Equations of motion and free fall
A mango falls faster and faster; the same rule applies to every falling object. Link to earlier classes : the equations of motion and graphs of classes 9-10 now extend to two-dimensional motion with vectors and calculus.
Constant acceleration means velocity changes by the same amount every second.
In the figure : 5, 20, 45 m in 1, 2, 3 s.
Rule 1: Equations of motion — For constant acceleration : v = u + at; s = ut + ½at²; v² = u² + 2as; s_n = u + a(2n − 1) ÷ 2 (in the nth second).
- Easy : u = 10, a = 2, t = 5 ⇒ v = 20; s = 75 m.
- Medium : Braking : u = 20, a = −5 ⇒ s = u²/2|a| = 40 m; t = 4 s.
- Exam (CET) : v² = u² + 2as has no t; use it when t is unknown? (1) this equation (2) v = u + at (3) s = ut (4) none → answer 1)
Rule 2: Free fall — a = g = 10 m/s² downward. Dropped from rest : v = gt, h = ½gt², v² = 2gh, t = √(2h/g).
- Easy : From 80 m : t = √(2×80/10) = 4 s; v = 40 m/s.
- Medium : Distances in 1, 2, 3 s are 5, 20, 45 m; per second 5 : 15 : 25 = 1 : 3 : 5.
- Exam (CET) : Dropped stone in 2 s falls? (1) 20 m (2) 10 m (3) 40 m (4) 5 m → answer 1) Trick : 5t².
Rule 3: Ball thrown upward — a = −g. Maximum height H = u²/2g; time up u/g; total time 2u/g; returns with speed u (downward).
- Easy : u = 20 ⇒ H = 400/20 = 20 m; time up 2 s; total 4 s.
- Medium : At the top v = 0 but a = 10 m/s².
- Exam (CET) : Acceleration at the top? (1) g (2) 0 (3) −g/2 (4) infinite → answer 1)
Derivation : v² = u² + 2as
- Step 1 : v = u + at ⇒ t = (v − u) ÷ a.
- Step 2 : Average velocity = (u + v) ÷ 2 ⇒ s = (u + v) t ÷ 2.
- Step 3 : s = (u + v)(v − u) ÷ 2a = (v² − u²) ÷ 2a.
- Step 4 : v² = u² + 2as.
v = u + at; s = ut + ½at²; v² = u² + 2as; H = u²/2g.
Going up a = −g; fix a sign convention; the equations hold only for constant a.
Memory clue : "Falling : 5t²".
Spot the difference
| Equation | Missing | Use when |
|---|---|---|
| v = u + at | s | s unknown |
| s = ut + ½at² | v | v unknown |
| v² = u² + 2as | t | t unknown |
Practice questions
- 1Three equations for constant acceleration?
- 2a in free fall?
- 3v at the top?
- 4u = 0, a = 4, t = 6 : s?
- 5u = 15, v = 35, t = 5 : a?
- 6Dropped from 45 m : t?
- 7Thrown up at u = 30 : H and total time?
- 8Distance in the 3rd second of free fall?
Show answers (try them yourself first)
- 1v = u + at; s = ut + ½at²; v² = u² + 2as
- 2g = 10 m/s²
- 30
- 4s = 72 m
- 5a = 4 m/s²
- 6t = 3 s
- 7H = 45 m; 6 s
- 825 m
Motion graphs, average and instantaneous velocity
Total journey time gives a bus's average speed; the speedometer shows instantaneous speed.
Instantaneous velocity is the average velocity over a very small time; dx/dt.
In the figure : Triangle 25 m + rectangle 50 m = 75 m.
Motion graphs, average and instantaneous velocity
In the app you can move this figure with your finger.
Open in the appRule 1: x – t and v – t graphs — Slope of x – t = velocity; slope of v – t = acceleration; area under v – t = displacement.
- Easy : x 0 → 20 m in 4 s ⇒ velocity = 5 m/s.
- Medium : v 0 → 10 m/s (5 s), then constant 10 (5 s) ⇒ s = ½·5·10 + 5·10 = 75 m.
