Ever wondered how your phone's GPS calculates arrival time? Or how a car's airbag knows exactly when to deploy? The three equations of motion (v = u + at, s = ut + ½at², v² = u² + 2as) are the foundation of classical mechanics. They describe how moving objects behave when acceleration is constant — whether it's a ball falling under gravity, a car braking to a stop, or a rocket launching upward. Nearly every NEET kinematics problem boils down to picking the right one of these three.
🧠 The Big Idea — In Plain Words
These three equations connect the five key variables of motion: u (initial velocity), v (final velocity), a (acceleration), t (time), and s (displacement). If you know any three, you can find the fourth. They only work when acceleration is constant (like gravity g = 9.8 m/s²). If acceleration changes, you need calculus.
The first equation (v = u + at) comes straight from the definition of acceleration. The second (s = ut + ½at²) comes from the area under a velocity-time graph. The third (v² = u² + 2as) is derived by eliminating t from the first two — it's useful when you don't know the time.
Key variables: u = initial velocity (m/s), v = final velocity (m/s), a = acceleration (m/s²), t = time (s), s = displacement (m)
Fig 1.1: Velocity vs Time graph. The animation advances as the lecture progresses through each concept.
🔍 Step-by-Step: First Equation — v = u + at
1
Acceleration = how fast velocity changes. By definition, a = (change in velocity) / (time taken) = (v - u)/t. This is true only if a is constant during the motion.
2
Rearrange to get the formula. Multiply both sides by t: a·t = v - u. Then add u to both sides: v = u + at. This says: final velocity = starting velocity + how much you speed up each second × how many seconds.
v = u + at
🔍 Step-by-Step: Second Equation — s = ut + ½at²
1
Displacement = area under the v-t graph. For a trapezium-shaped area (rectangle + triangle): rectangle area = u × t, triangle area = ½ × (v - u) × t = ½ × (at) × t = ½at².
2
Add the two areas.s = u·t + ½·a·t². The first term (ut) is how far you'd go at constant speed u. The second term (½at²) adds the extra distance due to acceleration.
s = ut + ½at²
🔍 Step-by-Step: Third Equation — v² = u² + 2as
1
Eliminate time from the first two equations. From v = u + at, we get t = (v - u)/a. Substitute this into s = ut + ½at².
2
Do the algebra. s = u·(v-u)/a + ½·a·[(v-u)/a]² = (uv - u²)/a + (v-u)²/(2a) = (2uv - 2u² + v² + u² - 2uv)/(2a) = (v² - u²)/(2a).
3
Rearrange. 2as = v² - u² → v² = u² + 2as. This is useful when you don't know t — for example, finding how fast something hits the ground when dropped from a height.
v² = u² + 2as
🔍 Common Mistake
These equations work ONLY for constant acceleration. Also: pick a sign convention (e.g., upward = positive) and stick to it. If a ball is thrown upward, a = -g (negative because gravity pulls downward). Many students forget the minus sign and get the wrong answer.
🎯 NEET Pattern — Kinematics
• Free fall: u = 0, a = g = 10 m/s² (NEET usually rounds to 10). v = gt, h = ½gt², v² = 2gh.
• Braking problems: v = 0, a negative (deceleration). Find stopping distance: s = u²/(2a).
• NEET classic: "A ball thrown upward with speed u. Maximum height? Time to reach top?" → H = u²/(2g), t = u/g.
• Displacement in nth second: s_n = u + a(2n-1)/2. This comes from s(n) - s(n-1).
Example 1: Equations of Motion
A car starts from rest and accelerates uniformly at 2 m/s² for 5 seconds. Find its final velocity and distance covered.
a) v = 10 m/s, s = 25 m b) v = 5 m/s, s = 25 m c) v = 10 m/s, s = 50 m d) v = 20 m/s, s = 25 m
Solution: u = 0, a = 2, t = 5. v = u + at = 0 + 2×5 = 10 m/s. s = ut + ½at² = 0 + ½×2×25 = 25 m. Option (a).
