1. Chapter at a Glance
Kinematics describes motion — position, velocity and acceleration — without asking what causes it. It covers motion in a straight line, graphs, vectors, projectile motion, relative velocity and uniform circular motion. Almost every mechanics problem in JEE starts with kinematics, and one or two direct questions appear every year (often a projectile or a graph question).
- Equations of motion and the distance covered in the nth second.
- Reading x–t, v–t and a–t graphs (slope and area).
- Projectile motion: time of flight, maximum height, range, trajectory.
- Relative velocity (river–boat, rain–man) and circular motion.
2. Motion in a Straight Line
- Distance is the total path length (scalar); displacement is the change in position (vector). |displacement| ≤ distance.
- Average speed = total distance/total time; average velocity = displacement/time.
- Instantaneous velocity v = dx/dt; acceleration a = dv/dt = v(dv/dx).
Equations of motion (constant acceleration)
For free fall take a = g (≈ 9.8 m s⁻², often 10 m s⁻² in problems) downward. A body thrown up with speed u rises to H = u²/2g in time u/g and returns to the same point after 2u/g.
Motion graphs
| Graph | Slope gives | Area gives |
|---|---|---|
| x–t | Velocity | — |
| v–t | Acceleration | Displacement (signed area); distance = total unsigned area |
| a–t | Jerk | Change in velocity |
With variable acceleration, integrate: v = ∫a dt, x = ∫v dt. To find distance when the velocity changes sign, split the time interval at the moments where v = 0.
3. Vectors in Kinematics
- Resultant of A and B at angle θ: R = √(A² + B² + 2AB cosθ), direction tan α = B sinθ/(A + B cosθ).
- Dot product A·B = AB cosθ (scalar); cross product |A × B| = AB sinθ (vector, perpendicular to both).
- Resolve a vector into components: Ax = A cosθ, Ay = A sinθ. Motion along x and y can be treated independently.
4. Projectile Motion
A body thrown at speed u at angle θ with the horizontal moves with constant horizontal velocity u cosθ and constant downward acceleration g.
| Quantity | Formula |
|---|---|
| Time of flight | T = 2u sinθ/g |
| Maximum height | H = u² sin²θ/2g |
| Horizontal range | R = u² sin2θ/g (maximum u²/g at 45°) |
| Trajectory | y = x tanθ − gx²/(2u² cos²θ) |
| Horizontal projection from height h | t = √(2h/g), range = u√(2h/g) |
Angles θ and (90° − θ) give the same range. Also tanθ = 4H/R.
5. Relative Velocity and Circular Motion
- Velocity of A relative to B: vAB = vA − vB.
- River–boat (river speed vr, boat speed vb in still water, width d): shortest time d/vb by heading straight across (drift = vrd/vb); shortest path (zero drift, needs vb > vr) by heading upstream at sin α = vr/vb, time d/√(vb² − vr²).
- Rain–man: hold the umbrella along the rain's velocity relative to the man.
- Uniform circular motion: v = ωr, centripetal acceleration ac = v²/r = ω²r towards the centre. In non-uniform circular motion there is also a tangential acceleration at = dv/dt, and a = √(ac² + at²).
6. Formula Master Table
| Situation | Formula |
|---|---|
| Constant acceleration | v = u + at; s = ut + ½at²; v² = u² + 2as |
| nth second | sn = u + a(2n − 1)/2 |
| Vertical throw | H = u²/2g; T = 2u/g |
| Average speed (equal distances) | 2v₁v₂/(v₁ + v₂) |
| Projectile | T = 2u sinθ/g; H = u² sin²θ/2g; R = u² sin2θ/g |
| Relative velocity | vAB = vA − vB |
| Circular motion | v = ωr; ac = v²/r = ω²r |
7. Mind Map
- Equations of motion
- Free fall
- Graphs: slope, area
- Resultant, components
- Dot, cross product
- T, H, R
- Trajectory
- From a height
- River–boat, rain–man
- ac = v²/r
8. Tips and Tricks
- Choose one direction as positive and stick to it; g is then −10 m s⁻² for an upward-positive axis.
- For objects released from a moving balloon or lift, they start with the balloon's velocity, not zero.
- Complementary angles give equal ranges; the one closer to 90° goes higher and stays longer.
- In relative-motion problems, "freeze" one body and give the other the relative velocity.
- Using s = ut + ½at² to find distance when the velocity reverses — split the motion.
- Taking the average of two speeds as the average speed.
- Forgetting that the velocity at the top of a projectile path is not zero, only its vertical part.
9. Important One-liners
- Area under a v–t graph gives displacement.
