Kinematics for NEET

Kinematics describes motion without asking what causes it. For NEET it covers two NCERT chapters, motion in a straight line and motion in a plane, and it sits under almost everything else in mechanics. If your projectile and graph questions are slow, your laws-of-motion and energy questions will be slow too.

Questions here are rarely long. They test whether you know when a formula applies, whether you can read a graph, and whether you can split a vector into parts without mixing up sin and cos.

The basic quantities

• Distance is the path length (scalar). Displacement is the straight-line change in position (vector). Displacement can be zero when distance is not.
• Average speed = total distance ÷ total time. Average velocity = total displacement ÷ total time. For a round trip, average velocity is zero but average speed is not.
• Instantaneous velocity v = dx/dt and acceleration a = dv/dt = d²x/dt².

When position is given as a function of time, differentiate. If x = 3t² − 2t + 5 (in metres), then v = 6t − 2 and a = 6 m s⁻². At t = 2 s, v = 10 m s⁻¹.

Equations of motion and their condition

These hold only for constant acceleration along a straight line:

• v = u + at
• s = ut + ½at²
• v² = u² + 2as
• Distance in the nth second: sₙ = u + (a/2)(2n − 1)

The derivation is worth knowing in outline. On a velocity–time graph with constant acceleration, the line rises from u to v over time t. The area under it is a trapezium: s = ½(u + v)t. Substituting v = u + at gives s = ut + ½at².

Free fall and vertical throws

Take g = 10 m s⁻² unless told otherwise, and fix a sign convention before you start (upward positive is common).

• Dropped from height h: time t = √(2h/g), final speed v = √(2gh).
• Thrown up at u: time to the top = u/g, maximum height = u²/2g, and it returns to the same point with speed u after 2u/g.
• From rest, distances in successive equal seconds are in the ratio 1 : 3 : 5 : 7… (Galileo's law of odd numbers).

Reading motion graphs

On a position–time graph, a straight line means uniform velocity. A curve bending upward means increasing velocity. A line parallel to the time axis means the body is at rest.

Vectors in a nutshell

• Addition (parallelogram law): for vectors A and B at angle θ, R = √(A² + B² + 2AB cos θ). R is largest (A + B) at θ = 0 and smallest (A − B) at θ = 180°.
• Resolution: a vector A at angle θ to the x-axis has components A cos θ and A sin θ.
• Unit vector: Â = A/|A|, magnitude 1.
• Dot product: A·B = AB cos θ, a scalar. If A·B = 0, the vectors are perpendicular.
• Cross product: |A × B| = AB sin θ, a vector perpendicular to both, direction by the right-hand rule. î × ĵ = k̂, and A × B = −(B × A).

Relative velocity

The velocity of A relative to B is v_AB = v_A − v_B. In one dimension: two cars at 60 and 40 km h⁻¹ in the same direction have relative speed 20 km h⁻¹; in opposite directions, 100 km h⁻¹.

In two dimensions, subtract as vectors. A standard case is a boat crossing a river. Heading straight across gives the shortest time, t = width ÷ boat speed (relative to water), but the current carries the boat downstream while it crosses.

Projectile motion

A projectile has constant horizontal velocity u cos θ and constant downward acceleration g. The two motions are independent.

• Range is maximum at 45°, and R_max = u²/g. At 45°, H = R/4.
• Complementary angles (30° and 60°) give the same range but different heights and times.
• At the top, velocity is horizontal (u cos θ), not zero. Acceleration is g downward everywhere.
• Thrown horizontally from height h: time = √(2h/g), range = u√(2h/g).

Uniform circular motion

Speed is constant, but direction changes, so there is acceleration. It points to the centre: a = v²/r = ω²r, where v = ωr and ω = 2π/T. The velocity is along the tangent at every point.

Worked numericals

Example 1: stone thrown up from a tower

A stone is thrown upward at 20 m s⁻¹ from the top of a 25 m tower. When does it hit the ground, and how fast? (g = 10 m s⁻²)

• Take upward positive and the tower top as origin. The ground is at s = −25 m.
• −25 = 20t − 5t², so t² − 4t − 5 = 0, which gives (t − 5)(t + 1) = 0 and t = 5 s.
• v = 20 − 10 × 5 = −30 m s⁻¹, so the speed is 30 m s⁻¹ downward.
• Check with v² = u² + 2as: 400 + 2 × (−10) × (−25) = 900, so |v| = 30 ✓.

