Kinematics & Motion (1D & 2D)
Practice JEE Main 1D, 2D & Projectile Motion chapter-wise PYQs online. Relative velocity, trajectory equations, instant timer & step-by-step solutions.
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🎯 High-Yield Syllabus Breakdown
These high-frequency concepts account for over 90% of all questions asked in Kinematics & Motion (1D & 2D):
- 1D Motion with Constant & Variable Acceleration (v = dx/dt, a = v dv/dx)
- Projectile Motion: Range R = u² sin 2θ / g, Max Height H = u² sin² θ / (2g)
- Equation of Trajectory: y = x tan θ [1 - x / R]
- Relative Motion in 1D & 2D (Rain-Man & River-Boat Problems)
- Motion on Inclined Plane & Projectile on Incline
⚡ Essential Formulas & Speed Shortcuts
Formulas and time-saving shortcuts to solve Kinematics questions in under 45 seconds:
Complementary Angles: Launch angles θ and (90° - θ) produce the same horizontal range R.
Stopping Distance: s ∝ v² (Doubling initial speed quadruples stopping distance).
Trajectory factored form: y = x tan θ (1 - x/R).
📖 Sample Solved Exam Questions
Preview authentic questions from this chapter with full step-by-step solutions:
💡 View Step-by-Step Solution & Answer
Correct Answer: -1.5 m/s
- Step 1: Velocity is v = dx/dt = 6t² - 18t + 12.
- Step 2: Acceleration is a = dv/dt = 12t - 18.
- Step 3: Set acceleration to zero: 12t - 18 = 0 ⇒ 12t = 18 ⇒ t = 1.5 s.
- Step 4: Substitute t = 1.5 into velocity: v = 6(1.5)² - 18(1.5) + 12 = 6(2.25) - 27 + 12 = 13.5 - 27 + 12 = -1.5 m/s.
⚡ Speed Tip: Differentiate twice: a = 12t - 18 = 0 gives t = 1.5 s. Plug into v(t) to get -1.5 m/s.
💡 View Step-by-Step Solution & Answer
Correct Answer: 4
- Step 1: Formula for maximum height: H = (u² sin² θ) / (2g).
- Step 2: Formula for horizontal range: R = (u² sin 2θ) / g = (2u² sin θ cos θ) / g.
- Step 3: Given H = R ⇒ (u² sin² θ) / (2g) = (2u² sin θ cos θ) / g.
- Step 4: Cancel u²/g and sin θ: sin θ / 2 = 2 cos θ ⇒ sin θ / cos θ = 4 ⇒ tan θ = 4.
⚡ Speed Tip: Standard Projectile Relation: R = 4H cot θ ⇒ When R = H, tan θ = 4 directly.
💡 View Step-by-Step Solution & Answer
Correct Answer: 3/4
- Step 1: Height after nth bounce is h_n = e^(2n) · h₀. For the first bounce, h₁ = e² h₀.
- Step 2: Here h₀ = 80 m and h₁ = 45 m.
- Step 3: e² = h₁ / h₀ = 45 / 80 = 9 / 16.
- Step 4: Take the square root: e = √(9/16) = 3/4.
⚡ Speed Tip: Bounce Height Formula: e = √(h₁ / h₀) = √(45/80) = √(9/16) = 3/4.
❓ Frequently Asked Questions (FAQs)
What is the relation between Maximum Range and Maximum Height?
At θ = 45°, Range is maximum R_max = u²/g and H = u²/(4g) = R_max / 4.