- A body thrown up takes the same time to go up as to come down (no air resistance).
- Range is maximum at 45° for projection on level ground.
- Uniform circular motion has constant speed but changing velocity.
- The horizontal velocity of a projectile stays constant.
- Two balls dropped and thrown horizontally from the same height reach the ground together.
10. Solved Numericals: Average to Advanced
Five average-level and five advanced-level problems. Try each first, then click to reveal the answer and full working. Take g = 10 m s⁻².
Average level
A car starts from rest and accelerates at 2 m s⁻² for 10 s. It then brakes with a retardation of 4 m s⁻² until it stops. Find the total distance covered.
Show answer & solution
Given: u = 0, a₁ = 2 m s⁻², t₁ = 10 s, a₂ = −4 m s⁻². Concept: apply the equations of motion to each stage.
- Stage 1: s₁ = ½ × 2 × 10² = 100 m; speed reached v = 2 × 10 = 20 m s⁻¹.
- Stage 2: 0 = 20² − 2 × 4 × s₂ ⇒ s₂ = 400/8 = 50 m.
- Total distance = 100 + 50 = 150 m.
Answer: 150 m
A ball is thrown vertically upward at 20 m s⁻¹. Find the maximum height and the time after which it returns to the hand.
Show answer & solution
Concept: at the top v = 0; the motion is symmetric.
- H = u²/2g = 400/20 = 20 m.
- Time to rise = u/g = 2 s.
- Total time = 2 × 2 = 4 s.
Answer: 20 m; 4 s
A cyclist rides 60 km to a town at 30 km h⁻¹ and returns by the same road at 60 km h⁻¹. Find the average speed and average velocity for the whole trip.
Show answer & solution
Concept: average speed = total distance/total time.
- Time going = 60/30 = 2 h; time returning = 60/60 = 1 h.
- Average speed = 120 km/3 h = 40 km h⁻¹ (same as 2v₁v₂/(v₁ + v₂)).
- Net displacement = 0, so average velocity = 0.
Answer: 40 km h⁻¹; zero
A body starts with 5 m s⁻¹ and moves with a uniform acceleration of 2 m s⁻². How far does it travel during the 6th second?
Show answer & solution
Concept: sn = u + (a/2)(2n − 1).
- s₆ = 5 + (2/2)(2 × 6 − 1).
- = 5 + 11 = 16 m.
Answer: 16 m
A stone tied to a string moves in a horizontal circle of radius 0.5 m, making 2 revolutions per second. Find its speed and centripetal acceleration.
Show answer & solution
Concept: ω = 2πf, v = ωr, ac = ω²r.
- ω = 2π × 2 = 4π rad s⁻¹.
- v = 4π × 0.5 = 2π ≈ 6.28 m s⁻¹.
- ac = (4π)² × 0.5 = 8π² ≈ 79 m s⁻².
Answer: 6.28 m s⁻¹; about 79 m s⁻²
Advanced level
A ball is projected at 40 m s⁻¹ at 30° above the horizontal. Find its time of flight, maximum height and horizontal range.
Show answer & solution
Given: u = 40 m s⁻¹, θ = 30°. Concept: split into u cosθ (horizontal) and u sinθ (vertical).
- u sinθ = 20 m s⁻¹; u cosθ = 34.64 m s⁻¹.
- T = 2u sinθ/g = 40/10 = 4 s.
- H = (u sinθ)²/2g = 400/20 = 20 m.
- R = u cosθ × T = 34.64 × 4 ≈ 138.6 m.
Answer: 4 s; 20 m; about 138.6 m
A river 100 m wide flows at 3 m s⁻¹. A swimmer can swim at 5 m s⁻¹ in still water. (a) How should she swim to cross along the shortest path, and how long does it take? (b) What is the least time to cross, and how far downstream does she land?
Show answer & solution
Concept: her velocity relative to ground = her velocity relative to water + river velocity.
- (a) For zero drift, her upstream component must cancel the river: 5 sin α = 3 ⇒ sin α = 3/5, α ≈ 37° upstream from the straight-across direction.
- Across component = 5 cos α = 4 m s⁻¹, so time = 100/4 = 25 s.
- (b) Least time: swim straight across at full 5 m s⁻¹: t = 100/5 = 20 s.
- Drift = river speed × time = 3 × 20 = 60 m downstream.
Answer: (a) 37° upstream, 25 s; (b) 20 s, 60 m downstream
A stone is thrown horizontally at 15 m s⁻¹ from the top of an 80 m tower. Find (a) the time to reach the ground, (b) the horizontal distance, and (c) the speed and direction just before landing.
