Scalars and vectors, position and displacement
A delivery rider told "go 5 km" is lost; told "go 5 km towards the north-east" he can reach the address. Link to earlier classes : motion along a line used only + and − signs; in a plane, direction needs more than a sign.
A scalar has magnitude only (distance, speed, mass, time, temperature, energy). A vector has magnitude and direction and follows the rules of vector addition (displacement, velocity, acceleration, force, momentum). A vector is written in bold or with an arrow; its magnitude is |A| or A.
Rule 1: Position and displacement vectors — The position vector of a point (x, y) is r = xî + yĵ, drawn from the origin; the displacement from r₁ to r₂ is Δr = r₂ − r₁. Two vectors are equal if they have the same magnitude and direction, wherever they are drawn.
- Easy : Position changes from (2, 3) to (7, 15) ⇒ Δr = (5î + 12ĵ) m, magnitude √(25 + 144) = 13 m.
- Medium : A drone flies 30 m east then 40 m north. Path length = 70 m; |Δr| = √(30² + 40²) = 50 m.
- Exam : A rider goes half-way round a circular track of radius 7 m. Path length = πr = 22 m (approx.); displacement = the diameter = 14 m.
Rule 2: Magnitude and direction — The magnitude of a displacement is never more than the path length. A quantity with a direction is a vector only if it also obeys vector addition (electric current has a direction but is a scalar, since currents add as numbers).
- Easy : Mass, temperature and time are scalars.
- Medium : Force, velocity and momentum are vectors.
- Exam : Speed is a scalar, velocity a vector; a car going round a circle at constant speed has a changing velocity.
a vector needs both "how much" and "which way"; displacement ≤ path length.
Treating any quantity that "has a direction" as a vector, and adding vector magnitudes like numbers.
Memory clue : "Scalar — only a size; vector — a size and an arrow."
Spot the difference
| Property | Scalar | Vector |
|---|---|---|
| Described by | magnitude | magnitude and direction |
| Examples | distance, speed, mass | displacement, velocity, force |
| Adds by | ordinary arithmetic | vector rules |
Multiplying a vector by a number; adding and subtracting vectors
A boat crosses a river while the current pushes it sideways; where it lands depends on both. Link to earlier classes : parallelogram of forces from earlier classes is now used for every vector.
Adding vectors means finding one vector, the resultant, that has the same effect as the others together.
In the figure : B is drawn starting at the head of A; the resultant R joins the tail of A to the head of B.
Rule 1: Multiplication by a real number — λA has magnitude |λ| times that of A. It points the same way if λ is positive and the opposite way if λ is negative. −A has the same length as A but opposite direction.
- Easy : 3 × (4 m east) = 12 m east.
- Medium : −2 × (4 m east) = 8 m west.
- Exam : A velocity of 5 m s⁻¹ towards north, multiplied by 0.4, becomes 2 m s⁻¹ towards north.
Rule 2: Adding and subtracting graphically — Head-to-tail (triangle) method : place the tail of B at the head of A; R = A + B runs from the tail of A to the head of B. Parallelogram method : draw both from the same point; the diagonal is R. Addition is commutative (A + B = B + A) and associative ((A + B) + C = A + (B + C)). Subtraction : A − B = A + (−B). The zero vector 0 has zero magnitude, and A + (−A) = 0.
- Easy : 6 m and 8 m in the same direction ⇒ resultant 14 m; in opposite directions ⇒ 2 m.
- Medium : 6 m east and 8 m north (at 90°) ⇒ R = √(6² + 8²) = 10 m.
- Exam : Two vectors of 5 units each at 120° to each other ⇒ R² = 25 + 25 + 2 × 25 × cos 120° = 50 − 25 = 25 ⇒ R = 5 units.
Rule 3: Magnitude of the resultant — For vectors A and B with angle θ between them : R² = A² + B² + 2AB cos θ. Hence the resultant lies between the difference and the sum of the magnitudes.
- Easy : θ = 0° gives R = A + B (largest); θ = 180° gives R = |A − B| (smallest).
- Medium : A = 3, B = 4 at 90° ⇒ R = 5.
- Exam : Can vectors of 3, 4 and 9 units add to zero? The largest possible sum of 3 and 4 is 7 < 9, so no.
A + B = B + A; A − B = A + (−B); the resultant is drawn from the first tail to the last head.
Adding magnitudes directly (3 + 4 = 7 even for perpendicular vectors), and drawing the second vector from the tail instead of the head.
