1
Distance and Displacement
Key Definitions
- Distance — scalar quantity; total path length covered; always ≥ 0
- Displacement — vector quantity; shortest distance from initial to final position; can be negative
- Kinematics = Kine (motion) + Matics (math) → mathematical study of motion
Relation Between Distance and Displacement
- Distance ≥ |Displacement| (always)
- Distance = |Displacement| only when motion is in a straight line without reversal
- Distance < Displacement → Never possible
Standard Results
| Path/Shape | Distance | Displacement | Ratio (dist/disp) |
|---|
| Semicircle of radius R | πR | 2R | π/2 |
| Full circle of radius R | 2πR | 0 | ∞ |
| Quarter circle of radius R | πR/2 | R√2 | π/(2√2) |
| Body diagonal of cube (side a) | √3 a | √3 a | 1 |
For angle θ: Displacement = 2R sin(θ/2)
When a particle moves through angle θ on a circle of radius R
Q: A particle covers 1/4th of a circle of radius R. Find displacement.
AB² = OA² + OB² = R² + R² = 2R² → AB = R√2
Ans: R√2 ✓
Q: A wheel of radius R rolls on road in half rotation. Find displacement of point P on circumference.
Horizontal = πR (half circumference), Vertical = 2R (diameter); d = √(πR)² + (2R)² = R√(π² + 4)
Ans: R√(π² + 4) ✓
Q: A particle makes 120° angle with its initial position on a circle of radius R. Find displacement.
|d| = 2R sin(θ/2) = 2R sin(60°) = 2R × (√3/2) = R√3
Ans: R√3 ✓
Definitions
- Speed = Distance / Time → scalar quantity
- Average Speed = Total distance / Total time (for large time interval Δt)
- Instantaneous Speed = speed at a particular instant → |vinst|
- Velocity = Displacement / Time → vector quantity
- Average Velocity = Total displacement / Total time
- Instantaneous Velocity = ds/dt → slope of s-t graph at that instant
- Average speed ≥ |Average velocity| (always)
Average Speed — Important Formulae
| Situation | Average Speed Formula |
|---|
| Same distance at v₁ and v₂ | vavg = 2v₁v₂/(v₁ + v₂) (Harmonic Mean) |
| Same time at v₁ and v₂ | vavg = (v₁ + v₂)/2 (Arithmetic Mean) |
| Equal 3 distances at v₁, v₂, v₃ | vavg = 3v₁v₂v₃/(v₁v₂ + v₂v₃ + v₃v₁) |
⚡ Key Trick — A→B→A (Round Trip)
- A to B at v₁, B to A at v₂ → Average velocity = 0 (displacement = 0)
- Average speed = 2v₁v₂/(v₁ + v₂)
Q: A man covers half distance at v₁ and remaining half at v₂. Find average speed.
t₁ = d/v₁, t₂ = d/v₂; avg speed = 2d/(d/v₁ + d/v₂) = 2v₁v₂/(v₁ + v₂)
Ans: 2v₁v₂/(v₁ + v₂) ✓
Q: A man covers first half time with v₁ and second half time with v₂. Find average speed.
d = v₁t + v₂t = (v₁ + v₂)t; avg speed = 2(v₁+v₂)t/2t = (v₁+v₂)/2
Ans: (v₁ + v₂)/2 ✓
Q: Man covers 3 equal distances with v₁, v₂, v₃. Find average speed.
avg speed = 3d/(d/v₁ + d/v₂ + d/v₃) = 3v₁v₂v₃/(v₁v₂ + v₂v₃ + v₃v₁)
Ans: 3v₁v₂v₃/(v₁v₂ + v₂v₃ + v₃v₁) ✓
Definitions
- Acceleration = rate of change of velocity = Δv/Δt → vector quantity; unit m/s²
- Average acceleration = Δv/Δt (for large time)
- Instantaneous acceleration = dv/dt = slope of v-t graph at that instant
- If v and a have same sign → speed increases
- If v and a have opposite signs → speed decreases
Graph Interpretations
- s-t graph: slope = velocity; concave upward → a positive; concave downward → a negative
- v-t graph: slope = acceleration; area under curve = displacement
- a-t graph: area under curve = change in velocity
- θ < 90° with positive x-axis → slope positive → velocity positive
- θ > 90° with positive x-axis → slope negative → velocity negative
Q: s = t³ − 6t². Find velocity at t = 1, 2, 3 sec.
v = ds/dt = 3t² − 12t; at t=1: v = 3−12 = −9 m/s; at t=2: v = 12−24 = −12 m/s; at t=3: v = 27−36 = −9 m/s
Ans: −9, −12, −9 m/s ✓
Q: s = t³ − 6t². Find time when body is at rest (v = 0).