- Exam (CET) : A horizontal x – t line means? (1) object at rest (2) constant a (3) v increases (4) a increases → answer 1)
Rule 2: Average and instantaneous velocity — Average v = Δx ÷ Δt (displacement ÷ time); average speed = total distance ÷ time; instantaneous v = dx/dt.
- Easy : x = 3t² + 2t : from t = 2 to 3 s : average v = (33 − 16)/1 = 17 m/s; at t = 2, v = 14.
- Medium : Going 60 km/h, returning 40 km/h ⇒ average speed = 2·60·40/(60+40) = 48 km/h; average velocity 0.
- Exam (CET) : After returning to the start, average velocity? (1) 0 (2) equals speed (3) infinite (4) negative → answer 1)
Rule 3: Reading graphs — A horizontal v – t line ⇒ a = 0; a rising line ⇒ a > 0; a curve ⇒ a changes. An x – t curve bending upward ⇒ v increases.
- Easy : v – t line from (0, 0) to (4, 12) ⇒ a = 3 m/s².
- Medium : A parabolic x – t ⇒ constant a.
- Exam (CET) : Area under a v – t graph? (1) displacement (2) a (3) force (4) time → answer 1)
Derivation : Area under v – t = s
- Step 1 : A straight line from u to v forms a trapezium.
- Step 2 : Area = ½ (u + v) × t.
- Step 3 : Put v = u + at : ½ (2u + at) t.
- Step 4 : s = ut + ½at² = displacement.
v = dx/dt; a = dv/dt; area under v – t = displacement.
Average speed ≠ average velocity; for equal distances speed is the harmonic mean, not (v₁ + v₂)/2.
Memory clue : "x – t slope = v; v – t slope = a".
Spot the difference
| Graph | Slope | Area |
|---|---|---|
| x – t | velocity | — |
| v – t | acceleration | displacement |
| a – t | — | change in velocity |
Practice questions
- 1What does the x – t slope give?
- 2Average speed versus velocity?
- 3Formula for instantaneous velocity?
- 4x = 5t² : v at t = 3?
- 5v – t : 0 → 8 m/s in 2 s : s?
- 6Equal distances at 30 and 60 km/h : average speed?
- 7v = 4t : displacement 0 to 3 s?
- 8x = t³ : a at t = 2?
Show answers (try them yourself first)
- 1Velocity
- 2Speed = distance ÷ time; velocity = displacement ÷ time
- 3dx/dt
- 430 m/s
- 58 m
- 640 km/h
- 718 m
- 812 m/s²
Motion in two dimensions
A ball hit slantwise moves forward and upward at once — two motions together.
2-D motion is the combined result of two independent motions along x and y.
In the figure : Velocity components v_x and v_y.
Rule 1: Position, velocity, acceleration vectors — r = x î + y ĵ; v = dr/dt = v_x î + v_y ĵ; a = dv/dt. Motions along the two directions are independent.
- Easy : r = 3t î + 4t² ĵ ⇒ v = 3î + 8t ĵ; t = 1 ⇒ |v| = √(9 + 64) = 8.54 m/s.
- Medium : a = dv/dt = 8ĵ m/s².
- Exam (CET) : Path of r = 2t î + t² ĵ? (1) parabola (2) straight line (3) circle (4) point → answer 1) Trick : x = 2t, y = t² ⇒ y = x²/4.
Rule 2: Displacement and average velocity — Δr = (x₂ − x₁) î + (y₂ − y₁) ĵ; average velocity = Δr ÷ Δt.
- Easy : (1, 2) → (4, 6) ⇒ Δr = 3î + 4ĵ; |Δr| = 5 m.
- Medium : In 5 s ⇒ average velocity = 1 m/s (direction tanθ = 4/3).
- Exam (CET) : Magnitude of Δr for (0,0) → (6,8)? (1) 10 (2) 14 (3) 8 (4) 6 → answer 1)
Rule 3: 2-D motion with constant acceleration — v = u + a t (vector); r = u t + ½ a t². Solve each component separately.
