Physics · Kinematics · Chapter notes
Motion in a Straight Line and in a Plane · Class 11 Notes
Class 11 kinematics notes: equations of motion, motion graphs, projectiles, and a full treatment of relative velocity including rain-umbrella, boat-and-river and pursuit, with a JEE Advanced part.
In short
Kinematics describes motion without asking what caused it. Three equations cover constant acceleration, graphs turn them into something you can read at a glance, and projectile motion is just those equations applied separately to two perpendicular directions. Relative velocity then handles every problem where two things move at once.
Contents
- 1Distance, Displacement, Speed, Velocity
- 2The Equations of Motion
- 3Graphs of Motion Slopes and areas
- 4Vectors in a Plane Components and addition
- 5Projectile Motion Range, height, flight time
- 6Relative Velocity: The Core Idea
- 7Rain and Umbrella Tilt angle and the vertical twist
- 8Boat and River Minimum time vs shortest path
- 9Dog and Cat (Pursuit) Closing speed, the square problem
- 10Variable (Non-Uniform) Acceleration
- 11Projectile on an Inclined Plane
- 12Advanced Graph Transformations
- 13General Minimum Distance of Approach
- ·Motion: One-Page Formula Sheet
Distance, Displacement, Speed, Velocity
Rectilinear motion is the simplest kind, yet its definitions underpin all of mechanics. The first thing JEE tests is whether you keep the scalar and vector quantities apart.
| Quantity | Type | Meaning |
|---|---|---|
| Distance | Scalar | Total path length; never decreases |
| Displacement | Vector | Straight-line change in position; can be zero or negative |
| Speed | Scalar | Distance / time; always ≥ 0 |
| Velocity | Vector | Displacement / time; sign gives direction |
| Acceleration | Vector | Rate of change of velocity |
Average velocity = total displacement / total time. Instantaneous velocity is the limit as the interval shrinks to zero, v = dx/dt, which is the slope of the position-time graph. Average speed and average velocity are equal only for motion in a straight line without reversal.
A body goes 60 m east in 20 s, then 20 m west in 10 s. Average speed = 80/30 ≈ 2.67 m/s. Net displacement = 40 m east, so average velocity = 40/30 ≈ 1.33 m/s east. They differ because the motion reversed.
The Equations of Motion
When acceleration a is constant, three equations connect u, v, a, t and s. Each omits one variable, so pick the one missing the quantity you neither know nor want.
s = ut + ½at² (no v)
v² = u² + 2as (no t)
s in n-th second: sₙ = u + ½a(2n − 1)
Fix a positive direction first, then give every vector its correct sign. For a body thrown up, taking up as positive: u is +, but a = −g. Mixing signs is the biggest source of errors in this chapter.
A ball is thrown up at u = 20 m/s (g = 10, up positive so a = −10). At the top v = 0, so 0 = 20 − 10t gives t = 2 s. And 0 = 400 − 20s gives s = 20 m. The ball rises 20 m in 2 s.
Graphs of Motion
Graphs turn a word problem into a picture. JEE loves extracting information from slopes and areas.
Position-time (x-t): the slope is the velocity. A curve means acceleration; a horizontal line means rest.
Velocity-time (v-t): the slope is the acceleration, and the area under the graph is the displacement. This area rule is one of the most useful shortcuts in kinematics.
A body accelerates from rest to 20 m/s in 5 s. Displacement = area of the triangle = ½ × 5 × 20 = 50 m. Acceleration = slope = 20/5 = 4 m/s².
Vectors in a Plane
Every quantity becomes a vector with two components. The key simplification: the two perpendicular directions are independent, so x-motion and y-motion do not affect each other.
Aₓ = A cosθ, Aₖ = A sinθ
A vector of magnitude A at angle θ splits into A cosθ horizontally and A sinθ vertically. Working in components turns one vector problem into two independent scalar problems, one per axis.
A velocity 3î + 4ĵ m/s has speed √(9 + 16) = 5 m/s and direction tanθ = 4/3, so θ ≈ 53° above the x-axis.
Projectile Motion
The classic plane-motion problem: horizontal velocity stays constant while vertical motion is free fall. Treat the axes independently and combine.
H = u²sin²θ / 2g
R = u²sin2θ / g (max at θ = 45°)
Launch at u = 20 m/s, θ = 30° (g = 10). T = 2(20)(0.5)/10 = 2 s; H = 400(0.25)/20 = 5 m; R = 400(0.866)/10 ≈ 34.6 m. Maximum range for a given speed is always at 45°.
Relative Velocity: The Core Idea
Relative velocity is the velocity of one body as measured by an observer moving with another. It is pure vector subtraction, and the key to every rain, boat and pursuit problem.
1. Write each real velocity as a vector.
2. Form the relative velocity by subtracting the observer's velocity.
3. Read off magnitude and direction, or set a component to zero for a special condition (zero drift, rain appearing vertical).
Rain and Umbrella
A person walking in vertically falling rain must tilt the umbrella forward, because in their frame the rain gains a backward horizontal velocity equal and opposite to the walking velocity.
