Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

4 m/s², 2 m/s

2 m/s², 4 m/s

6 m/s², 2 m/s

2 m/s², 6 m/s

Step-by-Step Solution

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

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