Memory clue : "Head to tail, then close the triangle." For 90° use Pythagorean triples : 3-4-5, 5-12-13, 8-15-17.
Spot the difference
| Property | Triangle method | Parallelogram method |
|---|---|---|
| Drawing | head of A to tail of B | both from one point |
| Resultant | joins first tail to last head | diagonal from the common point |
| Best for | adding many vectors | two vectors |
Resolution of vectors, unit vectors and analytical addition
Pulling a wheeled suitcase by its handle at an angle : one part of your pull moves it forward, another part lifts it up. Link to earlier classes : trigonometry of class 10 (sin, cos, tan) now splits every vector.
Resolving a vector means writing it as the sum of two perpendicular components. A unit vector has magnitude 1 and shows only direction : î, ĵ, k̂ along the x, y, z axes.
In the figure : A is split into Aₓ along the x-axis and A_y along the y-axis; θ is measured from the x-axis.
Rule 1: Components — Aₓ = A cos θ, A_y = A sin θ, with A = Aₓî + A_yĵ. Magnitude A = √(Aₓ² + A_y²) and direction tan θ = A_y ÷ Aₓ. The unit vector along A is n̂ = A ÷ |A|.
- Easy : A = 10 units at 30° ⇒ Aₓ = 10 cos 30° = 8.66; A_y = 10 sin 30° = 5.
- Medium : The unit vector along 3î + 4ĵ is (3î + 4ĵ) ÷ 5 = 0.6î + 0.8ĵ.
- Exam : A 20 N force acts at 60° above the horizontal. Horizontal part = 20 cos 60° = 10 N; vertical part = 20 sin 60° = 17.3 N.
Rule 2: Analytical addition — Add the components separately : Rₓ = Aₓ + Bₓ, R_y = A_y + B_y. Then R = √(Rₓ² + R_y²) and tan θ = R_y ÷ Rₓ.
- Easy : (2î + 3ĵ) + (4î + 1ĵ) = 6î + 4ĵ.
- Medium : A = 3î + 4ĵ and B = 2î − 6ĵ ⇒ R = 5î − 2ĵ; |R| = √29 = 5.39; angle = tan⁻¹(−2 ÷ 5) = −21.8°.
- Exam : A − B for the same vectors = (3 − 2)î + (4 + 6)ĵ = î + 10ĵ; magnitude = √101 = 10.05.
Aₓ = A cos θ, A_y = A sin θ when θ is measured from the x-axis; add components separately.
Using cos for the y-component when the angle is measured from the y-axis, and losing signs of components in the 2nd, 3rd and 4th quadrants.
Memory clue : "Cos goes along the angle, sin goes across." The component next to the angle uses cos, the one opposite uses sin.
Spot the difference
| Method | Graphical | Analytical (components) |
|---|---|---|
| Needs | scale drawing | trigonometry |
| Accuracy | limited by drawing | exact |
| Works for many vectors | clumsy | easy |
Velocity and acceleration in a plane
A cricket ball rises, curves and falls; at every instant it moves in a slightly different direction. Link to earlier classes : v = dx ÷ dt and a = dv ÷ dt of the previous chapter now apply to each component.
In a plane, position is r = xî + yĵ. Average velocity = Δr ÷ Δt; instantaneous velocity v = dr ÷ dt = vₓî + v_yĵ with vₓ = dx ÷ dt and v_y = dy ÷ dt. Acceleration a = dv ÷ dt = aₓî + a_yĵ. The velocity is always along the tangent to the path.
Rule 1: Component form — Motion in a plane is two independent straight-line motions along two perpendicular axes. Speed = |v| = √(vₓ² + v_y²).
- Easy : Position changes from (1, 2) m to (5, 5) m in 2 s ⇒ average velocity = (2î + 1.5ĵ) m s⁻¹, speed 2.5 m s⁻¹.
- Medium : Velocity changes from 3î to 3î + 8ĵ in 4 s ⇒ a = 2ĵ m s⁻².
- Exam : x = 2t, y = t² (m). Then vₓ = 2, v_y = 2t. At t = 3 s : v = 2î + 6ĵ, |v| = √40 = 6.32 m s⁻¹; direction tan θ = 3 ⇒ 71.6° from the x-axis; a = 2ĵ m s⁻².
Rule 2: Acceleration and direction — Acceleration need not point along velocity. It may change the speed, the direction, or both.
- Easy : Along the velocity ⇒ only the speed changes.
- Medium : Perpendicular to the velocity ⇒ only the direction changes (circular motion).