- Easy : u = 4î, a = 2ĵ, t = 3 ⇒ v = 4î + 6ĵ; |v| = 7.21.
- Medium : r = 12î + 9ĵ (from the origin).
- Exam (CET) : If a = 0 the motion is? (1) uniform along a straight line (2) a circle (3) a parabola (4) rest → answer 1)
Derivation : Components of v = dr/dt
- Step 1 : r = x î + y ĵ; î, ĵ are constant.
- Step 2 : dr/dt = (dx/dt) î + (dy/dt) ĵ.
- Step 3 : So v_x = dx/dt, v_y = dy/dt.
- Step 4 : |v| = √(v_x² + v_y²); tanθ = v_y/v_x.
v = dr/dt; a = dv/dt; |v| = √(v_x² + v_y²).
The components are independent but share the same t; do not forget the √ for the magnitude.
Memory clue : "x and y separate, t is common".
Spot the difference
| Quantity | Formula | Components |
|---|---|---|
| Velocity | dr/dt | dx/dt, dy/dt |
| Acceleration | dv/dt | dv_x/dt, dv_y/dt |
| Magnitude | √(v_x²+v_y²) | — |
Practice questions
- 1Write r, v, a in 2-D motion.
- 2Why are the components independent?
- 3Formula for |v|?
- 4r = 2t î + t² ĵ : v at t = 2?
- 5(0,0) → (5, 12) : |Δr|?
- 6u = 3î, a = 4ĵ, t = 2 : v?
- 7r = 3t î + (4t − 5t²) ĵ : a?
- 8v = 6î + 8ĵ : |v| and direction?
Show answers (try them yourself first)
- 1r = x î + y ĵ; v = dr/dt; a = dv/dt
- 2A force along one axis does not affect the other
- 3√(v_x² + v_y²)
- 4v = 2î + 4ĵ
- 513 m
- 6v = 3î + 8ĵ
- 7a = −10ĵ
- 810 m/s; 53.1°
Relative velocity
In a train, a neighbouring train moving at your speed seems still — relative velocity zero.
Relative velocity is the velocity of one object as seen from another.
In the figure : v_b boat, v_r river, v resultant.
Relative velocity
In the app you can move this figure with your finger.
Open in the appRule 1: Relative velocity — Velocity of A relative to B : v_AB = v_A − v_B. Same direction : subtract; opposite : add.
- Easy : Cars at 20 and 15 m/s in the same direction ⇒ relative velocity 5 m/s.
- Medium : In opposite directions ⇒ 20 + 15 = 35 m/s.
- Exam (CET) : Relative speed of head-on vehicles? (1) sum (2) difference (3) 0 (4) product → answer 1)
Rule 2: River–boat : shortest time — Point the boat perpendicular to the flow : t = d ÷ v_b; drift = v_r t; resultant speed = √(v_b² + v_r²).
- Easy : v_b = 4, v_r = 3, d = 200 m ⇒ t = 50 s; drift = 150 m; v = 5 m/s.
- Medium : A wider river ⇒ both t and drift increase.
- Exam (CET) : For minimum time the boat's heading is? (1) perpendicular to the flow (2) upstream (3) downstream (4) slanted → answer 1)
Rule 3: River–boat : shortest path — To land straight across, point slightly upstream : sinθ = v_r ÷ v_b; net speed = √(v_b² − v_r²).
- Easy : v_b = 5, v_r = 3 ⇒ sinθ = 0.6 ⇒ θ = 36.9°; net speed = 4 m/s.
- Medium : d = 200 m ⇒ t = 200/4 = 50 s.
- Exam (CET) : This is possible when? (1) v_b > v_r (2) v_b < v_r (3) v_b = 0 (4) always → answer 1)
Derivation : v_AB = v_A − v_B
- Step 1 : A is at x_A, B at x_B; relative to B, x_AB = x_A − x_B.
- Step 2 : Differentiate with t : dx_AB/dt = dx_A/dt − dx_B/dt.