v = 3t² − 12t = 0 → 3t(t−4) = 0 → t = 0 or t = 4 sec
Ans: t = 0 s and t = 4 s ✓
Q: s = t³ − 6t² + 9t. Find when body moves in negative direction.
v = 3t² − 12t + 9 < 0 → (t−1)(t−3) < 0 → body moves in negative direction for 1 < t < 3 sec
Ans: 1 < t < 3 sec ✓
4
Equations of Motion (Uniform Acceleration)
① v = u + at
② s = ut + ½at²
③ v² = u² + 2as
④ snth = u + a(2n − 1)/2 (displacement in nth second)
where u = initial velocity, v = final velocity, a = acceleration, s = displacement, t = time
Derivations — Key Steps
v = u + at: a = dv/dt → ∫dv = ∫a dt → [v]ᵤᵛ = a[t]₀ᵗ → v − u = at
s = ut + ½at²: v = ds/dt → ∫ds = ∫(u+at)dt → s = ut + ½at²
v² = u² + 2as: a = v·dv/ds → ∫a ds = ∫v dv → as = (v²−u²)/2 → v² = u² + 2as
Displacement in nth Second
- snth = sn − sn−1 = u + a(2n−1)/2
- Displacements in successive seconds are in ratio: 1 : 3 : 5 : 7 ... (when u = 0)
Convention
- Upward direction → positive; downward → negative
- Moving upward: v positive, a = −g (g ≈ 10 m/s²)
- Moving downward: v negative, a = −g, s = −ve
- At maximum height: v = 0
Time to reach max height: t = u/g
Maximum height: H = u²/2g
Total time of flight: T = 2u/g
Speed of return = speed of projection (symmetric)
⚡ Key Results — Must Remember
- Stone dropped from height h: time = √(2h/g), final speed = √(2gh)
- For two balls thrown simultaneously with vAB = const and aAB = 0 → separation increases linearly
- Water drops at interval T: when nth drop just falls, distance of kth drop from tap = ½g(n−k)²T²
Q: Ball thrown vertically up at u = 28 m/s. Find speed 1 second before reaching max height.
v = 0 at max; t_max = 28/10 = 2.8 s; at t = 1.8 s: v = 28 − 10(1.8) = 28 − 18 = 10 m/s
Ans: 10 m/s ✓
Q: Tap drops water droplets at 1 s interval. Find distance between 3rd and 5th drop when 6th drop just comes out.
3rd drop has fallen for 3 s: s₃ = ½(10)(9) = 45 m; 5th drop has fallen for 1 s: s₅ = ½(10)(1) = 5 m; distance = 45 − 5 = 40 m
Ans: 40 m ✓
Q: Two balls A (u=0) and B (u=3 m/s down) thrown simultaneously. Find separation after 3 s.
v_AB = v_A − v_B = 0 − (−3) = 3 m/s; a_AB = g − g = 0; separation = v_AB × t = 3 × 3 = 9 m
Ans: 9 m ✓
Core Concept
- Relative velocity of A with respect to B: VAB = VA − VB
- Relative acceleration: AAB = AA − AB
- Relative displacement: SAB = SA − SB
- Time does NOT depend on frame of reference
⚡ Police–Thief Problems
- Treat thief at rest, police moves with relative velocity VPT = VP − VT
- Relative distance SPT = initial gap between them
- Time to catch = SPT / VPT (if both have same acceleration)
- If accelerations differ → use s = ut + ½at² with relative values
Q: Thief runs at 10 m/s, police at 20 m/s. Gap = 100 m. Time to catch?
V_PT = 20 − 10 = 10 m/s; t = 100/10 = 10 s; distance from police start = 20×10 = 200 m
Ans: 10 s, police catches at 200 m from its start ✓
Q: Rain falls at 40 km/h vertically. Man walks at 80 km/h horizontally. Umbrella angle?