Show answer & solution
Concept: the horizontal velocity stays 15 m s⁻¹; vertical motion is free fall from rest.
- (a) 80 = ½ × 10 × t² ⇒ t² = 16 ⇒ t = 4 s.
- (b) Horizontal distance = 15 × 4 = 60 m.
- (c) Vertical velocity = gt = 40 m s⁻¹; speed = √(15² + 40²) = √1825 ≈ 42.7 m s⁻¹.
- Angle below horizontal: tan β = 40/15 ⇒ β ≈ 69.4°.
Answer: 4 s; 60 m; 42.7 m s⁻¹ at about 69.4° below the horizontal
A particle moves along the x-axis as x = t³ − 6t² + 9t (x in metres, t in seconds). Find its displacement and the distance covered from t = 0 to t = 4 s.
Show answer & solution
Concept: the particle turns back where v = 0; split the interval there.
- v = dx/dt = 3t² − 12t + 9 = 3(t − 1)(t − 3), so v = 0 at t = 1 s and 3 s.
- x(0) = 0, x(1) = 1 − 6 + 9 = 4 m, x(3) = 27 − 54 + 27 = 0, x(4) = 64 − 96 + 36 = 4 m.
- Displacement = x(4) − x(0) = 4 m.
- Distance = |4 − 0| + |0 − 4| + |4 − 0| = 12 m.
Answer: displacement 4 m; distance 12 m
A balloon is rising at 10 m s⁻¹. When it is 75 m above the ground, a packet is released from it. After how long does the packet reach the ground, and with what speed?
Show answer & solution
Concept: the packet starts with the balloon's upward velocity. Take upward as positive.
- u = +10 m s⁻¹, a = −10 m s⁻², displacement s = −75 m.
- −75 = 10t − 5t² ⇒ 5t² − 10t − 75 = 0 ⇒ t² − 2t − 15 = 0.
- (t − 5)(t + 3) = 0 ⇒ t = 5 s (reject the negative root).
- v = u + at = 10 − 50 = −40 m s⁻¹, i.e. 40 m s⁻¹ downward.
Answer: 5 s; 40 m s⁻¹
11. PYQ-pattern Questions with Solutions
These questions are original and modelled on concepts frequently tested in JEE Main.
The area under a velocity–time graph gives:
- Acceleration
- Displacement
- Speed
- Force
Show solution
Area = ∫v dt = displacement. The slope of a v–t graph gives acceleration.
Answer: (B)
For a particle with x = 3t² − t³, the acceleration becomes zero at:
- t = 0
- t = 1 s
- t = 2 s
- t = 3 s
Show solution
v = 6t − 3t², a = 6 − 6t = 0 ⇒ t = 1 s. (The velocity is zero at t = 2 s.)
Answer: (B)
Two projectiles are thrown with the same speed at 25° and 65°. The ratio of their ranges is:
- 1 : 1
- 25 : 65
- sin25° : sin65°
- 1 : 2
Show solution
R = u² sin2θ/g; sin50° = sin130°, so complementary angles give equal ranges.
Answer: (A)
Rain falls vertically at 12 m s⁻¹. A man walks east at 5 m s⁻¹. To keep dry he should hold his umbrella tilted:
- Towards east at tan⁻¹(5/12) from vertical
- Towards west at tan⁻¹(5/12) from vertical
- Towards east at tan⁻¹(12/5) from vertical
- Vertically
Show solution
Relative to the man, rain has 12 m s⁻¹ downward and 5 m s⁻¹ westward, so it appears to come from the east, slanting at tan⁻¹(5/12) to the vertical. The umbrella must be tilted towards the east (forward) by this angle.
Answer: (A)
A projectile's maximum height equals its horizontal range. Find tanθ, where θ is the angle of projection.
Show solution
tanθ = 4H/R = 4.
Answer: 4
A car moves on a circular track of radius 50 m at 10 m s⁻¹ and speeds up at 1.5 m s⁻². Find the magnitude of its total acceleration in m s⁻².
Show solution
ac = v²/r = 100/50 = 2 m s⁻²; at = 1.5 m s⁻². a = √(4 + 2.25) = 2.5 m s⁻².
Answer: 2.5
12. Quick Revision Summary
- v = u + at, s = ut + ½at², v² = u² + 2as; sn = u + a(2n − 1)/2.
- v–t graph: slope = a, area = displacement.
- Projectile: T = 2u sinθ/g, H = u² sin²θ/2g, R = u² sin2θ/g; Rmax at 45°.
- River: least time d/vb; zero drift needs sin α = vr/vb.
- Circular motion: ac = v²/r towards the centre.