- Step 3 : So v_AB = v_A − v_B.
- Step 4 : v_BA = −v_AB.
v_AB = v_A − v_B; t = d/v_b; sinθ = v_r/v_b.
Minimum time and minimum path differ; add for opposite directions, do not subtract.
Memory clue : "Same = minus; opposite = plus".
Spot the difference
| Case | Relative velocity |
|---|---|
| Same direction | v₁ − v₂ |
| Opposite | v₁ + v₂ |
| Perpendicular | √(v₁² + v₂²) |
Practice questions
- 1What is relative velocity?
- 2Relation between v_AB and v_BA?
- 3Heading for minimum time?
- 460 and 45 km/h same direction : relative?
- 5v_b = 6, v_r = 2, d = 120 : t and drift?
- 6v_b = 10, v_r = 6 : θ to go straight?
- 7Opposite 54 and 36 km/h (relative in m/s)?
- 8v_b = 5, v_r = 3 : time to go 200 m straight?
Show answers (try them yourself first)
- 1Velocity of one as seen from the other
- 2v_BA = −v_AB
- 3Perpendicular to the flow
- 415 km/h
- 5t = 20 s; drift 40 m
- 636.9° upstream
- 725 m/s
- 850 s
Projectile motion
A high cricket shot : the ball rises, travels forward and comes down along a curve.
A projectile is a thrown object acted on only by gravity; its path is a parabola.
In the figure : Blue path; H = height, R = range.
Projectile motion
In the app you can move this figure with your finger.
Open in the appRule 1: Oblique projectile — Thrown with u at angle θ : u_x = u cosθ (constant), u_y = u sinθ. T = 2u sinθ ÷ g; H = u² sin²θ ÷ 2g; R = u² sin2θ ÷ g.
- Easy : u = 20, θ = 30° ⇒ T = 2 s; H = 5 m; R = 34.64 m.
- Medium : θ = 45° ⇒ R_max = u²/g = 40 m; θ and (90° − θ) give equal R.
- Exam (CET) : Maximum range at which angle? (1) 45° (2) 30° (3) 60° (4) 90° → answer 1) Trick : sin2θ = 1.
Rule 2: Equation of the path — y = x tanθ − g x² ÷ (2u² cos²θ) ⇒ a parabola. At the top v = u cosθ (horizontal).
- Easy : θ = 45° : y = x − g x²/u².
- Medium : Speed at the top u cosθ = 17.3 m/s (u = 20, θ = 30°).
- Exam (CET) : Acceleration at the top? (1) g downward (2) 0 (3) g upward (4) u cosθ → answer 1)
Rule 3: Horizontal projectile — From height h with horizontal speed u : T = √(2h ÷ g); R = u T; v_y = gT on landing.
- Easy : h = 80, u = 15 ⇒ T = 4 s; R = 60 m.
- Medium : T depends on h only, not on u.
- Exam (CET) : Two balls from the same height, one dropped, one thrown horizontally; time of fall? (1) equal (2) thrown longer (3) dropped longer (4) unknown → answer 1)
Derivation : R = u² sin2θ ÷ g
- Step 1 : Vertical motion : at y = 0, T = 2u sinθ ÷ g.
- Step 2 : Horizontal motion : R = (u cosθ) × T.
- Step 3 : R = u cosθ × 2u sinθ ÷ g = u² (2 sinθ cosθ) ÷ g.
- Step 4 : R = u² sin2θ ÷ g.
T = 2u sinθ/g; H = u² sin²θ/2g; R = u² sin2θ/g; R_max = u²/g.
Horizontal velocity is constant, vertical changes; do not mix sin2θ and sin²θ; air resistance is ignored.
Memory clue : "R : double angle; H : squared".
Spot the difference
| Quantity | Formula | Maximum / special |
|---|---|---|
| T | 2u sinθ/g | max at 90° |
| H | u² sin²θ/2g | max at 90° |
| R | u² sin2θ/g | max at 45° |
Practice questions
- 1What is a projectile?
- 2Shape of the path?