Rain falls vertically at 10 m/s; a man walks at 5 m/s. The umbrella tilts at tanθ = 5/10 = 0.5, so θ ≈ 26.6° from the vertical. Apparent rain speed = √(100 + 25) ≈ 11.2 m/s.
If the rain itself falls at an angle and a man walks so it appears vertical, his velocity must match the horizontal component of the rain's velocity. If rain velocity = 3î − 4ĵ and he walks at 3î, the relative velocity is −4ĵ, purely vertical.
Boat and River
A boat of speed v_b relative to water crosses a river of current v_r. Two goals give different aiming directions; keep them separate.
Aim the boat straight across. The whole of v_b works on crossing; the current only carries you downstream.
To land directly opposite, aim upstream at angle θ so the upstream component cancels the current: v_b sinθ = v_r.
Only possible if v_b > v_r, and it always takes longer than the minimum-time crossing.
Width d = 100 m, v_b = 5 m/s, v_r = 3 m/s. Minimum time: t = 100/5 = 20 s, drift = 3 × 20 = 60 m. Shortest path: sinθ = 3/5, θ ≈ 36.9°, speed across = 4 m/s, t = 25 s, zero drift. The zero-drift crossing takes 5 s longer.
Zero drift is impossible, so the goal becomes minimum drift. Aim upstream at sinθ = v_b/v_r (measured from across). Minimising gives:
For v_b = 3, v_r = 5, d = 100: minimum drift = 100(4)/3 ≈ 133 m. A common JEE Advanced extension.
Dog and Cat (Pursuit)
Pursuit problems turn on the closing speed: the rate at which the separation shrinks equals the component of the relative velocity along the line joining the two bodies.
time to meet = separation / closing speed (when constant)
A dog at x = 0 runs at 4 m/s toward a cat at x = 12 m running toward it at 2 m/s. Closing speed = 4 + 2 = 6 m/s, so they meet in 12/6 = 2 s.
Four insects at the corners of a square of side a = 10 m each chase the next at v = 2 m/s. By symmetry each target moves perpendicular to the line joining it to its chaser, so the closing speed is just v. Time to meet = a/v = 10/2 = 5 s.
Variable (Non-Uniform) Acceleration
When acceleration is not constant, go back to calculus. Everything follows from v = dx/dt and a = dv/dt.
• a = f(t): integrate a dt to get v.
• a = f(v): use a = dv/dt, so t = ∫ dv/f(v).
• a = f(x): use the chain-rule form a = v dv/dx, so ∫ v dv = ∫ f(x) dx.
A particle decelerates as a = −kv. Then dv/v = −k dt, so v = u e−kt: the velocity decays exponentially and never quite reaches zero. Total distance = ∫v dt = u/k, finite despite never stopping.
For a = −ω²x (simple harmonic motion), a = v dv/dx = −ω²x gives v dv = −ω²x dx. Integrating: v² = u² − ω²x², the standard SHM velocity-position relation.
Projectile on an Inclined Plane
Launching up or down a slope is an Advanced staple. Tilt the axes along and perpendicular to the incline, so gravity splits into two components.
T = 2u sinθ / (g cosα)
R = 2u² sinθ cos(θ+α) / (g cos²α)
Rmax (up) = u² / [g(1 + sinα)]
For projection down the incline, flip the sign of α: gravity's down-slope component now aids the motion. Rmax down = u²/[g(1 − sinα)], larger than the up-slope range, and the optimum angle shifts to 45° + α/2.
Advanced Graph Transformations
Main asks you to read a slope; Advanced asks you to convert one graph into another. The bridge is always a = v dv/dx.
• From a v-x graph: a = v(dv/dx). A straight v-x line does NOT mean constant acceleration.
• From a v²-x graph: the slope is 2a, so a = ½ × slope. A straight line here DOES mean constant acceleration.
• Area under an a-x graph = ½(v² − u²).
If v = 10 − ½x (a straight v-x line), then dv/dx = −½. At x = 0, v = 10, so a = v(dv/dx) = 10(−½) = −5 m/s². At x = 20, v = 0, so a = 0. The acceleration is not constant even though the graph is a straight line.
General Minimum Distance of Approach
The square-of-insects problem uses symmetry. The general problem gives two particles on arbitrary lines and asks the closest they ever get. Use relative velocity.
t* = −(r₀ · vrel) / |vrel|²
dmin = |r₀ × vrel| / |vrel|
Freeze A: in A's frame, B travels in a straight line along v_rel. The minimum distance is the perpendicular distance from A to that line, which is the cross product magnitude |r₀ × v_rel| divided by |v_rel|. If t* is negative, the closest approach was in the past.
A at (10, 0) moving (0, 2); B at origin moving (−3, 0). Take A as reference: r₀ = (−10, 0), v_rel = (−3, −2). Cross product = (−10)(−2) − 0 = 20, |v_rel| = √13, so d_min = 20/√13 ≈ 5.55 m.
Motion: One-Page Formula Sheet
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