V_RM = V_R − V_M = −40ĵ − 80î; tan θ = 80/40 = 2 → θ = tan⁻¹(2) ≈ 63° from vertical (towards front)
Ans: θ = tan⁻¹(2) ≈ 63° from vertical ✓
Basic Setup
- Projectile motion = motion under gravity in a plane (2D)
- Initial velocity u at angle θ → ux = u cosθ (constant), uy = u sinθ
- Acceleration: ax = 0, ay = −g
- x-direction → uniform motion; y-direction → uniformly retarded motion
Time of Flight: T = 2u sinθ / g
Maximum Height: H = u² sin²θ / 2g
Horizontal Range: R = u² sin2θ / g
Equation of Trajectory: y = x tanθ − gx² / (2u² cos²θ)
y = x tanθ (1 − x/R)
Trajectory is a parabola. Velocity at any time: vx = u cosθ (const), vy = u sinθ − gt
| Property | Result | Condition |
|---|
| Maximum Range | Rmax = u²/g | θ = 45° |
| Relation: Rmax & Hmax | Rmax = 4Hmax | θ = 45° |
| Same Range two angles | θ and (90° − θ) | Same u |
| H₁, H₂ for complementary θ | R = 4√(H₁H₂) | Same u |
| Speed at any point | |v| = √(vx² + vy²) | — |
| Velocity at ground (same level) | vx = u cosθ, vy = −u sinθ | Ground to ground |
⚡ If H = R (important problem)
H = R → u² sin²θ/2g = u² sin2θ/g → sinθ/2 = 2cosθ → tanθ = 4 → θ = tan⁻¹(4)
Projectile from Height (Horizontal Projection)
When u is horizontal (uy = 0, θ = 0)
- ux = u, uy = 0, ax = 0, ay = −g
- Time to fall height H: T = √(2H/g)
- Range: R = u × T = u√(2H/g)
- Speed at ground: v = √(u² + 2gH)
- Equation of trajectory: y = gx²/(2u²) (parabola)
Q: Ball thrown from 20 m tower horizontally at 5 m/s. Find time, range and final speed.
t = √(2×20/10) = 2 s; R = 5×2 = 10 m; v_y = 10×2 = 20 m/s; speed = √(25+400) = √425 ≈ 20.6 m/s
Ans: t = 2 s, R = 10 m, speed = √425 m/s ✓
Q: Ball thrown at 30° from 20 m tower at 30 m/s. Find time and range.
u_x = 30cos30° = 15√3 m/s; u_y = 30sin30° = 15 m/s (upward); −20 = 15t − ½(10)t² → 5t²−15t−20=0 → t²−3t−4=0 → t = 4 s; R = 15√3 × 4 = 60√3 m
Ans: t = 4 s, R = 60√3 m ✓
Q: Ball thrown at u = 5 m/s, θ = 37°. Find T, H, R.
T = 2×5×sin37°/10 = 2×5×0.6/10 = 0.6 s; H = 25×0.36/20 = 0.45 m; R = 25×sin74°/10 = 25×0.96/10 = 2.4 m
Ans: T = 0.6 s, H = 0.45 m, R = 2.4 m ✓
Setup
- VSG = VSR + VRG (swimmer w.r.t. ground = swimmer w.r.t. river + river w.r.t. ground)
- Vertical component crosses river; horizontal component causes drift
- Width of river d; river speed Vr; swimmer speed Vm
| Case | Condition | Time | Drift |
|---|
| Minimum Time | Swim perpendicular to river (θ = 90°) | t = d/Vm | Drift = Vr × d/Vm |
| Zero Drift (Shortest path) | Vm sinθ = Vr; sinθ = Vr/Vm; only if Vm > Vr | t = d/(Vm cosθ) | 0 |
| Minimum Distance | VSG perpendicular to river bank | Longer | 0 |
Q: River flows at 3 km/h. Swimmer swims perpendicular at 4 km/h. Width = 16 km. Find time and drift.
t = 16/4 = 4 hours; drift = 3 × 4 = 12 km; actual distance = √(16² + 12²) = 20 km
Ans: t = 4 h, drift = 12 km, actual = 20 km ✓
Q: Man swims at 3 km/h in still water. River speed = 2 km/h. Width = 500 m = 0.5 km. To reach directly opposite — direction and time?
