1. Chapter at a Glance
Kinematics describes how things move — position, velocity and acceleration — without asking why. This chapter combines two NCERT chapters: Motion in a Straight Line and Motion in a Plane. You will learn the equations of motion, how to read motion graphs, how to add vectors, and how to handle projectiles, relative velocity and uniform circular motion. NEET asks one or two questions every year, most often on graphs and projectiles.
- Using the equations of motion, including free fall.
- Slope and area of x–t and v–t graphs.
- Resultant of two vectors and components.
- Projectile: range, maximum height, time of flight.
- Centripetal acceleration in uniform circular motion.
2. Motion in a Straight Line
- Distance is the length of the path (scalar, never negative). Displacement is the change in position (vector). Distance ≥ |displacement|.
- Average speed = total distance/total time; average velocity = displacement/time.
- Instantaneous velocity v = dx/dt; acceleration a = dv/dt. Retardation means acceleration opposite to velocity.
These hold only for constant acceleration. For a freely falling body, a = g ≈ 9.8 m s⁻² (take 10 m s⁻² in quick calculations) downward. A body dropped from height h reaches the ground in √(2h/g) with speed √(2gh). A body thrown up with speed u rises to u²/2g in time u/g.
Reading motion graphs
| Graph | Slope gives | Area gives | NEET Note |
|---|---|---|---|
| Position–time | Velocity | — | A straight line means uniform velocity |
| Velocity–time | Acceleration | Displacement | Area below the axis is negative displacement |
| Acceleration–time | — | Change in velocity | Horizontal line = uniform acceleration |
3. Scalars, Vectors and Motion in a Plane
- A vector has magnitude and direction and adds by the triangle or parallelogram law. Resultant of A and B at angle θ: R = √(A² + B² + 2AB cosθ).
- R is largest (A + B) when θ = 0° and smallest |A − B| when θ = 180°.
- Components: Ax = A cosθ, Ay = A sinθ; A = √(Ax² + Ay²).
- Unit vector  = A/|A|; î, ĵ, k̂ point along x, y, z.
- Motion in a plane can be split into two independent straight-line motions along x and y.
4. Projectile Motion
A body projected at speed u and angle θ to the horizontal moves under gravity alone. Its horizontal velocity u cosθ stays constant; its vertical motion has acceleration g downward. The path is a parabola.
| Quantity | Formula | NEET Note |
|---|---|---|
| Time of flight | T = 2u sinθ/g | Depends only on vertical component |
| Maximum height | H = u² sin²θ/2g | Largest at 90° |
| Range | R = u² sin2θ/g | Maximum u²/g at 45° |
| Equation of path | y = x tanθ − gx²/(2u²cos²θ) | Parabola |
| Horizontal throw from height h | t = √(2h/g); x = u√(2h/g) | Same time as a dropped body |
Projections at θ and (90° − θ) give the same range; the steeper one goes higher and stays in the air longer. At the highest point, velocity = u cosθ (not zero) and acceleration = g.
5. Relative Velocity and Circular Motion
- Velocity of A relative to B: vAB = vA − vB. Two bodies moving in the same direction: relative speed = difference; opposite directions: sum.
- Boat–river: to cross in the least time, head straight across (time = width/boat speed, drift = river speed × time). To land directly opposite, head upstream at sin α = vriver/vboat.
- Uniform circular motion: speed is constant, velocity changes direction. Angular speed ω = θ/t = 2π/T, v = ωr.
- Centripetal acceleration ac = v²/r = ω²r, always towards the centre.
6. Formula Master Table
| Situation | Formula |
|---|---|
| Uniform acceleration | v = u + at; s = ut + ½at²; v² = u² + 2as |
| Distance in nth second | sn = u + a(2n − 1)/2 |
| Free fall from height h | t = √(2h/g); v = √(2gh) |
| Resultant of two vectors | R = √(A² + B² + 2AB cosθ) |
| Projectile | T = 2u sinθ/g; H = u²sin²θ/2g; R = u² sin2θ/g |
| Relative velocity | vAB = vA − vB |
| Circular motion | v = ωr; ac = v²/r = ω²r |
7. Master Summary Table
| Concept | Key point | NEET Note |
|---|---|---|
| Distance vs displacement | Distance ≥ |displacement| | Round trip: displacement zero |
| Equations of motion | Valid only for constant a | Integrate if a varies with t |
| v–t graph | Slope = a, area = s | Most-asked graph in NEET |
| Vectors | Add by parallelogram law | Equal vectors at 120° give resultant equal to each |
| Projectile | Horizontal and vertical motions independent | Range same for θ and 90° − θ |
| Circular motion | ac = v²/r towards centre | Uniform speed, not uniform velocity |
8. Mind Map
- Equations of motion
- Free fall
- Graphs
- Resultant
- Components
- Unit vectors
- T, H, R
- Parabolic path
- Horizontal throw
- Boat–river
- ω = v/r
- ac = v²/r
9. Tips and Tricks
- A body thrown up returns with the same speed (no air resistance); time up = time down.
