Question

If \[ a=-kv \] represents the variation of acceleration of a particle with velocity, then the time taken to reduce the velocity from \(v\) to \(\dfrac{v}{2}\) is

• A. \[ \frac{\ln 2}{k} \]
• B. \[ \frac{1}{2k} \]
• C. \[ \frac{\ln 2}{2k} \]
• D. \[ \frac{2}{k} \]

Hint:

Whenever acceleration is proportional to velocity, \[ a=-kv, \] the velocity decreases exponentially: \[ v=v_0e^{-kt}. \] The time required to reduce the velocity by half is always \[ t=\frac{\ln 2}{k}. \]

Answer 1

The Correct Option is A

Concept: Acceleration is the rate of change of velocity with time. \[ a=\frac{dv}{dt} \] Given, \[ a=-kv. \] Therefore, \[ \frac{dv}{dt}=-kv. \] This is a first-order differential equation.

Step 1: Separate the variables. \[ \frac{dv}{v}=-k\,dt. \]

Step 2: Apply the limits. Initially, \[ v=v \] and after time \(t\), \[ v=\frac{v}{2}. \] Hence, \[ \int_{v}^{v/2}\frac{dv}{v} = -k\int_{0}^{t}dt. \]

Step 3: Integrate both sides. \[ \left[\ln v\right]_{v}^{v/2} = -kt. \] \[ \ln\left(\frac{v/2}{v}\right) = -kt. \] \[ \ln\left(\frac12\right) = -kt. \] \[ -\ln 2=-kt. \]

Step 4: Calculate the time. \[ t=\frac{\ln 2}{k}. \] Therefore, \[ \boxed{t=\frac{\ln 2}{k}} \] \[ \boxed{\text{Answer = (A)}} \]