Question

A particle moving in the positive x direction has initial velocity $v_0$. The particle undergoes retardation $kv^2$, where $v$ is instantaneous velocity. The velocity of the particle as a function of time is

• A. $v = \frac{v_0}{1+kv_0t}$
• B. $v = \frac{2v_0}{1+kt}$
• C. $v = \frac{v_0}{kt}$
• D. $v = \frac{v_0}{(1+k^2v_0^2t)}$

Answer 1

Answer: (A)

(1) $v = \frac{v_0}{1+kv_0t}$

Answer 2

The problem describes the motion of a particle undergoing retardation (deceleration) proportional to the square of its instantaneous velocity. We need to find the velocity as a function of time.

1. Define acceleration:
   Retardation means the acceleration is negative. Given retardation is $kv^2$, the acceleration $a$ is:
   $a = \frac{dv}{dt} = -kv^2$

2. Separate variables:
   To solve this differential equation, separate the variables $v$ and $t$:
   $\frac{dv}{v^2} = -k dt$

3. Integrate both sides:
   Integrate the left side with respect to $v$ from the initial velocity $v_0$ (at $t=0$) to the instantaneous velocity $v$ (at time $t$). Integrate the right side with respect to $t$ from $0$ to $t$.
   $\int_{v_0}^{v} \frac{dv}{v^2} = \int_{0}^{t} -k dt$

4. Perform integration:
   The integral of $1/v^2$ is $-1/v$.
   $\left[ -\frac{1}{v} \right]_{v_0}^{v} = -k [ t ]_{0}^{t}$
   $\left( -\frac{1}{v} \right) - \left( -\frac{1}{v_0} \right) = -k(t - 0)$
   $-\frac{1}{v} + \frac{1}{v_0} = -kt$

5. Rearrange to solve for $v$:
   $\frac{1}{v_0} - \frac{1}{v} = -kt$
   $\frac{1}{v} = \frac{1}{v_0} + kt$
   To combine the terms on the right, find a common denominator:
   $\frac{1}{v} = \frac{1 + kv_0t}{v_0}$
   Now, invert both sides to get $v$:
   $v = \frac{v_0}{1 + kv_0t}$

This expression gives the velocity of the particle as a function of time.