- For equal distances at speeds v₁ and v₂, average speed = 2v₁v₂/(v₁ + v₂); for equal times it is (v₁ + v₂)/2.
- Two equal vectors of magnitude A at 120° have a resultant A; at 90°, A√2; at 60°, A√3.
- tanθ = 4H/R for a projectile — handy when H and R are given.
- Saying velocity is zero at the top of a projectile path — only its vertical component is zero.
- Using the equations of motion when acceleration depends on time.
- Taking g as positive upward and downward in the same problem.
10. Important One-liners
- Slope of a position–time graph gives velocity.
- Area under an acceleration–time graph gives change in velocity.
- In the absence of air resistance, all bodies fall with the same acceleration.
- Maximum range of a projectile is at 45°.
- The horizontal velocity of a projectile remains constant.
- In uniform circular motion, acceleration is always perpendicular to velocity.
- A negative acceleration does not always mean the body is slowing down.
11. Solved Numericals: Average to Advanced
Five average and five advanced numericals. Take g = 10 m s⁻². Click the button under a question to see its answer and working.
Average level
A train moving at 72 km h⁻¹ is brought to rest uniformly in 10 s. Find the retardation and the distance covered while stopping.
Show answer & solution
Given: u = 72 km h⁻¹ = 20 m s⁻¹, v = 0, t = 10 s. Concept: equations of motion.
- a = (v − u)/t = (0 − 20)/10 = −2 m s⁻², so retardation = 2 m s⁻².
- s = (u² − v²)/2|a| = 400/4 = 100 m.
Answer: 2 m s⁻²; 100 m
A stone is dropped from the top of a 45 m tall building. Find the time taken to reach the ground and its speed on impact.
Show answer & solution
Concept: free fall from rest.
- t = √(2h/g) = √(90/10) = 3 s.
- v = gt = 10 × 3 = 30 m s⁻¹.
Answer: 3 s; 30 m s⁻¹
Forces of 3 N and 4 N act at a point. Find the resultant when the angle between them is (a) 90°, (b) 60°.
Show answer & solution
Concept: R = √(A² + B² + 2AB cosθ).
- (a) R = √(9 + 16 + 0) = 5 N.
- (b) R = √(9 + 16 + 2 × 3 × 4 × 0.5) = √37 ≈ 6.08 N.
Answer: 5 N; about 6.08 N
A ball is kicked at 20 m s⁻¹ at 45° to the ground. Find its range, maximum height and time of flight.
Show answer & solution
Concept: projectile formulas; sin 90° = 1, sin²45° = ½.
- R = u² sin2θ/g = 400 × 1/10 = 40 m.
- H = u² sin²θ/2g = 400 × 0.5/20 = 10 m.
- T = 2u sinθ/g = 2 × 20 × 0.707/10 ≈ 2.83 s.
Answer: 40 m; 10 m; about 2.83 s
A car goes round a circular track of radius 50 m at a steady 36 km h⁻¹. Find its centripetal acceleration.
Show answer & solution
Concept: ac = v²/r with v in m s⁻¹.
- v = 36 km h⁻¹ = 10 m s⁻¹.
- ac = 100/50 = 2 m s⁻² towards the centre.
Answer: 2 m s⁻²
Advanced level
A ball is thrown vertically upward at 30 m s⁻¹ from the top of a 35 m high tower. After how long does it hit the ground, and with what speed?
Show answer & solution
Concept: take upward as positive; the ball ends 35 m below the start, so s = −35 m.
- −35 = 30t − 5t² ⇒ 5t² − 30t − 35 = 0 ⇒ t² − 6t − 7 = 0.
- (t − 7)(t + 1) = 0 ⇒ t = 7 s.
- v = u − gt = 30 − 70 = −40 m s⁻¹, i.e. 40 m s⁻¹ downward.
Answer: 7 s; 40 m s⁻¹
A scooter accelerates uniformly from rest to 20 m s⁻¹ in 4 s, moves at this speed for 6 s, and then stops uniformly in 5 s (the graph drawn above). Find the total distance and average speed.
Show answer & solution
Concept: distance = area under the v–t graph (a trapezium).
- Accelerating part: ½ × 4 × 20 = 40 m.
- Constant part: 20 × 6 = 120 m.
- Braking part: ½ × 5 × 20 = 50 m.
- Total = 210 m; average speed = 210/15 = 14 m s⁻¹.
Answer: 210 m; 14 m s⁻¹
Car A moves north at 60 km h⁻¹ and car B moves east at 80 km h⁻¹. Find the velocity of B relative to A.
Show answer & solution
Concept: vBA = vB − vA = vB + (−vA).
- −vA is 60 km h⁻¹ south; vB is 80 km h⁻¹ east.
- Magnitude = √(80² + 60²) = 100 km h⁻¹.
- Direction: tan β = 60/80 ⇒ β ≈ 37° south of east.
