = u²sin2θ/g Max height H = u²sin²θ/2g Time of flight T = 2u sinθ/g

Maximum range: at θ = 45°, R_max = u²/g

Relative Velocity

Velocity of A relative to B: v_AB = v_A − v_B

If two objects move towards each other with speeds v₁ and v₂: relative speed = v₁ + v₂

If moving in same direction: relative speed = |v₁ − v₂|

Rain-man problem: v_rain + v_man = v_rain relative to man. Hold umbrella at angle tan⁻¹(v_man/v_rain) from vertical.

Displacement-Time & Velocity-Time Graphs

s-t graph: slope = velocity. Straight line → constant velocity. Curve → acceleration.

v-t graph: slope = acceleration. Area under curve = displacement.

Negative slope on v-t graph → deceleration. Object reverses when v crosses zero.

Sth — Displacement in nth Second

Displacement in the nth second (not in n seconds):

Sₙ = u + a(2n−1)/2

This is a frequently tested formula. The 1st second: S₁ = u + a/2. Note it is linear in n (not quadratic).

Formula Vault

All kinematic formulas — motion, projectile, and relative velocity.

1st Equation of Motion

v = u + at

Velocity-time relation

2nd Equation of Motion

s = ut + ½at²

Displacement-time relation

3rd Equation of Motion

v² = u² + 2as

Velocity-displacement

Average Velocity

v_avg = (u+v)/2

Only for uniform acceleration

Displacement in nth Second

Sₙ = u + a(2n−1)/2

Not displacement in n seconds

Projectile Range

R = u²sin2θ/g

Max at θ=45°: R = u²/g

Max Height

H = u²sin²θ/2g

At highest point, vᵧ = 0

Time of Flight

T = 2u sinθ/g

Total time in air

Free Fall — Max Height

H = u²/2g

When thrown straight up

Relative Velocity

v_AB = v_A − v_B

Vector subtraction

Worked Examples

5 problems — from equations of motion to projectile to relative velocity.

EasyCar decelerates from 20 m/s to rest in 4 s — find deceleration and distance▾

A car moving at 20 m/s decelerates uniformly and stops in 4 s. Find deceleration and distance covered.

u = 20 m/s, v = 0, t = 4 s. From v = u + at: 0 = 20 + 4a → a = −5 m/s²

Distance: s = ut + ½at² = 20(4) + ½(−5)(16) = 80 − 40 = 40 m

✓ Deceleration = 5 m/s², Distance = 40 m

MediumBall thrown up at 20 m/s — find max height and time to return▾

A ball is thrown vertically upward with 20 m/s. Find (a) max height and (b) total time of flight. (g = 10 m/s²)

At max height: v = 0. v² = u² − 2gH → H = u²/2g = 400/20 = 20 m

Time to top: t = u/g = 20/10 = 2 s. Total time = 2 × 2 = 4 s

✓ Max height = 20 m, Total time = 4 s

MediumProjectile at 30° with 40 m/s — find range▾

A body is projected at 40 m/s at 30° to the horizontal. Find the range. (g = 10 m/s²)

R = u²sin2θ/g = (40)²×sin60°/10 = 1600×(√3/2)/10

= 1600 × 0.866 / 10 = 1385.6/10 = 138.6 m ≈ 80√3 m

✓ Range = 80√3 m ≈ 138.6 m

EAPCET LevelDisplacement in the 5th second for a = 3 m/s², u = 2 m/s▾

A body starts with u = 2 m/s and uniform acceleration a = 3 m/s². Find displacement in the 5th second specifically.

Use Sₙ = u + a(2n−1)/2. Here n = 5.