Example 2: distance from a v–t graph

A car accelerates uniformly from rest to 20 m s⁻¹ in 10 s, moves steadily for 20 s, then decelerates uniformly to rest in 5 s. Find the distance and average speed.

• Triangle: ½ × 10 × 20 = 100 m. Rectangle: 20 × 20 = 400 m. Triangle: ½ × 5 × 20 = 50 m.
• Distance = 550 m in 35 s. Average speed = 550/35 ≈ 15.7 m s⁻¹.

Example 3: ball thrown horizontally off a cliff

A ball is thrown horizontally at 15 m s⁻¹ from a 45 m cliff.

• Time: 45 = ½ × 10 × t², so t = 3 s.
• Horizontal distance = 15 × 3 = 45 m.
• At impact, vₓ = 15 and v_y = 10 × 3 = 30 m s⁻¹. Speed = √(225 + 900) = √1125 ≈ 33.5 m s⁻¹, at tan⁻¹(2) below the horizontal.

Example 4: crossing a river

A river is 400 m wide and flows at 3 m s⁻¹. A boat moves at 4 m s⁻¹ relative to the water and heads straight across.

• Time = 400/4 = 100 s (the current does not change this).
• Drift downstream = 3 × 100 = 300 m.
• Speed relative to the bank = √(4² + 3²) = 5 m s⁻¹.

Practice MCQs

1. A body starts from rest with uniform acceleration. The distances covered in the 1st, 2nd and 3rd seconds are in the ratio: (a) 1 : 2 : 3 (b) 1 : 4 : 9 (c) 1 : 3 : 5 (d) 1 : 1 : 1
2. A stone is dropped from rest. The distance it falls in the 3rd second is (g = 10 m s⁻²): (a) 15 m (b) 25 m (c) 30 m (d) 45 m
3. x = 3t² − 2t + 5 (x in m, t in s). The velocity at t = 2 s is: (a) 8 m s⁻¹ (b) 10 m s⁻¹ (c) 12 m s⁻¹ (d) 14 m s⁻¹
4. Two projectiles are launched with the same speed at 30° and 60°. The ratio of their maximum heights is: (a) 1 : 1 (b) 1 : 2 (c) 1 : 3 (d) 3 : 1
5. A projectile is launched at 60° with speed 20 m s⁻¹. Its speed at the highest point is: (a) 0 (b) 10 m s⁻¹ (c) 10√3 m s⁻¹ (d) 20 m s⁻¹
6. A projectile launched at 45° reaches a maximum height of 10 m. Its range is: (a) 10 m (b) 20 m (c) 40 m (d) 80 m
7. The angle between A = 2î + 3ĵ and B = 3î − 2ĵ is: (a) 0° (b) 45° (c) 60° (d) 90°
8. A particle moves in a circle of radius 2 m at a constant speed of 4 m s⁻¹. Its acceleration is: (a) 0 (b) 2 m s⁻² (c) 8 m s⁻² (d) 16 m s⁻²

Answers

1. (c) Galileo's odd-number rule for motion from rest.
2. (b) sₙ = 0 + (10/2)(2 × 3 − 1) = 25 m.
3. (b) v = 6t − 2 = 10 m s⁻¹.
4. (c) H ∝ sin²θ: (1/4) : (3/4) = 1 : 3.
5. (b) Only u cos θ remains: 20 × ½ = 10 m s⁻¹.
6. (c) At 45°, R = 4H = 40 m.
7. (d) A·B = 6 − 6 = 0, so they are perpendicular.
8. (c) a = v²/r = 16/2 = 8 m s⁻², towards the centre.

What to do next

• Write the four equations of motion with the condition "constant acceleration" next to them on your formula sheet.
• Sketch x–t, v–t and a–t graphs for a body thrown upward and caught again. Check the signs.
• Solve 30 previous-year NEET kinematics questions, timed at about 90 seconds each.
• Revise vector resolution before laws of motion for NEET, where you will use it constantly.

If units and dimensions still slow you down, go back to units, dimensions and errors.