Which one of the following graphs represents the velocity-time (v - t) graph of a small spherical body falling in a viscous liquid?
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For objects falling in viscous media, the net force decreases as velocity increases due to drag. Consequently, the acceleration (slope of \( v - t \)) must decrease monotonically, leveling off to a horizontal asymptote when terminal velocity is achieved.
Step 1: Understanding the Question:
The question asks for the velocity-time (\( v - t \)) graph of a small spherical body released from rest and falling through a viscous liquid. Step 2: Key Formula or Approach:
As the body falls, it experiences gravity downward, and buoyancy and viscous drag (described by Stokes' Law) upward. The equation of motion leads to a velocity profile:
\[ v(t) = v_T \left( 1 - e^{-\frac{t}{\tau}} \right) \]
where \( v_T \) is the terminal velocity and \( \tau \) is the characteristic relaxation time. Step 3: Detailed Explanation:
1. At \( t = 0 \), the body starts from rest, so \( v(0) = 0 \).
2. The acceleration is given by the derivative of velocity:
\[ a(t) = \frac{dv}{dt} = \frac{v_T}{\tau} e^{-\frac{t}{\tau}} \]
At \( t = 0 \), the acceleration is maximum (\( a = g_{\text{effective}} \)), meaning the slope of the \( v - t \) graph is maximum at the origin.
3. As \( t \to \infty \), \( e^{-\frac{t}{\tau}} \to 0 \). The velocity asymptotically approaches a constant value called the terminal velocity \( v_T \), and the acceleration/slope becomes zero.
4. Therefore, the theoretical curve should be concave downwards throughout, with its slope decreasing continuously.
- While option (A) contains an initial S-shape (inflection point) which is a minor graphical error in the drawing, it is the only option that correctly starts from zero and asymptotically stabilizes at a constant terminal velocity. Thus, (A) is the intended correct representation. Step 4: Final Answer:
The velocity-time graph is represented by option (A).
