Motion in a Plane: NEET notes.

01 · NCERT §3.2

Scalars and vectors

• A scalar is fully given by a number with its unit: distance, mass, temperature, time, volume. Scalars combine by ordinary algebra.
• A vector has a magnitude and a direction and combines by the triangle law (equivalently the parallelogram law): displacement, velocity, acceleration and force are vectors. In print a vector is bold (v); by hand it is written with an arrow over it.
• The position vector r of a point is the arrow from a chosen origin O to that point. If an object moves from P to P′, its displacement is the vector PP′ = r′ − r, which depends only on the start and end points.
• The path length between two points is never less than the magnitude of the displacement between them: walking 40 m east and then 30 m north covers 70 m, but the displacement is only 50 m.
• A and B are equal vectors when both their sizes and their directions agree. A vector may be shifted parallel to itself without changing it (a free vector); some applications, like a force acting at a point, care about the line of action as well (a localised vector).
• The magnitude of a vector is written |A| or simply A; it is never negative.

02 · NCERT §3.3

Multiplying a vector by a number

• Multiplying a vector A by a positive number λ gives λA: same direction, magnitude λ|A|. A 5 m/s velocity east doubled is 10 m/s east.
• Multiplying by a negative number reverses the direction: −1 × (5 m/s east) is 5 m/s west, and −1.5A points opposite to A with 1.5 times its length.
• If λ carries a dimension, the product has a new dimension: a constant velocity (m/s) times a time (s) gives a displacement (m).
• −A has the same magnitude as A and the opposite direction; this is what makes subtraction possible.

03 · NCERT §3.4

Adding and subtracting vectors graphically

• Triangle (head-to-tail) method: place the tail of B at the head of A; the arrow from the tail of A to the head of B is R = A + B.
• Vector addition is commutative, A + B = B + A, and associative, (A + B) + C = A + (B + C).
• Adding A and −A gives the null (zero) vector 0, of zero magnitude and no fixed direction. A + 0 = A, λ0 = 0 and 0A = 0. A displacement that returns to its start is the null vector.
• Subtraction is addition of the reversed vector: A − B = A + (−B).
• Parallelogram method: draw A and B from a common tail and complete the parallelogram; the diagonal from that tail is A + B. It gives the same result as the triangle method.
• Rain falling vertically at 8 m/s in a wind blowing from east to west at 6 m/s moves at 10 m/s, at an angle whose tangent is 6/8 (about 37°) from the vertical, drifting west. To meet it, the umbrella is tilted about 37° from the vertical towards the east.

04 · NCERT §3.5

Resolving vectors into components

• Any vector A in a plane can be written as λa + μb along two chosen non-parallel vectors a and b; λa and μb are its components along them.
• Unit vectors î, ĵ, k̂ have magnitude 1, point along +x, +y and +z, and have no dimension or unit. Any vector A equals |A| n̂, where n̂ is the unit vector along A.
• A vector of magnitude A at angle θ to the x-axis has components Ax = A cos θ and Ay = A sin θ, so A = Ax î + Ay ĵ. A component can be positive, negative or zero.
• From the components back to the vector: A = √(Ax² + Ay²) and tan θ = Ay/Ax. A ball thrown at 20 m/s at 30° has components 17.3 m/s and 10.0 m/s.
• In three dimensions, Ax = A cos α, Ay = A cos β, Az = A cos γ, where α, β, γ are the angles with the axes, and A = √(Ax² + Ay² + Az²).
• The position vector of a point (x, y, z) is r = x î + y ĵ + z k̂.

05 · NCERT §3.6

Adding vectors analytically

• To add vectors, add their components: if R = A + B, then Rx = Ax + Bx, Ry = Ay + By (and Rz = Az + Bz). The same works for any number of vectors and for subtraction.
• For two vectors of magnitudes A and B with angle θ between them, R² = A² + B² + 2AB cos θ (the law of cosines).
• The direction of R from A is given by tan α = B sin θ / (A + B cos θ); the law of sines, R/sin θ = A/sin β = B/sin α, relates the sides and angles of the triangle.
• The resultant lies between |A − B| (θ = 180°) and A + B (θ = 0°); at θ = 90° it is √(A² + B²).
• 40 m and 30 m displacements at 60° to each other give R = √(1600 + 900 + 1200) ≈ 60.8 m, at about 25° to the 40 m leg.
• Worked with components, a boat heading north across a current adds its velocity through the water to the water's velocity; the ground velocity is the vector sum.