2. Projectile Motion — Time of Flight, Height & Range
💡 Why this matters in real life
When you throw a basketball, hit a golf ball, or fire a cannonball, it follows a parabolic path — this is projectile motion. Understanding this helps athletes optimize their throws, engineers design artillery, and even helps us understand how water fountains work. The key insight: horizontal and vertical motions are independent — gravity only affects the vertical component, while horizontal motion stays constant (ignoring air resistance).
🧠 The Big Idea — In Plain Words
Think of projectile motion as two independent motions happening at the same time: (1) Horizontal: constant speed (no horizontal force, so no acceleration). (2) Vertical: constant downward acceleration g (like free fall). The object is launched with speed u at angle θ. The horizontal component u_x = u·cosθ stays constant. The vertical component u_y = u·sinθ changes due to gravity. We can calculate three things: Time of flight T (total time in air), Maximum height H (highest point), and Range R (horizontal distance covered).
Key variables: u = launch speed, θ = launch angle, g = gravity, T = time of flight, H = max height, R = range
Fig 2.1: Parabolic path — animation advances as the lecture covers launch, components, height, range, and formulas.
🔍 Step-by-Step: Time of Flight T
1
Focus on vertical motion only. The total time in air depends only on the vertical component. At the highest point, vertical velocity becomes zero. The time to go up = time to come down (symmetry, if launch and landing are at same height).
2
Time to reach maximum height. Use v = u + at vertically: at top, v_y = 0. 0 = u·sinθ - g·t_up. So t_up = u·sinθ/g. Total time T = 2 × t_up: T = 2u·sinθ / g
🔍 Step-by-Step: Maximum Height H
1
Use v² = u² + 2as vertically. At highest point, v_y = 0, u_y = u·sinθ, a = -g, s = H: 0 = (u·sinθ)² - 2gH
2
Solve for H.H = u²·sin²θ / (2g). The height depends on the square of sinθ. At θ = 90° (straight up), sin²θ = 1, H = u²/(2g) — the maximum possible height for a given speed.
🔍 Step-by-Step: Range R
1
Horizontal distance = horizontal speed × time. Horizontal speed u_x = u·cosθ is constant (no horizontal force). Time = total time of flight T: R = (u·cosθ) × T = u·cosθ × (2u·sinθ/g) = u²·2sinθ·cosθ/g
2
Use 2sinθ·cosθ = sin(2θ).R = u²·sin(2θ) / g. Range is maximum when sin(2θ) = 1, i.e., 2θ = 90°, so θ = 45°. At 45°, R_max = u²/g. Also: θ and 90°-θ give the same range (since sin(2(90-θ)) = sin(180-2θ) = sin(2θ)).
T = 2u·sinθ/g H = u²·sin²θ/(2g) R = u²·sin(2θ)/g
🔍 Common Mistake
Students forget that these formulas assume level ground (launch and landing at same height). For inclined launches or uneven terrain, you need to solve using the general equations. Also: T, H, and R all depend on u² — doubling launch speed quadruples both range and height.
🎯 NEET Pattern — Projectile
• Max range at θ = 45°. Range same for θ and (90° - θ) — e.g., 30° and 60° give equal ranges.
• Vertical height at any time: y = (u·sinθ)t - ½gt². Horizontal distance: x = (u·cosθ)t.
• Equation of trajectory: y = x·tanθ - gx²/(2u²·cos²θ). NEET gives this and asks for range/H.
• For projectile on an incline, modify g to components parallel/perpendicular to the incline.
Example 2: Projectile
A projectile is launched at 30° with speed 20 m/s. Find maximum height and range. (g = 10 m/s²)
a) H = 5 m, R = 34.6 m b) H = 10 m, R = 40 m c) H = 5 m, R = 20 m d) H = 2.5 m, R = 34.6 m
Solution: u·sin30° = 20×0.5 = 10 m/s. H = (10)²/(2×10) = 100/20 = 5 m. u·cos30° = 20×0.866 = 17.32 m/s. T = 2×10/10 = 2 s. R = 17.32×2 = 34.64 m. Option (a).
3. Work-Energy Theorem
💡 Why this matters in real life
Fig 3.1: Force does work on a block, increasing its KE. Animation advances through work, KE, theorem, springs, and summary.