Answer: 100 km h⁻¹, about 37° south of east
A ball thrown at 20 m s⁻¹ must land 20 m away on level ground. Find the two possible angles of projection and the maximum height in each case.
Show answer & solution
Concept: R = u² sin2θ/g; complementary angles give equal range.
- 20 = 400 sin2θ/10 ⇒ sin2θ = 0.5 ⇒ 2θ = 30° or 150°.
- θ = 15° or 75°.
- H(15°) = 400 × sin²15°/20 ≈ 20 × 0.067 ≈ 1.34 m.
- H(75°) = 400 × sin²75°/20 ≈ 20 × 0.933 ≈ 18.66 m.
Answer: 15° (1.34 m high) or 75° (18.66 m high)
A boat can move at 4 m s⁻¹ in still water. It crosses a 200 m wide river flowing at 2 m s⁻¹. Find (a) the least crossing time and the drift, (b) the heading and time for reaching the point directly opposite.
Show answer & solution
Concept: boat velocity relative to ground = its velocity in water + river velocity.
- (a) Head straight across: t = 200/4 = 50 s.
- Drift = 2 × 50 = 100 m downstream.
- (b) Head upstream at α with sin α = 2/4 ⇒ α = 30° from the straight-across line.
- Speed across = 4 cos30° ≈ 3.46 m s⁻¹, so t = 200/3.46 ≈ 57.7 s.
Answer: (a) 50 s, 100 m; (b) 30° upstream, about 57.7 s
12. NEET-pattern Questions with Detailed Answers
These questions are original and modelled on concepts repeatedly asked in NEET UG.
The slope of a velocity–time graph gives:
- Displacement
- Acceleration
- Speed
- Distance
Show solution
Slope = Δv/Δt = acceleration. Displacement is the area under the v–t graph, not its slope.
Answer: (B)
At the highest point of its path, a projectile has:
- Zero velocity and zero acceleration
- Zero velocity but acceleration g
- Horizontal velocity and acceleration g downward
- Vertical velocity only
Show solution
Only the vertical component of velocity becomes zero; the horizontal part u cosθ remains. Gravity still acts, so acceleration is g downward.
Answer: (C)
Two vectors of equal magnitude P have a resultant also of magnitude P. The angle between them is:
- 60°
- 90°
- 120°
- 180°
Show solution
P² = P² + P² + 2P² cosθ ⇒ cosθ = −½ ⇒ θ = 120°. At 60° the resultant would be P√3; at 90°, P√2; at 180°, zero.
Answer: (C)
A particle's position is x = 4t² − 2t³ (SI units). Its velocity is zero at:
- t = 0 only
- t = 4/3 s only
- t = 0 and t = 4/3 s
- t = 2 s
Show solution
v = dx/dt = 8t − 6t² = 2t(4 − 3t). v = 0 at t = 0 and t = 4/3 s. At t = 2 s, x = 0 but v ≠ 0.
Answer: (C)
A body starting from rest with uniform acceleration covers distances in the 1st, 2nd and 3rd seconds in the ratio:
- 1 : 2 : 3
- 1 : 4 : 9
- 1 : 3 : 5
- 1 : 1 : 1
Show solution
sn = (a/2)(2n − 1) for u = 0, giving 1 : 3 : 5 (Galileo's law of odd numbers). 1 : 4 : 9 is the ratio of total distances after 1, 2, 3 s.
Answer: (C)
A stone tied to a 0.5 m string whirls in a horizontal circle at 4 rev s⁻¹. Its centripetal acceleration is about:
- 16 m s⁻²
- 79 m s⁻²
- 316 m s⁻²
- 32 m s⁻²
Show solution
ω = 2π × 4 = 8π rad s⁻¹; a = ω²r = 64π² × 0.5 ≈ 316 m s⁻².
Answer: (C)
For a projectile, the range equals four times the maximum height. The angle of projection is:
- 30°
- 45°
- 60°
- tan⁻¹(4)
Show solution
tanθ = 4H/R = 4H/(4H) = 1 ⇒ θ = 45°. (If H = R, then tanθ = 4.)
Answer: (B)
A ball is dropped from a height and another is thrown horizontally from the same height at the same instant. Which statement is correct?
- The dropped ball lands first
- The thrown ball lands first
- Both land together
- It depends on their masses
Show solution
Both have zero initial vertical velocity and the same vertical acceleration g, so they take the same time √(2h/g). Horizontal motion does not affect vertical motion, and mass does not matter.
Answer: (C)
13. Quick Revision Summary
- v = u + at, s = ut + ½at², v² = u² + 2as (constant a only).
- v–t graph: slope = acceleration, area = displacement.
- R = √(A² + B² + 2AB cosθ); resolve vectors into components.
- Projectile: T = 2u sinθ/g, H = u²sin²θ/2g, R = u² sin2θ/g; maximum range at 45°.
- Uniform circular motion: ac = v²/r = ω²r towards the centre.