A point moves in a straight line under the retardation av² . If the initial velocity is u, the distance covered in 't' seconds is 
a) aut
b) 1/a ln(aut)
c) 1/a ln(1+aut)
d) aln(aut)

Dear Student,

Please find below the solution to the asked query:

$$\begin{gathered} According to given information, \\ a=-a{V}^2\Rightarrow \frac{dV}{dt}=-a{V}^2\Rightarrow {\int }_u^V\frac{dV}{V^2}=-a{\int }_0^t dt\Rightarrow \frac{-1}{V}+\frac{1}{u}=-a\left(t-0\right) \\ \Rightarrow \frac{1}{V}=at+\frac{1}{u}\Rightarrow V=\frac{u}{aut+1} \\ So, the dis\tan ce covered is, \\ \frac{dS}{dt}=V\Rightarrow \frac{dS}{dt}=\frac{u}{aut+1}\Rightarrow {\int }_0^{s}dS={\int }_0^t\frac{u}{aut+1}dt \\ \Rightarrow S-0=u{\int }_0^t\frac{1}{aut+1} dt\Rightarrow S=u{\left[\frac{\ln \left(aut+1\right)}{au}\right]}_0^t \\ \Rightarrow S=\frac{1}{a}\left[\ln \left(\frac{aut+1}{1}\right)\right]\Rightarrow S=\frac{1}{a}\ln \left(aut+1\right) \end{gathered}$$

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