- Exam : Gravity on a thrown ball acts downwards while the ball moves sideways and upwards; both speed and direction change.
v = dr ÷ dt is tangent to the path; a = dv ÷ dt; treat the x and y motions separately.
Thinking acceleration always points along the motion, and mixing x and y data in one formula.
Memory clue : "Velocity touches the path; acceleration says where the velocity is being pulled."
Spot the difference
| Quantity | Average | Instantaneous |
|---|---|---|
| Velocity | Δr ÷ Δt | dr ÷ dt (tangent) |
| Acceleration | Δv ÷ Δt | dv ÷ dt |
Motion in a plane with constant acceleration
A ball kicked off a cliff edge moves forward and drops at the same time, with a steady downward acceleration. Link to earlier classes : the three equations of motion of the previous chapter are now applied to x and y separately.
When the acceleration a is constant, the velocity and position at time t follow the vector form of the kinematic equations.
Rule 1: Vector equations — v = v₀ + at and r = r₀ + v₀t + ½at². In components : x = x₀ + v₀ₓt + ½aₓt²; y = y₀ + v₀_yt + ½a_yt²; vₓ = v₀ₓ + aₓt; v_y = v₀_y + a_yt.
- Easy : From the origin, v₀ = 4î m s⁻¹ and a = 3ĵ m s⁻². At t = 2 s : x = 8 m, y = ½ × 3 × 4 = 6 m ⇒ r = 8î + 6ĵ, |r| = 10 m; v = 4î + 6ĵ, speed = 7.21 m s⁻¹.
- Medium : r₀ = (2, 1) m, v₀ = (1, 2) m s⁻¹, a = (2, −1) m s⁻². At t = 4 s : x = 2 + 4 + ½ × 2 × 16 = 22 m; y = 1 + 8 − ½ × 16 = 1 m; v = (1 + 8, 2 − 4) = (9, −2) m s⁻¹.
- Exam : When does v_y become zero in the previous case? 2 − t = 0 ⇒ t = 2 s.
v = v₀ + at and r = r₀ + v₀t + ½at² hold as vector equations for constant acceleration.
Using these equations for circular motion, where the direction of acceleration keeps changing.
Memory clue : "Two one-dimensional problems, one clock." Solve x and y separately; the same t connects them.
Spot the difference
| Property | Along x | Along y |
|---|---|---|
| Acceleration | aₓ | a_y |
| Velocity | vₓ = v₀ₓ + aₓt | v_y = v₀_y + a_yt |
| Position | x = x₀ + v₀ₓt + ½aₓt² | y = y₀ + v₀_yt + ½a_yt² |
Projectile motion
A basketball thrown at the hoop, water from a hose, a cricket ball hit in the air — each follows a curved path. Link to earlier classes : free fall of the previous chapter is combined with steady sideways motion.
A projectile is an object in flight after being thrown, acted on by gravity alone (air resistance ignored). Its acceleration is g downwards; sideways, there is no acceleration.
In the figure : The path is a parabola; at the top the velocity is horizontal; H is the maximum height and R the range.
Rule 1: Equations of a projectile — With launch speed v₀ at angle θ₀ : x = (v₀ cos θ₀) t; y = (v₀ sin θ₀) t − ½gt²; vₓ = v₀ cos θ₀ (constant); v_y = v₀ sin θ₀ − gt. The path is y = x tan θ₀ − g x² ÷ (2 v₀² cos² θ₀).
- Easy : v₀ = 20 m s⁻¹ at 30° : v₀ₓ = 20 cos 30° = 17.3 m s⁻¹; v₀_y = 20 sin 30° = 10 m s⁻¹.
- Medium : After 1 s : x = 17.3 m; y = 10 − 4.9 = 5.1 m; v_y = 10 − 9.8 = 0.2 m s⁻¹.
- Exam : For the same throw, the velocity at the top is vₓ = 17.3 m s⁻¹ horizontally; v_y = 0 but a = 9.8 m s⁻² downwards.
Rule 2: Height, time and range — H = (v₀ sin θ₀)² ÷ 2g; time to top tₘ = v₀ sin θ₀ ÷ g; time of flight T = 2v₀ sin θ₀ ÷ g; range R = v₀² sin 2θ₀ ÷ g. The range is greatest at 45°, and angles θ₀ and (90° − θ₀) give the same range.
- Easy : v₀ = 20 m s⁻¹, θ₀ = 30° : T = 2 × 10 ÷ 9.8 = 2.04 s; H = 100 ÷ 19.6 = 5.10 m; R = 400 × sin 60° ÷ 9.8 = 35.3 m.
