Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

🚆 Train Crossing Concept in 59 Seconds | Doubtify JEE Shorts 📌 Question: Ever wondered how long one train takes to pass another or a platform? The Train Crossing Concept helps solve such questions in seconds using speed-distance fundamentals.

  🚆 Train Crossing Time Ratio – Relative Motion | Speed, Distance & Time | JEE Physics | Doubtify JEE 💡 Question: A passenger train of length 60 m travels at a speed of 80 km/hr . Another freight train of length 120 m travels at a speed of 30 km/hr . Find the ratio of times taken by the passenger train to completely cross the freight train when: (i) They are moving in the same direction , and (ii) They are moving in opposite directions . 🖼️ Question Image:

  📈 Acceleration as a Function of Distance – Motion in One Dimension | JEE Physics | Doubtify JEE 🧮 Question: The distance x covered by a particle in one-dimensional motion varies with time t as: x² = at² + 2bt + c . If the acceleration of the particle depends on x as x⁻ⁿ , where n is an integer, what is the value of n ? 🖼️ Question Image: ✅ Detailed Concept & Solution: This is a JEE-level conceptual problem involving the relation between distance, velocity, and acceleration — with a twist of functional dependence. We are given: x² = at² + 2bt + c — (1) To find acceleration and its dependency on x , let’s differentiate both sides of equation (1): Differentiate w.r.t. time t : 2x(dx/dt) = 2at + 2b ⇒ v = dx/dt = (at + b)/x — (2) Now differentiate velocity v w.r.t time to find acceleration: a = dv/dt = d/dt[(at + b)/x] Use quotient rule: a = [a·x - (at + b)(dx/dt)] / x² Plug dx/dt = (at + b)/x from (2): a = [a·x - (at + b)·((at + b)/x)] / x² ⇒ ...

  🚴 Velocity-Displacement Graph of a Bicycle | Find the Acceleration-Displacement Graph | Motion in 1D – Doubtify JEE 📌 Question : The velocity-displacement graph describing the motion of a bicycle is shown in the figure. What will the corresponding acceleration-displacement graph look like? 🎯 Concept Tested: This question tests your graphical understanding of motion — a frequently asked and often misunderstood area in JEE Mains and Advanced . It focuses on converting one type of graph (v-s) into another (a-s) using calculus concepts. 🔍 How to Approach: If you're given a velocity-displacement (v-s) graph , and you need to find the acceleration-displacement (a-s) graph , you must apply: a = v ⋅ d v d s a = v \cdot \frac{dv}{ds} This is derived from the chain rule: a = d v d t = d v d s ⋅ d s d t = v ⋅ d v d s a = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = v \cdot \frac{dv}{ds} Depending on the slope of the v-s graph: If velocity increases with displaceme...

  ⏱️ Acceleration-Time Graph – Find Time When Initial Velocity is Regained | JEE Physics | Doubtify JEE ❓ Question: The acceleration-time graph of a particle is as shown. At what time does the particle acquire its initial velocity again?

🎯 Question: A balloon was moving upwards with a uniform velocity of 10 m/s. An object of finite mass is dropped from the balloon when it was at a height of 75 m from the ground level. The height of the balloon from the ground, when the object strikes the ground, was around: (Take g = 10 m/s²)