S₅ = 2 + 3×(2×5−1)/2 = 2 + 3×9/2 = 2 + 13.5 = 15.5 m

Verify: Total in 5 s = 2(5)+½(3)(25) = 10+37.5 = 47.5. In 4 s = 2(4)+½(3)(16) = 8+24 = 32. S₅ = 47.5−32 = 15.5 ✓

✓ Displacement in 5th second = 15.5 m

Trap QuestionTwo stones from same point at different times — when does stone 2 catch stone 1?▾

Stone A is thrown up at 30 m/s. 2 seconds later, stone B is thrown up from same point at 40 m/s. When does B catch A? (g = 10)

Let B be thrown at t=0. Then at time t, A has been travelling for (t+2) seconds.

Position of A: sₐ = 30(t+2) − ½(10)(t+2)²

Position of B: s_b = 40t − ½(10)t²

Set sₐ = s_b: 30(t+2)−5(t+2)² = 40t−5t²
30t+60−5(t²+4t+4) = 40t−5t²
30t+60−5t²−20t−20 = 40t−5t²
10t+40 = 40t → 30t = 40 → t = 4/3 s

✓ B catches A at t = 4/3 s after B is thrown (or 10/3 s after A was thrown)

Mistake DNA

The 4 kinematics errors that EAPCET distractors are designed around.

📏

Using Displacement Formulas for Distance

When a body reverses direction, displacement ≠ distance. Students use s = ut+½at² for total distance when the body changes direction.

❌ Wrong

Ball thrown up, comes
back: "total distance =
displacement = 0" ✗

✓ Correct

Split into two parts:
up (H) + down (H)
Distance = 2H ✓
Displacement = 0 ✓

When velocity changes sign, split the motion. Compute distance as sum of magnitudes of each segment.

🎯

Applying Equations of Motion When Acceleration is NOT Constant

v = u+at only works for uniform (constant) acceleration. Non-uniform motion requires calculus.

❌ Wrong

a = 3t (varies with time)
Still using s = ut+½at² ✗

✓ Correct

Integrate: v = ∫a dt
s = ∫v dt ✓
Or: use given v(t) directly

The 3 equations of motion are ONLY for constant acceleration. If a depends on t or x, use integration or work-energy methods.

🎪

Projectile: Forgetting Horizontal Velocity is Constant

Some students apply g to horizontal component too, or forget that horizontal speed never changes.

❌ Wrong

At max height:
horizontal speed = 0 ✗
(it only slows vertically)

✓ Correct

At max height:
vₓ = ucosθ (unchanged) ✓
vᵧ = 0 only ✓

Gravity acts only in vertical direction. Horizontal velocity = ucosθ throughout the flight, from launch to landing.

⏱️

Sₙ Formula Confusion — nth Second vs n Seconds

Sₙ = u + a(2n−1)/2 gives displacement IN the nth second, not the total displacement after n seconds.

❌ Wrong

"Find displacement in
3rd second" solved using
s = ut+½at² with t=3 ✗
(total after 3s, not in 3rd)

✓ Correct

Use Sₙ = u+a(2n−1)/2
with n=3 ✓
Or: s₃−s₂ total method ✓

The question "in the nth second" means the interval from (n−1) to n seconds. Two methods: use Sₙ formula, or compute (total in n s) − (total in (n−1) s).

Chapter Intelligence

Kinematics appears directly and as a tool in every mechanics chapter.

EAPCET Topic Weightage (2019–2024)

Equations of motion

Projectile motion

Relative velocity

v-t and s-t graphs

Sₙ — nth second

High-Yield PYQ Patterns

Range at two complementary angles Displacement in nth second Two-body meeting problem Projectile max height/range Area under v-t graph = distance Relative velocity of trains

Exam Strategy

Projectile: complementary angles (θ and 90°−θ) give the SAME range — a very common MCQ. R(30°) = R(60°) = u²sin60°/g.
For "when do they meet" problems: set both position equations equal. The algebra resolves faster than intuition.
v-t graph area = displacement (signed). If v goes negative, those areas subtract. Always check sign of v.
Free fall time and distance: t = √(2h/g) and v = √(2gh). Memorise these — saves the two-step substitution.
Kinematics is the prerequisite for Laws of Motion, Work-Energy, Circular Motion, and Gravitation. Every chapter uses v=u+at or projectile components.