Question

If a particle is moving in such a way that it's average acceleration turns out to be different for a number of different time intervals, the particle is said to have variable acceleration. The acceleration can vary in magnitude, or in direction or both. In such cases we find acceleration at any instant, called the instantaneous acceleration. It is defined as $\overrightarrow{a}=\underset{\Delta t=0}{\text{lim}}\frac{\Delta \overrightarrow{v}}{\Delta t}=\frac{d\overrightarrow{v}}{dt}$
That is acceleration of a particle at time $t$ is the limiting value of $\frac{\Delta v}{\Delta t}$at time $t$ as $\Delta t$ approaches zero. The direction of the instantaneous acceleration $\overrightarrow{a}$ is the limiting direction of the vector in velocity $\Delta v$.

A particle travels according to the equation such that it's acceleration $a=-kv$ where $k$ is constant of proportionality. Find the distance covered when it's velocity falls from $v_0$ to $v$.

• A. $x=\frac{-1}{k}[\frac{1}{v_0}-\frac{1}{v}]$
• B. $x=\frac{v_0-v}{k}$
• C. $x=k{\log }_e\frac{v_0}{v}$
• D. $x=\frac{1}{k}{\log }_e\frac{v_0}{v}$

Answer 1

Answer: (B)

$x=\frac{v_0-v}{k}$

Let $a=-kv$, where k is a constant of proportionality.
Integrating with respect to time, we get,
${\int }_{t_0}^{t}adt=-k{\int }_{t_0}^{t}vdt$
$v-v_0=-k\Delta x$
or Distance = $\frac{v_0-v}{k}$