- Medium : v₀ = 14 m s⁻¹ at 45° : R = 196 ÷ 9.8 = 20 m; H = (14 sin 45°)² ÷ 19.6 = 98 ÷ 19.6 = 5 m.
- Exam : For v₀ = 20 m s⁻¹, 30° and 60° give the same range 35.3 m, but 60° reaches H = (20 sin 60°)² ÷ 19.6 = 15.3 m, three times the height of the 30° throw.
Rule 3: Horizontal projection — An object projected horizontally from height h with speed u takes t = √(2h ÷ g) to land and travels a horizontal distance u t.
- Easy : h = 4.9 m ⇒ t = √(2 × 4.9 ÷ 9.8) = 1 s.
- Medium : With u = 6 m s⁻¹ it lands 6 × 1 = 6 m away.
- Exam : Landing speed = √(6² + 9.8²) = 11.5 m s⁻¹ (v_y = g t = 9.8 m s⁻¹ downwards).
horizontal velocity stays constant; vertical motion is free fall; the path is a parabola; R is maximum at 45°.
Using v₀ instead of its components, and believing that the acceleration is zero at the top (it is still g).
Memory clue : "Across — nothing changes; up and down — gravity rules; one clock links both." For range remember "sin 2θ — 45° is best, 30° and 60° tie."
Spot the difference
| Property | Horizontal motion | Vertical motion |
|---|---|---|
| Acceleration | 0 | g downwards |
| Velocity | constant | changes by g every second |
| Position | x = v₀ cos θ₀ · t | y = v₀ sin θ₀ · t − ½gt² |
| At the top | vₓ = v₀ cos θ₀ | v_y = 0 |
Uniform circular motion
The tip of a ceiling fan blade, a toy car on a round track, a seat on a giant wheel — each moves at constant speed in a circle, yet is always being pulled inwards. Link to earlier classes : circles of class 9 and the idea of changing velocity of this chapter join here.
In uniform circular motion an object moves along a circle of radius R with constant speed. The velocity keeps changing direction, so there is an acceleration towards the centre, called the centripetal acceleration.
In the figure : The velocity v is tangent to the circle; the acceleration a_c points to the centre.
Rule 1: Angular speed and relations — Angular speed ω = Δθ ÷ Δt (rad s⁻¹); v = ωR; ω = 2π ÷ T = 2πν (T = period, ν = frequency). Centripetal acceleration a_c = v² ÷ R = ω²R = 4π²ν²R, always towards the centre.
- Easy : R = 2 m, T = 4 s : v = 2πR ÷ T = π = 3.14 m s⁻¹; ω = 2π ÷ 4 = 1.57 rad s⁻¹; a_c = v² ÷ R = 4.93 m s⁻².
- Medium : A wheel of radius 0.5 m turns at 120 rpm. ν = 2 Hz; ω = 4π = 12.6 rad s⁻¹; v = ωR = 6.28 m s⁻¹; a_c = ω²R = 79.0 m s⁻².
- Exam : A satellite in a circle of radius 6.8 × 10⁶ m with period 93 min (5580 s) : v = 2π × 6.8 × 10⁶ ÷ 5580 = 7.66 × 10³ m s⁻¹; a_c = v² ÷ R = 8.6 m s⁻².
Rule 2: Speed constant, velocity not — The magnitude of a_c is constant but its direction changes every instant, so the equations for constant acceleration do not apply. a_c is at right angles to v, so it changes only direction, not speed.
- Easy : A car going round a bend of radius 50 m at 10 m s⁻¹ : a_c = 100 ÷ 50 = 2 m s⁻².
- Medium : Doubling the speed on the same bend makes a_c four times larger (v²).
- Exam : For the same speed, halving the radius doubles a_c.
v = ωR; ω = 2π ÷ T; a_c = v² ÷ R = ω²R, towards the centre.
Saying that acceleration is zero because the speed is constant, and pointing a_c along the velocity.
Memory clue : "v along the tangent, a to the centre, 90° between them." Use "2-pi-nu" for ω and "a = ω v" as a quick check (since a_c = ω²R = ω × ωR).
Spot the difference
| Straight-line quantity | Circular counterpart | Link |
|---|---|---|
| distance x | angle θ | s = Rθ |
| speed v | angular speed ω | v = ωR |
| time period | T = 2π ÷ ω | ν = 1 ÷ T |