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NEET PHYSICSMOTION IN A STRAIGHT LINEMedium

Question

A body is falling freely in a resistive medium. The motion of the body is described by dvdt=(42v)\frac{dv}{dt} = (4 - 2v), where vv is the velocity of the body at any instant (in m s1\text{m s}^{-1}). The terminal velocity in this case refers to the velocity the body approaches as time tt \to \infty. The initial acceleration and terminal velocity of the body, respectively, are:

A

4 m/s², 2 m/s

B

2 m/s², 4 m/s

C

6 m/s², 2 m/s

D

2 m/s², 6 m/s

Step-by-Step Solution

  1. Analyze the Equation: The acceleration of the body is given by the derivative of velocity with respect to time: a=dvdt=42va = \frac{dv}{dt} = 4 - 2v .
  2. Calculate Initial Acceleration: 'Falling freely' in this context implies the body starts from rest (dropped) at t=0t=0. Thus, initial velocity v=0v = 0. Substituting this into the equation: ainitial=42(0)=4 m/s2a_{\text{initial}} = 4 - 2(0) = 4 \text{ m/s}^2
  3. Calculate Terminal Velocity: Terminal velocity (vTv_T) is the constant velocity reached when the resistive force balances the driving force, resulting in zero acceleration. As tt \to \infty, dvdt0\frac{dv}{dt} \to 0. Setting acceleration to zero: 0=42vT0 = 4 - 2v_T 2vT=42v_T = 4 vT=2 m/sv_T = 2 \text{ m/s}

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 LINEfallingfreelyresistivemediummotion

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