06 · NCERT §3.7

Velocity and acceleration in a plane

• The position vector is r = x î + y ĵ; as the object moves from r to r′ in Δt, the displacement is Δr = r′ − r.
• Average velocity is Δr/Δt, a vector along Δr. Instantaneous velocity is the limit v = dr/dt, with components vx = dx/dt and vy = dy/dt.
• The instantaneous velocity is always along the tangent to the path, in the direction of motion. Its magnitude is v = √(vx² + vy²) and its angle to x is given by tan θ = vy/vx.
• Average acceleration is Δv/Δt; instantaneous acceleration is a = dv/dt, with ax = dvx/dt = d²x/dt² and ay = dvy/dt.
• In one dimension velocity and acceleration lie on the same line; in a plane the angle between v and a can be anything from 0° to 180°.
• Example: r = 4.0t î + 1.5t² ĵ (metres, t in s) gives v = 4.0 î + 3.0t ĵ and a = 3.0 ĵ m/s²; at t = 1 s the speed is 5.0 m/s at about 37° to the x-axis.

07 · NCERT §3.8

Constant acceleration in a plane

• If a is constant, v = v₀ + a t and r = r₀ + v₀ t + ½ a t², as vectors.
• In components: x = x₀ + v₀ₓ t + ½ aₓ t² and y = y₀ + v₀ᵧ t + ½ aᵧ t², with vₓ = v₀ₓ + aₓ t and vᵧ = v₀ᵧ + aᵧ t.
• Motion in a plane with constant acceleration is two independent one-dimensional motions, along x and along y, sharing the same clock t.
• Example: starting at the origin with v₀ = 4.0 î m/s and a = 3.0 ĵ m/s², after 2 s the object is at 8.0 î + 6.0 ĵ m, 10 m from the start, moving at 4.0 î + 6.0 ĵ m/s (about 7.2 m/s).
• The velocity turns steadily towards the direction of a, while the component perpendicular to a never changes.

08 · NCERT §3.9

Projectile motion

• A projectile is an object that, once thrown, moves under gravity alone; air resistance is neglected. Its motion is a uniform horizontal motion and a uniformly accelerated vertical motion, independent of each other.
• Launched at speed v₀ at angle θ₀ above the horizontal: aₓ = 0, aᵧ = −g; x = (v₀ cos θ₀) t and y = (v₀ sin θ₀) t − ½ g t².
• The horizontal velocity v₀ cos θ₀ never changes; the vertical velocity vᵧ = v₀ sin θ₀ − g t falls by g every second and is zero at the top.
• The path is y = (tan θ₀) x − g x² / [2 (v₀ cos θ₀)²], a parabola.
• Time to the top tₘ = v₀ sin θ₀ / g; time of flight T_f = 2 v₀ sin θ₀ / g; maximum height hₘ = (v₀ sin θ₀)² / 2g; horizontal range R = v₀² sin 2θ₀ / g.
• The range is largest at θ₀ = 45°, Rₘ = v₀²/g. Angles 45° + α and 45° − α (complementary angles such as 30° and 60°) give the same range but different heights and times.
• A ball thrown at 20 m/s at 30° (g = 9.8 m/s²) rises 5.1 m, stays up 2.04 s and lands 35.3 m away.
• Thrown horizontally from a height h, a body falls for t = √(2h/g) whatever its horizontal speed: from 4.9 m it takes 1.0 s, and at 10 m/s it lands 10 m away moving at 14 m/s.

09 · NCERT §3.10

Uniform circular motion

• Uniform circular motion is motion on a circle at constant speed. The velocity is along the tangent and keeps changing direction, so the body is accelerated.
• The acceleration points towards the centre (centripetal acceleration) and has magnitude a_c = v²/R.
• a_c has constant magnitude but its direction keeps turning, always towards the centre, so it is not a constant vector; the constant-acceleration equations do not apply.
• Angular speed ω = Δθ/Δt (rad/s); v = R ω and a_c = ω² R.
• The time period T is the time for one revolution and the frequency ν = 1/T is revolutions per second; v = 2πRν, ω = 2πν and a_c = 4π²ν²R.
• Jogging at 5.0 m/s on a circle of radius 20 m: ω = 0.25 rad/s, one lap takes 2π/0.25 ≈ 25 s, and a_c = 25/20 = 1.25 m/s² towards the centre.
• If the speed is changing as well, the acceleration also has a component along the tangent and no longer points at the centre.