- 3Formula for R_max?
- 4u = 30, θ = 30° : T?
- 5u = 40, θ = 45° : R?
- 6h = 45, u = 10 horizontal : T, R?
- 7u = 20, θ = 60° : H?
- 8θ = 30° and 60° : ratio of R?
Show answers (try them yourself first)
- 1A thrown object acted on only by gravity
- 2Parabola
- 3u²/g
- 4T = 3 s
- 5R = 160 m
- 6T = 3 s; R = 30 m
- 7H = 15 m
- 81 : 1
Uniform circular motion
A child on a merry-go-round moves at a steady speed, yet the direction keeps changing.
Uniform circular motion is movement along a circle at constant speed.
In the figure : v is along the tangent; ω = 2π/T.
Rule 1: Angular quantities — θ = s ÷ r; ω = dθ/dt = 2π ÷ T = 2π f; v = ω r; T = 1 ÷ f. Unit of ω is rad/s.
- Easy : r = 0.5 m, f = 2 Hz ⇒ ω = 4π = 12.57 rad/s; v = 6.28 m/s.
- Medium : Second hand of a clock : ω = 2π/60 = 0.1047 rad/s.
- Exam (CET) : Angle in one revolution? (1) 2π rad (2) π rad (3) 1 rad (4) 360 rad → answer 1)
Rule 2: What makes it uniform — Speed is constant but direction changes, so velocity (a vector) changes ⇒ there is acceleration. v is tangent, perpendicular to r.
- Easy : m = 2 kg moving at 5 m/s in a circle : kinetic energy stays 25 J.
- Medium : Velocity changes, speed does not.
- Exam (CET) : In uniform circular motion what is constant? (1) speed (2) velocity (3) acceleration (4) momentum → answer 1)
Rule 3: Earth's motion — About its axis : T = 24 h ⇒ ω = 2π/86400 = 7.27 × 10⁻⁵ rad/s. At the equator v = ω R = 465 m/s (R = 6400 km).
- Easy : ω = 7.27 × 10⁻⁵ rad/s.
- Medium : v = 465 m/s.
- Exam (CET) : Earth's T (about its axis)? (1) 24 h (2) 12 h (3) 365 d (4) 1 h → answer 1)
Derivation : v = ω r
- Step 1 : Arc s = r θ (θ in radians).
- Step 2 : Differentiate with t : ds/dt = r dθ/dt.
- Step 3 : ds/dt = v and dθ/dt = ω.
- Step 4 : v = ω r.
ω = 2π/T = 2πf; v = ω r; T = 1/f.
Speed is constant but velocity changes; use θ in radians; do not mix f and ω.
Memory clue : "ω = 2π × f".
Spot the difference
| Quantity | Formula | Unit |
|---|---|---|
| ω | 2π/T = 2πf | rad/s |
| v | ω r | m/s |
| T | 1/f | s |
| f | 1/T | Hz |
Practice questions
- 1What is ω?
- 2Relation of T and f?
- 3What is uniform circular motion?
- 4f = 5 Hz : ω?
- 5r = 2 m, ω = 3 rad/s : v?
- 6T = 0.5 s : f and ω?
- 7Minute hand : ω?
- 8v = 20 m/s, r = 5 m : T?
Show answers (try them yourself first)
- 1Angular velocity dθ/dt
- 2T = 1/f
- 3Motion in a circle at constant speed
- 4ω = 31.4 rad/s
- 5v = 6 m/s
- 6f = 2 Hz; ω = 12.57 rad/s
- 7ω = 1.75 × 10⁻³ rad/s
- 8T = 1.57 s
Centripetal acceleration and force
On a turning bus we lean outward; the real force acts inward.
Centripetal force is the force toward the centre needed for circular motion.
In the figure : a_c points to the centre, v along the tangent.
Rule 1: Centripetal acceleration — a_c = v² ÷ r = ω² r, directed to the centre; perpendicular to v.
- Easy : v = 10, r = 5 ⇒ a_c = 20 m/s².
