Kinematics & Vectors

Kinematics is where JEE Advanced sets its first traps: relative motion, river-boat and rain-man problems, projectile on inclined planes, and constraint relations that couple two bodies through a string or wedge. The maths is light — the challenge is choosing the right frame and resolving vectors along the correct axes.

Key results

• Relative velocity: v(A rel B) = v(A) − v(B). To cross a river of width d in minimum time, head straight across (t = d/v_boat); for shortest path, aim upstream so the resultant is perpendicular.
• Projectile: range R = u²sin2θ/g, max height H = u²sin²θ/2g, time of flight T = 2u sinθ/g. On an incline (angle α), range up the plane = 2u²sin(θ−α)cosθ / (g cos²α).
• Constraint relation: components of velocity of two points along the connecting string are equal (string inextensible). Differentiate the length equation to link accelerations.

Quantity	Formula	Note
Range (max)	u²/g at θ=45°	flat ground
H = R/4	when θ=45°	height-range link
Two angles, same R	θ and 90°−θ	complementary
River min-time	d/v_boat	head straight

Exam Tricks & Tips

• 🎯 Two complementary angles give the same range — if a problem offers 30° and 60°, ranges are equal; use it to skip algebra.
• 🎯 Work in the frame of one body for relative-motion problems; collision/closest-approach becomes a straight-line geometry question.
• 🎯 For projectile on an incline, rotate axes so x is along the plane and g has two components (g sinα, g cosα). Landing means displacement perpendicular to plane = 0.
• 🎯 Constraint trick: the velocity of a block equals the rate at which the string shortens on its side — project velocities onto the string direction.
• 🎯 Average velocity ≠ average speed unless motion is 1-D without reversal; Advanced loves this distinction in graph questions.
• ❌ Common mistake: adding times for up-river and down-river trips using the same speed — the effective speed differs (v−u vs v+u), so compute each leg separately.

Expected exam pattern

1–2 questions, often a multi-concept projectile-plus-relative-motion or a constraint problem feeding into dynamics. Frequently paired with graphs (v–t, a–t) in the "list-match" or integer format.

Quick recap

Pick a smart frame, resolve vectors along natural axes, and use constraint relations to couple bodies. Remember complementary-angle range equality and the incline range formula. Distinguish average speed from average velocity.

Cover the answer, recall, then check. 12 cards on the must-know kinematics facts for JEE Advanced.

Q1. Two projectiles launched at 25° and 65° with equal speed — how do their ranges compare?
A1. Equal. Angles are complementary (sum 90°), so u²sin2θ/g is identical.

Q2. River width d, boat speed v — condition for minimum crossing time?
A2. Head straight across (velocity perpendicular to bank); t = d/v, independent of current.

Q3. Relation between max height H and range R at θ = 45°.
A3. H = R/4.

Q4. Define relative velocity of A with respect to B.
A4. v(A rel B) = v(A) − v(B); B is treated as stationary.

Q5. On a projectile, when is speed minimum?
A5. At the top, where vertical velocity is zero; speed = u cosθ (horizontal component).

Q6. Why resolve along/perpendicular to an inclined plane for projectile-on-incline?
A6. So gravity splits into g sinα (along) and g cosα (perpendicular); landing means perpendicular displacement returns to zero.

Q7. Constraint idea for a string over a pulley.
A7. The inextensible string means the sum of segment lengths is constant; differentiate to relate the blocks' velocities and accelerations.

Q8. A body moves in a semicircle of radius R. Distance vs displacement?
A8. Distance = πR, displacement = 2R (the diameter).

Q9. Two angles giving the same range on flat ground.
A9. θ and (90° − θ).

Q10. Rain falls vertically at speed v_r; man walks at v_m. Umbrella angle?
A10. tanθ = v_m/v_r from the vertical, pointing into the walking direction (relative velocity of rain).

Q11. Average velocity for a full round trip returning to start?
A11. Zero — net displacement is zero, though average speed is not.

Q12. Slope of a position-time graph and of a velocity-time graph?
A12. Slopes give velocity and acceleration respectively; area under v–t gives displacement.

Kinematics is the grammar of every mechanics problem — before forces, before energy, you must describe motion cleanly. JEE Advanced rarely asks a bare "find velocity" question; it hides kinematics inside relative-motion, projectile-on-incline, river-boat and constraint problems where a wrong frame or a dropped vector component quietly kills the whole solution.

