Kinematics: NEET PYQ, 2018–2026

Answer: (D)

For a body dropped from rest, h = ½gt², so t = √(2h/g) = √(2 × 45 / 10) = √9 = 3 s.

Answer: (C)

The angular speeds are ω₁ = 2π/3 and ω₂ = 2π/5. For motion in the same direction, the faster particle must gain one full turn (2π) on the slower one. Time = 2π/(ω₁ − ω₂) = 2π/[2π(1/3 − 1/5)] = 1/(2/15) = 7.5 s. By then the faster particle has made 2.5 turns and the slower one 1.5 turns, a difference of exactly one turn.

Answer: (A)

At latitude λ the person moves in a circle of radius r = R cos λ = 6400 km × 0.5 = 3.2 × 10⁶ m, about the Earth's axis. The centripetal acceleration is a = ω²r = (7.27 × 10⁻⁵)² × 3.2 × 10⁶ ≈ 1.7 × 10⁻² m/s², directed perpendicular to the axis. The line from the person to the Earth's centre makes angle λ = 60° with this horizontal direction. The component along it is a cos 60° ≈ 8.5 × 10⁻³ m/s².

Answer: (A)

In uniform circular motion the particle covers equal distances along the arc in equal times, so its speed (the magnitude of velocity) is constant. Velocity is tangent to the circle, and that tangent direction rotates continuously, so velocity changes. The change in velocity is what produces the centripetal acceleration directed toward the centre. A student who treats speed and velocity as the same thing will pick the first distractor, and one who thinks curved motion must change speed will pick the last.

Answer: (D)

The velocity–time graph is a triangle with the peak at speed v. The rising slope is 2, so t₁ = v/2. The falling slope has magnitude 3, so t₂ = v/3. The total time is t₁ + t₂ = v(1/2 + 1/3) = 5v/6 = 10 s, so v = 12 m s⁻¹. Check: t₁ = 6 s and t₂ = 4 s, which are not equal. The distance travelled is ½ × 12 × 10 = 60 m, and the average speed of 6 m s⁻¹ is half the peak speed.

Answer: (A)

Take v₁ = 10î and v₂ = 10(cos120° î + sin120° ĵ) = −5î + 5√3 ĵ. Then Δv = v₂ − v₁ = −15î + 5√3 ĵ, so |Δv| = √(225 + 75) = √300 = 10√3 m/s. The same result follows from |Δv|² = v² + v² − 2v² cos120° = 3v². The speed is |v| = 10 m/s both before and after, so the change in speed is zero. The magnitude of the change in a vector is not the change in its magnitude.

Answer: (B)

The horizontal and vertical components of the launch velocity are perpendicular, so the launch speed is the hypotenuse of the right triangle they form. Using Pythagoras, u = √(u_x² + u_y²) = √(225 + 400) = √625 = 25 m s⁻¹.

Answer: (D)

The area of a parallelogram with adjacent sides A and B equals the magnitude of the vector product |A × B|. For vectors in the x–y plane, A × B = (A_xB_y − A_yB_x) k̂ = (2·1 − 3·4) k̂ = −10 k̂. The magnitude is 10 m². The scalar product (11), half the area (5) and a sign slip (14) are the tempting errors.

Answer: (A)

The first fall takes t₀ = √(2h/g) = 2 s and the impact speed is 20 m/s. Each rebound at speed e·v lasts 2ev/g, so the rebound times are 2 s, 1 s, 0.5 s, …, a geometric series. The total time is 2 + 2(1 + ½ + ¼ + …) = 2 + 4 = 6 s. Rebound heights are 5 m, 1.25 m, … (ratio ¼). The total distance is 20 + 2(5 + 1.25 + …) = 20 + 2(5)/(1 − ¼) = 20 + 13.3 = 33.3 m. The average speed is 33.3/6 ≈ 5.6 m/s.

Answer: (D)

Acceleration is the rate of change of velocity, a = dv/dt, so the slope of the velocity–time graph at any instant gives the acceleration. The area under the graph gives the displacement.