🧠 The Big Idea — In Plain Words
Work = force × distance (in the direction of force). When you push an object and it moves, you're doing work. That work becomes the object's kinetic energy (energy of motion). The theorem W = ΔKE says: if you do 100 J of work on an object, its KE increases by 100 J (assuming no other forces like friction). This is why a hammer swung faster (more KE) drives a nail deeper (more work done on the nail).
Key variables: W = work (J), F = force (N), s = displacement (m), θ = angle between F and s, KE = ½mv², m = mass (kg), v = speed (m/s)
🔍 Step-by-Step: Deriving W = ΔKE
1
Definition of work. W = F·s·cosθ. For simplicity, consider force F in the direction of motion (θ = 0°): W = F·s
2
Use Newton's Second Law: F = ma.W = m·a·s
3
Use the third equation of motion: v² = u² + 2as → as = (v² - u²)/2. Substitute: W = m·(v² - u²)/2 = ½mv² - ½mu² = KE_final - KE_initial = ΔKE
W = ΔKE = ½mv² - ½mu²
🔍 Step-by-Step: Work Done by Gravity (Conservative Force)
1
Lifting an object against gravity. Force needed = mg (upward). Work done = mg·h (where h = vertical height). This work becomes gravitational potential energy stored in the object: PE = mgh
2
Conservative vs non-conservative forces. Gravity is conservative — work done depends only on start and end points, not the path. Friction is non-conservative — work depends on the path length. For conservative forces, total mechanical energy (KE + PE) is conserved if no other forces act.
Students think work is always F×s. It's actually F·s·cosθ — if you push against a wall and it doesn't move (s = 0), you do zero work, even though you feel tired! Also: work-energy theorem works even when forces are not constant — you can integrate to find the total work.
🎯 NEET Pattern — Work-Energy
• "A body of mass m slides down a frictionless incline of height h. Find speed at bottom." → Use energy conservation: mgh = ½mv² → v = √(2gh).
• With friction: work done by friction = μmg·d = loss in mechanical energy.
• Power P = W/t = F·v (if F constant). NEET asks: "A motor lifts a mass at constant speed. Find power." → P = mg·v.
• Spring force: Work done = ½kx² (stored as spring PE). F = -kx (Hooke's law).
Example 3: Work-Energy Theorem
A 2 kg block is pushed by a constant force of 10 N over a distance of 5 m on a frictionless surface, starting from rest. Find final speed.
a) 5 m/s b) 7.07 m/s c) 10 m/s d) 3.5 m/s
Solution: W = F·s = 10×5 = 50 J. ΔKE = 50 = ½×2×v² → v² = 50 → v = 7.07 m/s. Option (b).
4. Gravitation — g, Orbital Velocity & Escape Velocity
💡 Why this matters in real life
Why does the Moon orbit the Earth? Why do astronauts feel weightless in the ISS? How fast must a rocket go to escape Earth's gravity? These are all answered by Newton's law of gravitation. The three key formulas are: g (acceleration due to gravity at height h), v₀ (orbital velocity — speed needed to stay in orbit), and vₑ (escape velocity — speed needed to break free from Earth's gravity forever).
Fig 4.1: Earth with gravity g, orbiting satellite, escape velocity. Animation advances with the lecture.
🧠 The Big Idea — In Plain Words
Newton's universal law: every mass attracts every other mass with force F = Gm₁m₂/r². The acceleration due to gravity g is the gravitational force per unit mass. On Earth's surface, g₀ = 9.8 m/s². As you go higher, g decreases (g ∝ 1/r²). At height h above surface: g = g₀R²/(R+h)².
Orbital velocity v₀ is the horizontal speed a satellite needs to keep falling around the Earth (it falls toward Earth, but the Earth curves away at the same rate!). For a circular orbit at height h: mv₀²/(R+h) = GMm/(R+h)² → v₀ = √(GM/(R+h)). Close to surface: v₀ ≈ √(gR) ≈ 7.9 km/s.
