JEE Advanced PYQ Motion in a Straight Line Set 2 v–x & a–t Graphs, Rocket Frame Problem скачать в хорошем качестве

JEE Advanced PYQ Motion in a Straight Line Set 2 v–x & a–t Graphs, Rocket Frame Problem

In this video we solve JEE Advanced Past Year Questions from Motion in a Straight Line – PYQ Set 2, focused on non‑uniform motion and graphs: Converting a v–x graph to an a–x graph using calculus Using area under a–t graph to find maximum velocity Handling a classic rocket + two balls problem in an accelerating frame (integer answer) PYQs discussed in this video: v–x graph → correct a–x graph (JEE 2005S) Velocity vs displacement is a straight line from ((0, v_0)) to ((x_0, 0)). Which acceleration vs displacement graph is correct? Key idea: [ a = v,\frac{dv}{dx} ] Linear v–x ⇒ linear a–x with positive slope and negative intercept. a–t graph triangle → maximum velocity (JEE 2004S) Acceleration decreases linearly from 10 m/s² at t = 0 to 0 at t = 11 s (straight line, negative slope). Find the maximum velocity attained by a body starting from rest. Key idea: change in velocity = area under a–t graph: [ \Delta v = \frac{1}{2}\times 11\times 10 = 55\ \text{m/s} ] Rocket in gravity‑free space with two balls (JEE Adv 2014, integer type) Rocket accelerates at 2 m/s² along +x, chamber length = 4 m. At same instant: Ball A thrown from left end in +x with 0.3 m/s (relative to rocket) Ball B thrown from right end in –x with 0.2 m/s (relative to rocket) Find time when they collide (official answer 8 s). Key idea: write kinematics inside rocket (using (s = ut + \tfrac12 at^2) along the chamber), relate positions, solve for t. Who should watch this? JEE Advanced aspirants who want to master: v–x and a–x graph conversions, Using area under a–t graphs, Kinematics in non‑inertial (accelerating) frames. JEE Main / CBSE 11th students looking to upgrade from basic graphs to Advanced‑level interpretations. Level: Conceptual + exam‑oriented, with step‑by‑step Hinglish explanations and diagrams. Teaching style Clean slides (simple graphs drawn) explanation: Concept → formula → application on PYQ No skipping of key reasoning steps Each solution highlights: The core idea (e.g. (a = v,dv/dx), area under a–t graph) How to recognise this pattern in future questions Playlist & related content This video is part of the playlist: JEE Advanced PYQ – Motion in a Straight Line (Full Series) Set 1: Basics – distance vs displacement, avg velocity/acceleration, valid x–t graphs Set 2: Non‑uniform motion & graphs (this video) Set 3: Constraints on acceleration & v–t triangle methods Set 4–5: Relative velocity in 1D & frames questions Set 6: Motion under gravity (concept check) Also see the theory playlist: Motion in a Straight Line | Class 11 Physics | CBSE & JEE (for full concept build‑up before PYQs) If these JEE Advanced PYQ walkthroughs are helping your preparation, you can subscribe to “CBSE & JEE Physics | Dr Kedar Pathak” and follow the full Motion in a Straight Line PYQ playlist in order.