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NEET PHYSICSMOTION IN A STRAIGHT LINEEasy

Question

A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 s is s₁ and that covered in the first 20 s is s₂, then

A

s₂ = 2s₁

B

s₂ = 3s₁

C

s₂ = 4s₁

D

s₂ = s₁

Step-by-Step Solution

  1. Identify Conditions: The particle starts from rest (u=0u = 0) and moves under a constant force, which implies constant acceleration (aa).
  2. Select Kinematic Equation: The distance covered (ss) in time (tt) under constant acceleration is given by the second equation of motion: s=ut+12at2s = ut + \frac{1}{2}at^2 Since u=0u = 0, the equation simplifies to s=12at2s = \frac{1}{2}at^2. Thus, distance is directly proportional to the square of time (st2s \propto t^2).
  3. Calculate Distances: For the first 10 seconds (t1=10t_1 = 10 s): s1=12a(10)2=50as_1 = \frac{1}{2}a(10)^2 = 50a For the first 20 seconds (t2=20t_2 = 20 s): s2=12a(20)2=200as_2 = \frac{1}{2}a(20)^2 = 200a
  4. Find Relationship: Divide s2s_2 by s1s_1: s2s1=200a50a=4\frac{s_2}{s_1} = \frac{200a}{50a} = 4 s2=4s1s_2 = 4s_1

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOTION IN A STRAIGHT LINE. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOTION IN A STRAIGHT LINEparticlestartsmotionactionconstant

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