Escape velocity vₑ is the minimum speed to escape Earth's gravity forever: you need enough KE to overcome the (negative) gravitational PE. ½mvₑ² = GMm/R → vₑ = √(2GM/R) = √(2gR) ≈ 11.2 km/s.
Key variables: G = 6.67×10⁻¹¹ N·m²/kg², M = mass of Earth, R = radius of Earth, h = height above surface
🔍 Step-by-Step: Variation of g with Height
1
Force on mass m at height h above Earth's surface: F = GMm/(R+h)². By F = mg: g_h = GM/(R+h)²
2
Compare with g on surface (h = 0): g₀ = GM/R². Ratio: g_h/g₀ = R²/(R+h)² = 1/(1 + h/R)². For h << R: g_h ≈ g₀(1 - 2h/R). So at 100 km height, g is about 3% less than at surface.
🔍 Step-by-Step: Orbital Velocity v₀
1
For a satellite in circular orbit, centripetal force = gravitational force.mv₀²/(R+h) = GMm/(R+h)² → v₀² = GM/(R+h)
2
At surface (approx): v₀ = √(GM/R). Using GM = gR²: v₀ = √(gR). With g = 9.8, R = 6.4×10⁶: v₀ ≈ 7.9 km/s. This is why the ISS orbits at ~7.7 km/s — it's slightly higher, so it needs a bit less speed.
v₀ = √(GM/(R+h)) Time period: T = 2π(R+h)/v₀ = 2π√((R+h)³/GM) (Kepler's third law!)
🔍 Step-by-Step: Escape Velocity vₑ
1
To escape Earth's gravity, KE must be at least equal to the gravitational PE (which is negative). Gravitational PE at surface = -GMm/R. For escape: final KE = 0, final PE = 0 (at r = ∞). So: ½mvₑ² + (-GMm/R) = 0.
2
Solve for vₑ.½mvₑ² = GMm/R → vₑ = √(2GM/R) = √(2gR) ≈ 11.2 km/s. Interestingly, vₑ = √2 × v₀. Escape velocity from Earth is about 11.2 km/s, from the Sun it's ~618 km/s, from a black hole it's > c (speed of light) — that's why light can't escape!
vₑ = √(2GM/R) = √(2gR) ≈ 11.2 km/s (vₑ = √2 × v₀)
🔍 Common Mistake
Students think g is constant everywhere — it's not! g decreases with height and also varies with depth (linearly g decreases to zero at Earth's centre). Also: orbital speed does NOT depend on the satellite's mass — a heavy satellite and a light one orbit at the same speed at the same height.
🎯 NEET Pattern — Gravitation
• Geostationary satellite: orbits at h ≈ 36,000 km, period = 24 h, v₀ ≈ 3.1 km/s. Appears stationary over one point on Earth.
• g at depth d: g_d = g₀(1 - d/R). At centre: d = R → g = 0.
• Kepler's laws: (1) Elliptical orbits with Sun at focus, (2) Equal area in equal time, (3) T² ∝ a³.
• Weightlessness: g inside a freely falling satellite is zero relative to the satellite (everything falls together). Not zero gravity — gravity is still acting!
Example 4: Escape Velocity
Earth's mass M = 6×10²⁴ kg, radius R = 6.4×10⁶ m, G = 6.67×10⁻¹¹. Find escape velocity.
SHM is everywhere: a vibrating guitar string, a child on a swing, a car's suspension system, the swaying of a tall building in the wind, and even the atoms vibrating in a crystal. The formulas for the spring-mass system (T = 2π√(m/k)) and the simple pendulum (T = 2π√(L/g)) are two of the most important in physics. The pendulum formula was historically used to measure g and to make accurate clocks.
Fig 5.1: Spring-mass and pendulum. Animation advances through spring, pendulum, energy, formulas, and summary.
🧠 The Big Idea — In Plain Words
SHM is motion where the restoring force is proportional to the displacement and acts in the opposite direction: F = -kx (Hooke's law for springs). This creates a sinusoidal oscillation. The time period T is the time for one complete oscillation — it depends on the system's properties (mass and spring constant for a spring; length and gravity for a pendulum).
