Question

A point moves in a straight line under the retardation a $v^{2}$, where ' $a$ ' is a positive constant and $v$ is speed. If the initial speed is $u$, the distance covered in ' $t$ ' seconds is:

(A) a ut

(B) $\frac{1}{a} \log (a u t)$

(C) $\frac{1}{a} \log (1+a u t)$

(D) $a \log (a u t)$

Answer 1

The retardation is given by $\frac{d v}{d t}=-a v 2$ integrating between proper limits

$$

\begin{array}{l}

\Rightarrow \quad-\int_{u}^{v} \frac{d v}{v^{2}}=\int_{0}^{t} a d t \\

\text { or } \quad \frac{1}{v}=a t+\frac{1}{u} \\

\Rightarrow \quad \frac{d t}{d x}=a t+\frac{1}{u} \Rightarrow d x=\frac{u d t}{1+a u t}

\end{array}

$$

integrating between proper limits

$$

\begin{array}{ll}

\Rightarrow & \int_{0}^{s} d x=\int_{0}^{1} \frac{u d t}{1+a u t} \\

\Rightarrow & S=\frac{1}{a} \ln (1+a u t)

\end{array}

$$