Core concept: position, velocity and acceleration are vectors linked by calculus — v = dr/dt, a = dv/dt — and every "hard" kinematics problem is really a smart choice of frame, axes, or constraint equation.

Building the ideas

Beginner — the equations and when they apply

The three SUVAT equations (v = u + at, s = ut + ½at², v² = u² + 2as) hold only for constant a. For variable acceleration you integrate: if a = a(t), then v = u + ∫a dt; if a = a(x), use the trick a = v dv/dx, so ∫v dv = ∫a dx. This last identity is the single most tested idea in variable-acceleration problems.

Intermediate — vectors and projectiles

Resolve independently: horizontal motion has a = 0, vertical has a = −g. Range R = u²sin2θ/g, max height H = u²sin²θ/(2g), time of flight T = 2u sinθ/g. The elegant result: for a given speed, complementary angles θ and 90°−θ give the same range. On an inclined plane, don't use these formulas blindly — rotate axes along and perpendicular to the incline so gravity splits into g sinα and g cosα.

Advanced — relative motion, the JEE workhorse

The velocity of A relative to B is v_AB = v_A − v_B. River-boat, rain-man and closest-approach problems all reduce to this. For closest approach of two particles, work in the frame of one: the other moves in a straight line, and the minimum distance is the perpendicular from the stationary particle to that line. This converts a messy time-minimisation into one geometry step.

Constraint relations: when bodies are linked by inextensible strings over pulleys, the sum of string segment lengths is constant, so the components of velocity along the string are equal. Differentiating the length constraint gives the acceleration relation — far safer than guessing "a₁ = 2a₂".

Worked example

A boat can move at 5 m/s in still water across a river 100 m wide flowing at 3 m/s. (a) To cross in shortest time, head straight across: t = 100/5 = 20 s, and drift = 3 × 20 = 60 m downstream. (b) To reach the point directly opposite, aim upstream at angle θ with 5 sinθ = 3, so sinθ = 0.6, cosθ = 0.8; effective cross-speed = 5 × 0.8 = 4 m/s, giving t = 100/4 = 25 s. Note the shortest-time and shortest-path strategies are different — a classic trap.

How JEE Advanced tests this

Multi-part problems combining relative velocity with a constraint; graph-based questions (read a v–t or a–t graph, extract displacement as area, or identify the moment of reversal); projectile problems on an incline or with a moving frame; and matching-type questions pairing a motion description to its graph.

Exam tricks & shortcuts

• Area under v–t = displacement; slope of x–t = velocity. Reading graphs beats re-deriving.
• For a = v dv/dx problems, decide early whether a depends on t, x or v — that dictates which integral form to use.
• Mnemonic for relative motion: "Sit on B, watch A."

Key formulas

• Equations (constant a): v = u + at; s = ut + ½at²; v² = u² + 2as; sₙ = u + ½a(2n−1).
• Projectile: T = 2u sinθ/g; H = u²sin²θ/2g; R = u²sin2θ/g (R max at 45°).
• Relative velocity: v⃗_AB = v⃗_A − v⃗_B.
• Vectors: resultant R = √(A²+B²+2AB cosθ); A⃗·B⃗ = AB cosθ; |A⃗×B⃗| = AB sinθ.
• Component form: A⃗ = Aₓî + A_yĵ + A_zk̂; |A⃗| = √(Aₓ²+A_y²+A_z²).

Resolve motion into independent x and y components; use relative velocity for river/rain problems.

Worked Example

Problem: A river 100 m wide flows at 3 m/s. A boat can move at 5 m/s relative to the water. (a) If the boat is steered straight across (perpendicular to the bank), find the crossing time and the downstream drift. (b) In which direction must it be steered to reach the point directly opposite, and how long does that take?

Solution:
(a) Steering straight across, the full 5 m/s is directed across the river:
Crossing time t = width / v_cross = 100 / 5 = 20 s.
Meanwhile the current carries the boat downstream:
Drift = 3 × 20 = 60 m.

(b) To land directly opposite, the boat's upstream velocity component must cancel the current. If the boat is angled θ upstream from the perpendicular:
5 sin θ = 3 → sin θ = 3/5 → θ = 37° (upstream of the straight crossing).
The across-stream component is then:
v_cross = 5 cos θ = 5 × (4/5) = 4 m/s.
Crossing time t = 100 / 4 = 25 s.

Answer: (a) 20 s with 60 m drift; (b) steer 37° upstream, crossing in 25 s.