For a spring-mass system: a heavier mass or weaker spring → slower oscillations (longer period). For a pendulum: longer string → slower oscillations. Surprisingly, the pendulum's period does NOT depend on the mass of the bob — a heavy bob and light bob swing at the same rate!
Key variables: k = spring constant (N/m), m = mass (kg), L = length of pendulum (m), T = time period (s), f = frequency = 1/T (Hz)
🔍 Step-by-Step: Spring-Mass System (T = 2π√(m/k))
1
Hooke's law: restoring force F = -kx. Newton's second law: F = ma. So ma = -kx, or: d²x/dt² + (k/m)x = 0
2
This is the SHM differential equation. It has the form d²x/dt² + ω²x = 0, where ω = √(k/m) is the angular frequency. The solution is x = A·sin(ωt + φ).
3
Time period T = 2π/ω.T = 2π√(m/k). Key insight: T is independent of amplitude! Whether you stretch the spring 1 cm or 5 cm, the oscillation time is the same (for small oscillations within elastic limit).
Spring: T = 2π√(m/k) ω = √(k/m) f = (1/2π)√(k/m)
🔍 Step-by-Step: Simple Pendulum (T = 2π√(L/g))
1
The restoring force is the component of gravity along the arc. For small angle θ (less than ~10°), sinθ ≈ θ (in radians). The arc displacement x = L·θ. Restoring force F = -mg·sinθ ≈ -mg·θ = -(mg/L)·x.
2
Compare with F = -kx for spring: the effective spring constant is k = mg/L. Using the spring formula T = 2π√(m/k): T = 2π√(m/(mg/L)) = 2π√(L/g)
3
Key results. T does NOT depend on: (1) the bob's mass, (2) the amplitude (for small angles). T DOES depend on: (1) length L (longer → slower), (2) gravity g (more gravity → faster). This is why pendulum clocks run slower at higher altitudes (g is slightly smaller).
Pendulum: T = 2π√(L/g) f = (1/2π)√(g/L)
🔍 Common Mistake
The pendulum formula T = 2π√(L/g) is valid ONLY for small amplitudes (θ < 10°). For larger swings, the period is slightly longer. Also: if the pendulum is on an accelerating lift, the effective g changes (g_eff = g ± a), changing the period.
🎯 NEET Pattern — SHM
• Velocity & acceleration in SHM: v = ω√(A² - x²), a = -ω²x. Max velocity at mean position (x=0): v_max = ωA. Max acceleration at extremes (x=±A): a_max = ω²A.
• Energy in SHM: KE = ½mω²(A² - x²), PE = ½mω²x², total E = ½mω²A² = constant. Energy oscillates between KE and PE.
• For a pendulum, if L is quadrupled, T doubles. If g is quadrupled, T halves.
• Springs in series: 1/k_eq = 1/k₁ + 1/k₂. In parallel: k_eq = k₁ + k₂.
Example 5: SHM
A spring-mass system has mass 0.5 kg and spring constant 50 N/m. Find its time period. (π² = 10)
a) 0.2 s b) 0.63 s c) 1.0 s d) 0.5 s
Solution: T = 2π√(m/k) = 2π√(0.5/50) = 2π√(0.01) = 2π×0.1 = 0.628 s. Option (b).
6. Kinetic Theory of Gases — PV = ⅓Nmv²
💡 Why this matters in real life
Why does a balloon expand when heated? Why does a tyre feel firm when inflated? The kinetic theory of gases explains macroscopic gas behaviour (pressure, temperature, volume) in terms of microscopic molecular motion. The result PV = ⅓Nmv² (or equivalently PV = nRT, the ideal gas law) is one of the most beautiful connections between the microscopic and macroscopic worlds. It tells us that temperature is really just the average kinetic energy of molecules!
Fig 6.1: Gas molecules bounce randomly. Animation advances through molecules, pressure, formulas, energy, distribution, and summary.
🧠 The Big Idea — In Plain Words
Imagine a box full of tiny gas molecules zipping around in all directions. They constantly bounce off the walls. Pressure is caused by these molecules hitting the walls — each collision transfers a tiny bit of momentum, and billions of collisions per second add up to a measurable force. The faster the molecules move (higher temperature), the harder they hit, and the higher the pressure.