- Medium : ω = 4, r = 0.5 ⇒ a_c = 8 m/s².
- Exam (CET) : v doubles, r same ⇒ a_c? (1) four times (2) double (3) half (4) same → answer 1)
Rule 2: Centripetal force — F = m v² ÷ r = m ω² r. It is not a separate force; tension, friction, gravity or a normal force supplies it. Centrifugal force is a pseudo force (in a rotating frame).
- Easy : m = 2 kg, v = 10, r = 5 ⇒ F = 40 N.
- Medium : For a stone whirled on a string, tension is the centripetal force.
- Exam (CET) : What supplies the centripetal force for the Moon? (1) gravity (2) tension (3) friction (4) magnetism → answer 1)
Rule 3: Vehicle on a curve — Flat road : friction = centripetal force ⇒ μ m g = m v² ÷ r ⇒ v_max = √(μ r g).
- Easy : μ = 0.5, r = 20 m ⇒ v_max = √(0.5 × 20 × 10) = 10 m/s.
- Medium : r four times ⇒ v_max doubles.
- Exam (CET) : v_max does not depend on? (1) mass (2) μ (3) r (4) g → answer 1)
Derivation : a_c = v² ÷ r
- Step 1 : Velocities v₁, v₂ (magnitude v) with angle Δθ between them; Δv = v Δθ.
- Step 2 : Arc Δs = r Δθ ⇒ Δθ = Δs ÷ r.
- Step 3 : a = Δv ÷ Δt = v (Δs ÷ r) ÷ Δt = v (Δs/Δt) ÷ r.
- Step 4 : Δs/Δt = v ⇒ a_c = v² ÷ r, toward the centre.
a_c = v²/r = ω²r; F = m v²/r; v_max = √(μ r g).
Centripetal force is not a new force; centrifugal force is not a real force; v_max does not depend on m.
Memory clue : "Inward = real; outward = apparent".
Spot the difference
| Quantity | Formula | Direction |
|---|---|---|
| a_c | v²/r = ω²r | to the centre |
| F | m v²/r | to the centre |
| v_max (flat) | √(μ r g) | — |
Practice questions
- 1Direction of centripetal acceleration?
- 2Formula for centripetal force?
- 3Is centrifugal force real?
- 4v = 6, r = 3 : a_c?
- 5m = 0.5, v = 4, r = 2 : F?
- 6μ = 0.4, r = 10 : v_max?
- 7m = 1 kg, r = 1 m, T = 2 s : F?
- 8Maximum tension 50 N, m = 1, r = 2 : v_max?
Show answers (try them yourself first)
- 1Toward the centre
- 2m v²/r
- 3No; a pseudo force
- 412 m/s²
- 54 N
- 6v_max = 6 m/s
- 7ω = π ⇒ F = 9.87 N
- 810 m/s
Quick revision of the whole lesson
| Concept | Key formula / fact | Unit / dimensions |
|---|---|---|
| Motion | v = u + at; s = ut + ½at²; v² = u² + 2as | m/s; m |
| Free fall | h = ½gt²; v = gt; H = u²/2g | m |
| Relative velocity | v_AB = v_A − v_B | m/s |
| River–boat | t = d/v_b; sinθ = v_r/v_b | s |
| Projectile | T = 2u sinθ/g; R = u² sin2θ/g | s; m |
| Angular velocity | ω = 2π/T; v = ω r | rad/s |
| Centripetal | a = v²/r; F = mv²/r | m/s²; N |
| Curve | v_max = √(μ r g) | m/s |
🎯 Most important points for the exam
- x – t slope = v; v – t slope = a; area = displacement.
- a = g = 10 downward; H = u²/2g.
- Average speed ≠ average velocity.
- Head-on : add; same direction : subtract.
- R_max at 45°; equal R for θ and 90° − θ.
- In a circle speed is constant, velocity changes.
- Centripetal force points inward; centrifugal is apparent.
Constant a ⇒ equations; 2-D ⇒ separate x, y; circle ⇒ ω, v = ω r, F = mv²/r.