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A particle of mass $m$ falls from rest through a resistive medium having resistive force, $F = -kv$, where $v$ is the velocity of the particle and $k$ is a constant. Which of the following graphs represents velocity ($v$) versus time ($t$)?

The correct answer is

Analyzing Particle Motion with Resistive Force

The problem involves a particle of mass $m$ falling under gravity and encountering a resistive force $F_r = -kv$, where $v$ is the velocity and $k$ is a constant. The negative sign indicates the force opposes the velocity (upwards if falling downwards).

Deriving the Equation of Motion

Using Newton's second law, the net force $F_{net}$ acting on the particle is:

$F_{net} = F_g + F_r$

Taking the downward direction as positive, the gravitational force $F_g = mg$. The equation becomes:

$ma = mg - kv$

Since acceleration $a = \frac{dv}{dt}$, we have the differential equation:

$m \frac{dv}{dt} = mg - kv$

Rearranging gives:

$\frac{dv}{dt} = g - \frac{k}{m}v$

Solving for Velocity vs. Time

This is a first-order linear differential equation. We can solve it by separation of variables or integrating factors. The initial condition is that the particle starts from rest, so $v(0) = 0$.

The solution to this differential equation with the initial condition $v(0)=0$ is:

$v(t) = \frac{mg}{k} \left( 1 - e^{-\frac{k}{m}t} \right)$

Let $V_T = \frac{mg}{k}$. This is the terminal velocity. The equation is:

$v(t) = V_T \left( 1 - e^{-\frac{k}{m}t} \right)$

Interpreting the Velocity-Time Graph

Key characteristics of the velocity-time graph based on the equation $v(t) = V_T (1 - e^{-\frac{k}{m}t})$:

  • Initial Velocity ($t=0$): $v(0) = V_T (1 - e^0) = V_T (1 - 1) = 0$. The graph starts at the origin (0,0).
  • Terminal Velocity ($t \to \infty$): As time increases, $e^{-\frac{k}{m}t}$ approaches 0. Thus, $v(t)$ approaches $V_T$. The graph asymptotically approaches a horizontal line at $v = V_T$.
  • Slope (Acceleration): The derivative $\frac{dv}{dt} = g e^{-\frac{k}{m}t}$. At $t=0$, the slope is $g$ (maximum initial acceleration). As $t$ increases, the slope decreases, approaching 0 as velocity approaches $V_T$.

The graph should show velocity increasing from zero, with its rate of increase (slope) decreasing over time, eventually leveling off at the terminal velocity $V_T$. This describes an exponential curve rising from the origin.

Conclusion

Comparing these characteristics with the given options:

  • Option 1 shows decreasing velocity.
  • Option 2 shows velocity increasing from zero exponentially towards a constant value (terminal velocity), with a decreasing slope. This matches our derived equation and characteristics.
  • Option 3 shows a linear increase in velocity ($v=gt$), implying constant acceleration and no resistance.
  • Option 4 shows velocity decreasing linearly.

Therefore, the graph representing velocity versus time is the one showing exponential increase towards a terminal velocity.

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