From this simple picture, we can derive: PV = ⅓Nmv², where N = number of molecules, m = mass of each molecule, v² = average of squared speeds. This leads to: (1) average KE = ³⁄₂kT (temperature is average molecular KE), (2) root-mean-square speed v_rms = √(3kT/m) = √(3RT/M), and (3) explanations for why gases expand when heated and why lighter molecules move faster (lighter → faster v_rms).
Key variables: P = pressure (Pa), V = volume (m³), N = number of molecules, m = mass per molecule (kg), v² = mean square speed (m²/s²), k = Boltzmann constant = 1.38×10⁻²³ J/K, T = temperature (K)
🔍 Step-by-Step: PV = ⅓Nmv²
1
Consider a molecule moving towards a wall. Take a cubical box of side L. A molecule with velocity v_x hits the right wall. Before collision: momentum = mv_x. After elastic collision (bouncing back): momentum = -mv_x. Change in momentum per collision = 2mv_x.
2
How many collisions per second? The molecule travels from one wall to the other (distance L) and back (another L) — that's 2L between collisions with the same wall. Time between collisions with the right wall = 2L/v_x. So collisions per second with that wall = v_x/(2L).
3
Force contributed by one molecule = momentum change per second. Force = (2mv_x) × (v_x/2L) = mv_x²/L. This is the force exerted on the wall by this single molecule.
4
Total force from all N molecules. Total force F = Σ(mv_x²/L) = (m/L)Σv_x². Pressure P = F/A = F/L² = (m/L³)Σv_x² = (m/V)Σv_x² (since V = L³).
5
By symmetry, (v_x²)_avg = (v_y²)_avg = (v_z²)_avg = (v²)_avg/3. (Because v² = v_x² + v_y² + v_z², and all directions are equivalent). So Σv_x² = N·(v²)_avg/3. Substituting: P = (m/V)·N·(v²)_avg/3 → PV = ⅓·N·m·(v²)_avg
PV = ⅓·N·m·v² (where v² = mean square speed) PV = nRT (ideal gas law)
🔍 Step-by-Step: Temperature and Molecular Speed
1
From PV = ⅓Nm·v² and ideal gas law PV = nRT = NkT. Equate: ⅓Nmv² = NkT → ½mv² = (3/2)kT. The average kinetic energy of a molecule is directly proportional to temperature! This is the fundamental link — temperature IS molecular kinetic energy.
2
Root-mean-square speed v_rms = √(v²). From ½m·v² = (3/2)kT: v_rms = √(3kT/m) = √(3RT/M) (where M = molar mass in kg/mol). Lighter molecules (H₂) move much faster than heavier ones (O₂) at the same temperature. For H₂ at 300K: v_rms ≈ 1930 m/s. For O₂: v_rms ≈ 480 m/s.
The kinetic theory equation PV = ⅓Nmv² assumes ideal gas assumptions: (1) negligible molecular size, (2) no intermolecular forces, (3) perfectly elastic collisions, (4) random motion. Real gases deviate from this at high pressure and low temperature (van der Waals corrections).
🎯 NEET Pattern — Kinetic Theory
• "Ratio of v_rms of H₂ to O₂ at same temperature" → √(M_O₂/M_H₂) = √(32/2) = 4:1 (H₂ is 4× faster).
• Pressure doubles if: (1) temperature doubles at constant volume, (2) number of molecules doubles, (3) volume halves.
• Degrees of freedom: Monatomic (f=3), diatomic (f=5 at room T), polyatomic (f=6). γ = 1 + 2/f.
• Mean free path λ = average distance between collisions. λ ∝ T/P. At low pressure, λ is large (vacuum).
• NEET: "At what temperature is v_rms of N₂ equal to that of O₂ at 300 K?" → T_N₂ = 300 × 28/32 = 262.5 K.
Example 6: Kinetic Theory
Find the v_rms of oxygen molecules at 300 K. (R = 8.314 J/mol·K, M = 32 g/mol = 0.032 kg/mol)