sinθ = 2/3; cosθ = √5/3; t = 0.5/(3cosθ) = 0.5/(3 × √5/3) = 0.5/√5 = 1/(2√5) hours
Ans: θ = sin⁻¹(2/3) upstream; t = 1/(2√5) h ✓
9
Circular Motion — Kinematics
Key Quantities
- Angular displacement (θ) = arc/radius = s/r; unit = radian; dimensionless
- Angular velocity (ω) = dθ/dt; unit = rad/s; related to frequency: ω = 2πf = 2π/T
- Angular acceleration (α) = dω/dt = d²θ/dt²; unit = rad/s²; dim [T⁻²]
- Anticlockwise → ω outward (positive); Clockwise → ω inward (negative)
Linear ↔ Angular Relations:
v = rω (linear speed = radius × angular speed)
aT = rα (tangential acceleration)
ac = v²/r = ω²r = vω (centripetal acceleration)
Total acceleration: |a| = √(aT² + ac²)
Equations of Motion for Circular Motion
- ωf = ωi + αt
- θ = ωit + ½αt²
- ωf² = ωi² + 2αθ
- θnth = ωi + α(2n−1)/2
| Type | UCM (Uniform) | NUCM (Non-uniform) |
|---|
| Speed | Constant | Variable |
| Angular velocity ω | Constant (α = 0) | Variable (α ≠ 0) |
| Tangential acc aT | 0 | ≠ 0 |
| Centripetal acc ac | ≠ 0 (constant magnitude) | ≠ 0 (variable) |
| KE | Constant | Variable |
| Work done by centripetal force | 0 | 0 |
10
Circular Dynamics — Forces in Circular Motion
Centripetal Force: Fc = mv²/r = mω²r = mac
Centrifugal Force = pseudo force in non-inertial frame; equal and opposite to centripetal force
Centripetal force is not a new force — it's the net inward force (tension, friction, gravity, normal reaction etc.)
Important Circular Motion Cases
① Bending of Cyclist (Horizontal circle):
N sinθ = mv²/r; N cosθ = mg → tan θ = v²/(rg) → v = √(rg tanθ)
② Banked Road (without friction):
V = √(Rg tanθ)
③ Banked Road (with friction):
Vmax = √[Rg(μ + tanθ)/(1 − μ tanθ)] | Vmin = √[Rg(μ − tanθ)/(1 + μ tanθ)]
④ Death Well (vertical wall circular motion):
μ × N ≥ mg → μ × mω²R ≥ mg → ωmin = √(g/μR)
⑤ Normal Reaction at different points (circular track):
| Point | Normal Reaction |
|---|
| A (bottom of valley) | NA = mg + mv²/R |
| B (side — horizontal) | NB = mv²/R |
| C (top of hill) | NC = mg − mv²/R |
| D (inner top of loop) | ND = mv²/R − mg |
11
Vertical Circular Motion (VCM)
Key Concept
- Non-uniform circular motion in vertical plane
- Two forces act: Tension T and Gravity mg
- Speed is maximum at bottom and minimum at top
At angle θ from bottom: TP = mv²/R + mg cosθ
Speed at any point: v = √(v₀² + 2gR(cosθ − 1))
Tension at bottom (A): TA = mv₀²/R + mg
Tension at top (C): TC = mv₀²/R − 5mg
| Condition | String/Rod | vmin at Top | vmin at Bottom |
|---|
| Just complete VCM | String (T ≥ 0 at top) | √(gR) | √(5gR) |
| Just complete VCM | Massless Rod | 0 | √(4gR) |
Difference in Tension — Fixed Results
- TA − TB = 3mg (bottom − side)
- TA − TC = 6mg (bottom − top)
- TD − TC = 3mg (side − top)
- TB − TD = 0 (two sides are equal)
- These differences do NOT depend on given initial velocity
Behaviour Based on v₀ (Speed at Bottom)
- v₀ > √(5gR) → completes full VCM
- v₀ = √(5gR) → just completes VCM (T = 0 at top)
- √(2gR) < v₀ < √(5gR) → doesn't complete; leaves circular path between 90° to 180°
- v₀ = √(2gR) → reaches top with v = 0; critical case
- v₀ < √(2gR) → oscillates (SHM-like motion, pendulum)
- v₀ << √(2gR) → performs SHM
Conical Pendulum
- String makes angle θ with vertical; bob moves in horizontal circle
- T cosθ = mg; T sinθ = mω²r = mv²/r
- Speed: v = √(rg tanθ)
- Time period: T = 2π√(l cosθ / g) where l = length of string
- h = l cosθ → T = 2π√(h/g)
Q: Particle performs VCM with string of length R. If v at bottom = √(5gR), find tension at top, bottom and side.
v_top = √(v₀²−4gR) = √(5gR−4gR) = √(gR); T_C = m(gR)/R − mg = 0; T_A = m(5gR)/R + mg = 6mg; T_B = m(5gR−2gR)/R = 3mg
Ans: T_top = 0, T_bottom = 6mg, T_side = 3mg ✓
Q: A ball swings in vertical circle of radius R. Prove T_A − T_C = 6mg regardless of speed.
T_A = mv_A²/R + mg; T_C = mv_C²/R − mg; v_A² = v_C² + 4gR; T_A − T_C = m(v_A²−v_C²)/R + 2mg = m(4gR)/R + 2mg = 4mg + 2mg = 6mg ✓
Ans: T_A − T_C = 6mg (independent of speed) ✓