Question

A point moves in a straight line under the retardation $a{v}^2$. If the initial velocity is $u$, the distance covered in '$t$' seconds is

• A. $aut$
• B. $\frac{1}{a}\ln \left(aut\right)$
• C. $\frac{1}{a}\ln (1+aut)$
• D. $a\ln \left(aut\right)$

Answer 1

The correct option is C $\frac{1}{a}\ln (1+aut)$

The retardation is given by
$\frac{dv}{dt}=-a{v}^2$
Integrating between proper limits, we get
$\underset{u}{\overset{v}{\int }}\frac{dv}{v^2}=-\underset{0}{\overset{t}{\int }}adt$
${[-\frac{1}{v}]}_u^v=-at$
$-\frac{1}{v}+\frac{1}{u}=-at\Rightarrow \frac{1}{v}=at+\frac{1}{u}$
​​​​​​​$\Rightarrow \frac{dt}{dx}=at+\frac{1}{u}$
​​​​​​​$\Rightarrow dx=\frac{udt}{1+aut}$
Integrating between proper limits, we get
$\underset{0}{\overset{s}{\int }}dx=\underset{0}{\overset{t}{\int }}\frac{udt}{1+aut}$
$\therefore s=\frac{1}{a}\ln (1+aut)$