Revise

Must-know facts

1. 01Displacement magnitude ≤ path length; equal only for straight-line motion in one direction.
2. 02Unit vectors î, ĵ, k̂ have magnitude 1 and no unit or dimension.
3. 03Ax = A cos θ, Ay = A sin θ, where θ is measured from the x-axis.
4. 04R² = A² + B² + 2AB cos θ; tan α = B sin θ / (A + B cos θ).
5. 05|A − B| ≤ |A + B| ≤ A + B.
6. 06A − B = A + (−B); A + (−A) = 0, the null vector.
7. 07Instantaneous velocity is tangent to the path.
8. 08In a plane the angle between v and a can be anything from 0° to 180°.
9. 09Constant a: v = v₀ + at, r = r₀ + v₀t + ½at², component by component.
10. 10Projectile: horizontal velocity v₀ cos θ₀ constant; vertical acceleration −g.
11. 11T_f = 2v₀ sin θ₀/g; hₘ = v₀² sin² θ₀/2g; R = v₀² sin 2θ₀/g.
12. 12Maximum range v₀²/g at 45°; θ and 90° − θ give equal ranges.
13. 13The trajectory of a projectile is a parabola: y = x tan θ₀ − gx²/(2v₀² cos² θ₀).
14. 14Uniform circular motion: a_c = v²/R = ω²R = 4π²ν²R, towards the centre.
15. 15v = Rω; ω = 2πν = 2π/T.
16. 16Average speed ≥ magnitude of average velocity.

Where marks go

Common traps

Reference

Formulas

Components

Ax = A cos θ, Ay = A sin θ

θ measured from the x-axis.

Magnitude and direction

A = √(Ax² + Ay²), tan θ = Ay/Ax

In 3D add Az² under the root.

Resultant of two vectors

R² = A² + B² + 2AB cos θ

Law of cosines; θ is the angle between A and B.

Direction of the resultant

tan α = B sin θ / (A + B cos θ)

α measured from A.

Law of sines

R/sin θ = A/sin β = B/sin α

Angles of the vector triangle.

Constant acceleration

v = v₀ + at, r = r₀ + v₀t + ½at²

Applies to each component separately.

Projectile position

x = v₀ cos θ₀ t, y = v₀ sin θ₀ t − ½gt²

Launch from the origin; g = 9.8 m/s².

Projectile path

y = x tan θ₀ − gx² / (2v₀² cos² θ₀)

A parabola.

Time of flight

T_f = 2v₀ sin θ₀ / g

Time to the top is half of it.

Maximum height

hₘ = v₀² sin² θ₀ / 2g

Reached when vᵧ = 0.

Horizontal range

R = v₀² sin 2θ₀ / g

Maximum v₀²/g at 45°.

Centripetal acceleration

a_c = v²/R = ω²R = 4π²ν²R

Directed towards the centre.

Angular speed

v = Rω, ω = 2πν = 2π/T

ω in rad/s, ν in s⁻¹ (Hz).

Glossary

Key terms

Scalar

A quantity fully given by a number and a unit.

Vector

A quantity with magnitude and direction that adds by the triangle law.

Position vector

The arrow from the chosen origin to the point.

Displacement

The change in position vector, from the starting point to the end point.

Null vector

The vector of zero magnitude, the sum of a vector and its reverse.

Unit vector

A dimensionless vector of magnitude 1 that only marks a direction.

Resolution

Splitting a vector into components along chosen directions.

Projectile

A body moving under gravity alone after being launched.

Trajectory

The path a body follows; for a projectile, a parabola.

Horizontal range

The horizontal distance from launch to return to the launch level.

Centripetal acceleration

The acceleration towards the centre of a circle, v²/R in size.

Angular speed

The rate at which the angle swept about the centre changes, in rad/s.

Frequency

Number of revolutions per second, the reciprocal of the time period.